Who is our nearest planetary neighbor, on average?
Clash Royale CLAN TAG#URR8PPP
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Assume that the planets have circular orbits centered on the sun. Assume that the radius of the orbits is 0.39, 0.723, 1, and 1.524 for Mercury, Venus, Earth, and Mars.
Assume also that there are no resonances, in other words, that for a given position of planet A planet B will be in every other position over a long period of time.
What is the closest planet to Earth on average?
mathematics physics
edited Mar 14 at 6:45
smci
39229
asked Mar 13 at 4:53
Dr XorileDr Xorile
14.1k33083
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|
show 2 more comments
$begingroup$
Assume that the planets have circular orbits centered on the sun. Assume that the radius of the orbits is 0.39, 0.723, 1, and 1.524 for Mercury, Venus, Earth, and Mars.
Assume also that there are no resonances, in other words, that for a given position of planet A planet B will be in every other position over a long period of time.
What is the closest planet to Earth on average?
mathematics physics
edited Mar 14 at 6:45
smci
39229
asked Mar 13 at 4:53
Dr XorileDr Xorile
14.1k33083
$endgroup$
2
$begingroup$
Though this is a somewhat mathematical puzzle, the answer is a little surprising...
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– Dr Xorile
Mar 13 at 5:10
6
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I assume you mean the closest planet besides Earth?
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– Deusovi♦
Mar 13 at 5:31
3
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Didn't I see someone promoting their paper on this today?
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– Jay
Mar 13 at 5:38
3
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@Deusovi It's not in the body of the question, but "neighbour" in the title clearly excludes the Earth.
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– yo'
Mar 13 at 14:31
1
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A great puzzle for pi day
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– Strawberry
Mar 14 at 12:02
|
show 2 more comments
$begingroup$
Assume that the planets have circular orbits centered on the sun. Assume that the radius of the orbits is 0.39, 0.723, 1, and 1.524 for Mercury, Venus, Earth, and Mars.
Assume also that there are no resonances, in other words, that for a given position of planet A planet B will be in every other position over a long period of time.
What is the closest planet to Earth on average?
mathematics physics
edited Mar 14 at 6:45
smci
39229
asked Mar 13 at 4:53
Dr XorileDr Xorile
14.1k33083
$endgroup$
Assume that the planets have circular orbits centered on the sun. Assume that the radius of the orbits is 0.39, 0.723, 1, and 1.524 for Mercury, Venus, Earth, and Mars.
Assume also that there are no resonances, in other words, that for a given position of planet A planet B will be in every other position over a long period of time.
What is the closest planet to Earth on average?
mathematics physics
mathematics physics
edited Mar 14 at 6:45
smci
39229
asked Mar 13 at 4:53
Dr XorileDr Xorile
14.1k33083
edited Mar 14 at 6:45
smci
39229
asked Mar 13 at 4:53
Dr XorileDr Xorile
14.1k33083
edited Mar 14 at 6:45
smci
39229
edited Mar 14 at 6:45
smci
39229
edited Mar 14 at 6:45
smci
39229
39229
asked Mar 13 at 4:53
Dr XorileDr Xorile
14.1k33083
asked Mar 13 at 4:53
Dr XorileDr Xorile
14.1k33083
asked Mar 13 at 4:53
Dr XorileDr Xorile
14.1k33083
14.1k33083
2
$begingroup$
Though this is a somewhat mathematical puzzle, the answer is a little surprising...
$endgroup$
– Dr Xorile
Mar 13 at 5:10
6
$begingroup$
I assume you mean the closest planet besides Earth?
$endgroup$
– Deusovi♦
Mar 13 at 5:31
3
$begingroup$
Didn't I see someone promoting their paper on this today?
$endgroup$
– Jay
Mar 13 at 5:38
3
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@Deusovi It's not in the body of the question, but "neighbour" in the title clearly excludes the Earth.
$endgroup$
– yo'
Mar 13 at 14:31
1
$begingroup$
A great puzzle for pi day
$endgroup$
– Strawberry
Mar 14 at 12:02
|
show 2 more comments
2
$begingroup$
Though this is a somewhat mathematical puzzle, the answer is a little surprising...
$endgroup$
– Dr Xorile
Mar 13 at 5:10
6
$begingroup$
I assume you mean the closest planet besides Earth?
$endgroup$
– Deusovi♦
Mar 13 at 5:31
3
$begingroup$
Didn't I see someone promoting their paper on this today?
$endgroup$
– Jay
Mar 13 at 5:38
3
$begingroup$
@Deusovi It's not in the body of the question, but "neighbour" in the title clearly excludes the Earth.
$endgroup$
– yo'
Mar 13 at 14:31
1
$begingroup$
A great puzzle for pi day
$endgroup$
– Strawberry
Mar 14 at 12:02
2
2
$begingroup$
Though this is a somewhat mathematical puzzle, the answer is a little surprising...
$endgroup$
– Dr Xorile
Mar 13 at 5:10
$begingroup$
Though this is a somewhat mathematical puzzle, the answer is a little surprising...
$endgroup$
– Dr Xorile
Mar 13 at 5:10
6
6
$begingroup$
I assume you mean the closest planet besides Earth?
$endgroup$
– Deusovi♦
Mar 13 at 5:31
$begingroup$
I assume you mean the closest planet besides Earth?
$endgroup$
– Deusovi♦
Mar 13 at 5:31
3
3
$begingroup$
Didn't I see someone promoting their paper on this today?
$endgroup$
– Jay
Mar 13 at 5:38
$begingroup$
Didn't I see someone promoting their paper on this today?
$endgroup$
– Jay
Mar 13 at 5:38
3
3
$begingroup$
@Deusovi It's not in the body of the question, but "neighbour" in the title clearly excludes the Earth.
$endgroup$
– yo'
Mar 13 at 14:31
$begingroup$
@Deusovi It's not in the body of the question, but "neighbour" in the title clearly excludes the Earth.
$endgroup$
– yo'
Mar 13 at 14:31
1
1
$begingroup$
A great puzzle for pi day
$endgroup$
– Strawberry
Mar 14 at 12:02
$begingroup$
A great puzzle for pi day
$endgroup$
– Strawberry
Mar 14 at 12:02
|
show 2 more comments
7 Answers
7
active
oldest
votes
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The closest planet to Earth on average is:
Mercury
The other answers didn't give any calculations, so I'll provide some numbers. Hopefully they are correct!
As other answers suggested, we can leave Earth stationary and just have the other planets do their orbits. Actually we only need to do half an orbit, because the other half will be exactly like the first half and not change the average in any way.
By law of cosines, we can find the distance between the Earth and another planet by looking at the triangle that is formed when you connect the Earth with the other planet and the Sun. Obviously the distance to the Sun is the radius of the orbits and the angle will be $theta$. The distance between the planets is then $sqrtr^2+R^2-2*r*R*cos(theta)$ where $r$ is the radius of Earth's orbit and $R$ is the radius of the other planet's orbit and $theta$ is the angle between them.
Now just find the integral as $theta$ goes from 0 to $pi$ and divide by $pi$.
Earth to Mars:
$fracint_0^pi sqrt1^2+1.524^2-2*1*1.524*cos(theta) dthetapi = 1.693AU$
Earth to Venus:
$fracint_0^pi sqrt1^2+0.723^2-2*1*0.723*cos(theta) dthetapi = 1.136AU$
Earth to Mercury:
$fracint_0^pi sqrt1^2+0.39^2-2*1*.0.39*cos(theta) dthetapi = 1.038AU$
answered Mar 13 at 6:41
AmorydaiAmorydai
1,533215
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The key in this calculation is “divide by pi”. Here you are assuming a uniform distribution of theta. That is, you assume the probability of the other planet being in any other theta is equal, and then you take the expected value. Depending on the other planets’ theta as a function of time, this need not be true. So here we are making the simplifying assumption that all planets travel at a constant speed.
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– darksky
Mar 13 at 11:20
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@darksky Well, we are assuming the orbits are circles, so we are leaving Kepler out of it! Or, rather, we are leaving Kepler in it - “equal areas during equal intervals of time” in a circle would mean constant speed.
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– Amorydai
Mar 13 at 14:27
3
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@darksky Yep, constant speed and circular orbits. Both are not true, but the corrections would be much less than 7% difference between Mercury and Venus
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– yo'
Mar 13 at 14:27
1
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@Amorydai, how did you calculate the integrals?
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– Dr Xorile
Mar 13 at 16:54
1
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How very dare you use the word "Obviously"?! When did puzzling start allowing witchcraft in answers? That formula is satanic... I mean, what sort of wizard-y demon-scratch is that un-figurable? Let's have a vote on a site-wide ban on pomped up NASA geniuses... (in English: "This made me feel dumb")
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– Brent Hackers
Mar 14 at 11:39
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show 5 more comments
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I must admit, I'm a bit rusty at calculus. So here's an attempt at an answer free of calculations, but with some more visual reasoning.
First, let's draw out the orbits of the planets. Because there is no resonance, let's assume Earth is still and they all rotate at the same speed. Also, for reference we'll draw a circle of distance 1 AU around Earth.
Now, notice that:
I've drawn a few dotted lines here. If we imagine these circles as pie charts, the left part of the charts represent the time spent more than 1 AU away from the Earth.
So:
Let's plot the distance away from the Earth over time, which looks vaguely like this:
Here, you should note:
The arrows along the bottom, telling you the time spent above the green line (1AU still) and the arrows in the middle, telling you that for the Venus and Mercury, (maximum distance - 1AU) is equal to (1AU - minimum distance).
Also, Mars is clearly out of the question. Goodbye Mars.
So:
It's between Mercury and Venus. From calculus or intuition, we know that the average distance of the planet is proportional to the area under the curve. And now it gets a bit non-technical.
The area under the curve is equal to 2π AU·rad + (bit above the curve) - (bits below the curve). But the bit above the curve is approximately the same shape as the bit below the curve (if we shift the right bit under the curve to the left side of the graph), and since they are the same height their area is probably proportional to their width. And since Venus' width of bit above the curve to width of bit below the curve ratio is bigger than that of Mercury, and the fact that those bits are taller than Mercury's bits, I estimate Venus' total area is probably more than that of Mercury's.
So my guess is that:
Mercury is on average closest to the Earth. (I'd love to know how accurate this argument is, but that maths is beyond me.)
NB: Click on images for slightly higher quality if they're a bit fuzzy.
answered Mar 13 at 6:02
boboquackboboquack
15.9k150119
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Heck, I so love this one. If you wanna teach mathematical/physical intuition, this is the way!
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– yo'
Mar 13 at 14:29
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Love this answer. Super helpful for gaining the intuition.
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– Dr Xorile
Mar 13 at 16:48
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One of my favorite answers in this whole site. I would actually recommend reading this before reading the accepted answer so it's easier to digest the technical math.
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– yushi
Mar 15 at 10:07
add a comment |
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The answer:
Mercury
Reasoning:
The average position of Earth (and indeed all planets in a circular orbit) is the middle of the Sun. Since Mercury's orbit is closest to the sun, it's the nearest on average to the Earth, and indeed all the other planets.
answered Mar 13 at 6:12
Matthew BarberMatthew Barber
84516
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Following that reasonning, wouldn't the answer be "All planets are equally close to the Earth on average, as all of their average positions fall at the same place (middle of the Sun)" ?
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– Soltius
Mar 13 at 10:54
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By this logic, the average closest planet to our moon is also Mercury.
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– BlueRaja - Danny Pflughoeft
Mar 13 at 17:30
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@BlueRaja That's only true if you assume that the orbits of the Moon and the Earth around the sun are independent, which they obviously aren't.
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– Matthew Barber
Mar 13 at 21:32
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@Soltius No, that does not follow, as you cannot calculate average distances entirely from average positions. Averaging the position of the Earth won't help you calculate the distance either. It's just something that illustrates why the distance from the sun of the other planet is the only thing that matters in the ordering.
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– Matthew Barber
Mar 13 at 22:03
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Others have done all the necessary calculations, so here's some hairy maths. I assume as per the question that all orbits are circular and that the planets move in such a way that the average distance equals the average over all angular differences. Then it turns out that the average distance from earth to a planet whose orbit has radius $r$ astronomical units (i.e., $r$ times the radius of the earth's orbit) is $frac2pi(1+r)E(frac4r1+r^2)$ astronomical units, where E is the so-called complete elliptic integral of the second kind, what Mathematica calls EllipticE.
So what we'd like to be true is that this is an increasing function of $r$. This does appear to be true, but proving it is not so trivial.
Rather than looking at the average over the whole orbit, let's look at just two antipodal points. So, suppose the angle between earth's position and the other planet's position is $theta$, so that the distance is $sqrt(r-costheta)^2+sin^2theta$; half-way around the orbit the other planet's position is $theta+pi$ and the distance is $sqrt(r+costheta)^2+sin^2theta$. The sum of these is $f(r,theta):=sqrt(r-costheta)^2+sin^2theta+sqrt(r+costheta)^2+sin^2theta$, our average is the average of this over all values of $theta$, and it will be an increasing function of $r$ if $f$ is for every $theta$. This will be true if it's true when we consider instead $g(r,u,v):=sqrt(r-u)^2+v^2+sqrt(r+u)^2+v^2$ and allow $u,v$ to take any value at all. (Which just corresponds to letting the earth's distance from the sun be something other than 1 unit.)
The derivative of this thing is $fracr+usqrt(r+u)^2+v^2+fracr-usqrt(r-u)^2+v^2$. Obviously this is positive when $r>u$. When $r<u$ it's $h(u+r,v)-h(u-r,v)$ where $h(p,q)=fracpsqrtp^2+q^2=costan^-1frac qp$. But this is obviously a decreasing function of $q/p$, hence an increasing function of $p$, which means that $h(u+r,v)>h(u-r,v)$, which means that $fracpartial gpartial r>0$, which means that $fracpartial fpartial r>0$, which means that $fracpartialint fpartial r>0$, which means that indeed the average distance is an increasing function of $r$.
I suspect there may be an easier more purely geometrical way to do this.
answered Mar 13 at 11:05
Gareth McCaughan♦Gareth McCaughan
67.3k3170261
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@boboquack gets at a nice intuition for why this is true. I think also looking at the pairs of points: in line with the earth they balance out (so 0 difference) and at right angles the further out the orbit the further out the distance. So a continuity argument says that the overall average is monotonic.
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– Dr Xorile
Mar 13 at 16:52
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It's
Mercury.
Because:
Fix the position of the Earth, and let the planets move in orbit. We want the average distance. If the Sun was an object to consider, the radius R would be the average. If there was another planet on Earth's orbit, it's average would be greater than R, as most of the orbit is at a further distance than R from the Earth (draw a circle radius R from the Earth - it cuts the other orbit before the halfway points).
As this is a continuous and monotonic increasing function, the planet closest on average is Mercury.
answered Mar 13 at 6:03
JonMark PerryJonMark Perry
20.7k64099
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Why is it continuous and monotonic? And how did you deal with Mars?
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– boboquack
Mar 13 at 6:52
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@boboquack; the orbit moves continuously and so therefore does the average function, which is a quadratic, and therefore has a monotonic differential. Larger orbits just get bigger (monotonic increasing remember!). Also see en.wikipedia.org/wiki/Orbit.
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– JonMark Perry
Mar 13 at 7:20
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Average distance to earth is not a quadratic function of orbit radius because for very large orbits it's approximately equal to the radius.
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– Gareth McCaughan♦
Mar 13 at 8:49
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@GarethMcCaughan; this doesn't change my argument much though. the average function depends on R and only approximates R locally.
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– JonMark Perry
Mar 13 at 9:07
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The function in question is monotone increasing, though. At least, I think it is, though I haven't tried to prove it; it's a pretty ugly function involving elliptic integrals.
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– Gareth McCaughan♦
Mar 13 at 9:58
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show 2 more comments
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Here is another way to deduce the same result without any calculations.
First the inner planets.
Imagine that instead of the Earth, there is a wall at 1 AU from the sun. The average distance of any inner planet's orbit to the wall is easily seen to be exactly 1 AU.
This is because you can pair up points on the left and right sides of the orbit. The average distance of those two points to the wall is the same as the distance of their midpoint to the wall.
If you now go back to measuring the distance to the Earth itself, the distances get larger.
Crucially, the larger the orbit, the higher the slopes of the line segments we are measuring, and the further away the average distance is from 1 AU.
This shows that amongst the inner planets, the innermost planet (Mercury) has the smallest average distance to Earth.
What about the outer planets?
Let's go back to the wall replacing the Earth. If an planet's orbit crosses the wall, then when it comes to measuring its distance, we might as well mirror the planet's position in the wall and measure the distance to its mirror image.
When we then do the same trick of pairing points in the two halves of the orbit, it is clear that the average distance to the wall becomes greater than 1 AU to start with. Combine that with the fact that when measuring the distance to Earth itself the slopes of the line segments are even larger than before, it is clear that the average distance to the planet is even greater compared to the inner planets.
answered Mar 14 at 10:32
Jaap ScherphuisJaap Scherphuis
16.6k12872
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I'd say the closest planet to Earth is Earth with an average distance of 0
edited Mar 13 at 10:20
Ahmed Ashour
9741313
answered Mar 13 at 9:21
user57862user57862
471
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Well, unfortunately for you, a neighbour is a well defined notion in geometry, and excludes you yourself...
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– yo'
Mar 13 at 14:31
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@yo' I disagree. In graph theory this might be true, but if you’re talking about metric spaces (which we are), then an epsilon neighborhood around a point always contains that point.
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– Santana Afton
Mar 13 at 22:10
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Well, in clustering theory, statistics etc. it's all pretty clear, and that's the context in which I see this puzzle.
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– yo'
Mar 14 at 0:59
1
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@SantanaAfton: a neighbourhood of a point in a topological/metric space contains the point itself, but you don’t speak of points being neighbours of each other in that setting. (That said, I think everyday usage is more to the point here than any of the technical mathematical senses.)
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– Peter LeFanu Lumsdaine
Mar 15 at 10:37
add a comment |
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7 Answers
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7 Answers
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$begingroup$
The closest planet to Earth on average is:
Mercury
The other answers didn't give any calculations, so I'll provide some numbers. Hopefully they are correct!
As other answers suggested, we can leave Earth stationary and just have the other planets do their orbits. Actually we only need to do half an orbit, because the other half will be exactly like the first half and not change the average in any way.
By law of cosines, we can find the distance between the Earth and another planet by looking at the triangle that is formed when you connect the Earth with the other planet and the Sun. Obviously the distance to the Sun is the radius of the orbits and the angle will be $theta$. The distance between the planets is then $sqrtr^2+R^2-2*r*R*cos(theta)$ where $r$ is the radius of Earth's orbit and $R$ is the radius of the other planet's orbit and $theta$ is the angle between them.
Now just find the integral as $theta$ goes from 0 to $pi$ and divide by $pi$.
Earth to Mars:
$fracint_0^pi sqrt1^2+1.524^2-2*1*1.524*cos(theta) dthetapi = 1.693AU$
Earth to Venus:
$fracint_0^pi sqrt1^2+0.723^2-2*1*0.723*cos(theta) dthetapi = 1.136AU$
Earth to Mercury:
$fracint_0^pi sqrt1^2+0.39^2-2*1*.0.39*cos(theta) dthetapi = 1.038AU$
answered Mar 13 at 6:41
AmorydaiAmorydai
1,533215
$endgroup$
$begingroup$
The key in this calculation is “divide by pi”. Here you are assuming a uniform distribution of theta. That is, you assume the probability of the other planet being in any other theta is equal, and then you take the expected value. Depending on the other planets’ theta as a function of time, this need not be true. So here we are making the simplifying assumption that all planets travel at a constant speed.
$endgroup$
– darksky
Mar 13 at 11:20
$begingroup$
@darksky Well, we are assuming the orbits are circles, so we are leaving Kepler out of it! Or, rather, we are leaving Kepler in it - “equal areas during equal intervals of time” in a circle would mean constant speed.
$endgroup$
– Amorydai
Mar 13 at 14:27
3
$begingroup$
@darksky Yep, constant speed and circular orbits. Both are not true, but the corrections would be much less than 7% difference between Mercury and Venus
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– yo'
Mar 13 at 14:27
1
$begingroup$
@Amorydai, how did you calculate the integrals?
$endgroup$
– Dr Xorile
Mar 13 at 16:54
1
$begingroup$
How very dare you use the word "Obviously"?! When did puzzling start allowing witchcraft in answers? That formula is satanic... I mean, what sort of wizard-y demon-scratch is that un-figurable? Let's have a vote on a site-wide ban on pomped up NASA geniuses... (in English: "This made me feel dumb")
$endgroup$
– Brent Hackers
Mar 14 at 11:39
|
show 5 more comments
$begingroup$
The closest planet to Earth on average is:
Mercury
The other answers didn't give any calculations, so I'll provide some numbers. Hopefully they are correct!
As other answers suggested, we can leave Earth stationary and just have the other planets do their orbits. Actually we only need to do half an orbit, because the other half will be exactly like the first half and not change the average in any way.
By law of cosines, we can find the distance between the Earth and another planet by looking at the triangle that is formed when you connect the Earth with the other planet and the Sun. Obviously the distance to the Sun is the radius of the orbits and the angle will be $theta$. The distance between the planets is then $sqrtr^2+R^2-2*r*R*cos(theta)$ where $r$ is the radius of Earth's orbit and $R$ is the radius of the other planet's orbit and $theta$ is the angle between them.
Now just find the integral as $theta$ goes from 0 to $pi$ and divide by $pi$.
Earth to Mars:
$fracint_0^pi sqrt1^2+1.524^2-2*1*1.524*cos(theta) dthetapi = 1.693AU$
Earth to Venus:
$fracint_0^pi sqrt1^2+0.723^2-2*1*0.723*cos(theta) dthetapi = 1.136AU$
Earth to Mercury:
$fracint_0^pi sqrt1^2+0.39^2-2*1*.0.39*cos(theta) dthetapi = 1.038AU$
answered Mar 13 at 6:41
AmorydaiAmorydai
1,533215
$endgroup$
$begingroup$
The key in this calculation is “divide by pi”. Here you are assuming a uniform distribution of theta. That is, you assume the probability of the other planet being in any other theta is equal, and then you take the expected value. Depending on the other planets’ theta as a function of time, this need not be true. So here we are making the simplifying assumption that all planets travel at a constant speed.
$endgroup$
– darksky
Mar 13 at 11:20
$begingroup$
@darksky Well, we are assuming the orbits are circles, so we are leaving Kepler out of it! Or, rather, we are leaving Kepler in it - “equal areas during equal intervals of time” in a circle would mean constant speed.
$endgroup$
– Amorydai
Mar 13 at 14:27
3
$begingroup$
@darksky Yep, constant speed and circular orbits. Both are not true, but the corrections would be much less than 7% difference between Mercury and Venus
$endgroup$
– yo'
Mar 13 at 14:27
1
$begingroup$
@Amorydai, how did you calculate the integrals?
$endgroup$
– Dr Xorile
Mar 13 at 16:54
1
$begingroup$
How very dare you use the word "Obviously"?! When did puzzling start allowing witchcraft in answers? That formula is satanic... I mean, what sort of wizard-y demon-scratch is that un-figurable? Let's have a vote on a site-wide ban on pomped up NASA geniuses... (in English: "This made me feel dumb")
$endgroup$
– Brent Hackers
Mar 14 at 11:39
|
show 5 more comments
$begingroup$
The closest planet to Earth on average is:
Mercury
The other answers didn't give any calculations, so I'll provide some numbers. Hopefully they are correct!
As other answers suggested, we can leave Earth stationary and just have the other planets do their orbits. Actually we only need to do half an orbit, because the other half will be exactly like the first half and not change the average in any way.
By law of cosines, we can find the distance between the Earth and another planet by looking at the triangle that is formed when you connect the Earth with the other planet and the Sun. Obviously the distance to the Sun is the radius of the orbits and the angle will be $theta$. The distance between the planets is then $sqrtr^2+R^2-2*r*R*cos(theta)$ where $r$ is the radius of Earth's orbit and $R$ is the radius of the other planet's orbit and $theta$ is the angle between them.
Now just find the integral as $theta$ goes from 0 to $pi$ and divide by $pi$.
Earth to Mars:
$fracint_0^pi sqrt1^2+1.524^2-2*1*1.524*cos(theta) dthetapi = 1.693AU$
Earth to Venus:
$fracint_0^pi sqrt1^2+0.723^2-2*1*0.723*cos(theta) dthetapi = 1.136AU$
Earth to Mercury:
$fracint_0^pi sqrt1^2+0.39^2-2*1*.0.39*cos(theta) dthetapi = 1.038AU$
answered Mar 13 at 6:41
AmorydaiAmorydai
1,533215
$endgroup$
The closest planet to Earth on average is:
Mercury
The other answers didn't give any calculations, so I'll provide some numbers. Hopefully they are correct!
As other answers suggested, we can leave Earth stationary and just have the other planets do their orbits. Actually we only need to do half an orbit, because the other half will be exactly like the first half and not change the average in any way.
By law of cosines, we can find the distance between the Earth and another planet by looking at the triangle that is formed when you connect the Earth with the other planet and the Sun. Obviously the distance to the Sun is the radius of the orbits and the angle will be $theta$. The distance between the planets is then $sqrtr^2+R^2-2*r*R*cos(theta)$ where $r$ is the radius of Earth's orbit and $R$ is the radius of the other planet's orbit and $theta$ is the angle between them.
Now just find the integral as $theta$ goes from 0 to $pi$ and divide by $pi$.
Earth to Mars:
$fracint_0^pi sqrt1^2+1.524^2-2*1*1.524*cos(theta) dthetapi = 1.693AU$
Earth to Venus:
$fracint_0^pi sqrt1^2+0.723^2-2*1*0.723*cos(theta) dthetapi = 1.136AU$
Earth to Mercury:
$fracint_0^pi sqrt1^2+0.39^2-2*1*.0.39*cos(theta) dthetapi = 1.038AU$
answered Mar 13 at 6:41
AmorydaiAmorydai
1,533215
answered Mar 13 at 6:41
AmorydaiAmorydai
1,533215
answered Mar 13 at 6:41
AmorydaiAmorydai
1,533215
answered Mar 13 at 6:41
AmorydaiAmorydai
1,533215
1,533215
$begingroup$
The key in this calculation is “divide by pi”. Here you are assuming a uniform distribution of theta. That is, you assume the probability of the other planet being in any other theta is equal, and then you take the expected value. Depending on the other planets’ theta as a function of time, this need not be true. So here we are making the simplifying assumption that all planets travel at a constant speed.
$endgroup$
– darksky
Mar 13 at 11:20
$begingroup$
@darksky Well, we are assuming the orbits are circles, so we are leaving Kepler out of it! Or, rather, we are leaving Kepler in it - “equal areas during equal intervals of time” in a circle would mean constant speed.
$endgroup$
– Amorydai
Mar 13 at 14:27
3
$begingroup$
@darksky Yep, constant speed and circular orbits. Both are not true, but the corrections would be much less than 7% difference between Mercury and Venus
$endgroup$
– yo'
Mar 13 at 14:27
1
$begingroup$
@Amorydai, how did you calculate the integrals?
$endgroup$
– Dr Xorile
Mar 13 at 16:54
1
$begingroup$
How very dare you use the word "Obviously"?! When did puzzling start allowing witchcraft in answers? That formula is satanic... I mean, what sort of wizard-y demon-scratch is that un-figurable? Let's have a vote on a site-wide ban on pomped up NASA geniuses... (in English: "This made me feel dumb")
$endgroup$
– Brent Hackers
Mar 14 at 11:39
|
show 5 more comments
$begingroup$
The key in this calculation is “divide by pi”. Here you are assuming a uniform distribution of theta. That is, you assume the probability of the other planet being in any other theta is equal, and then you take the expected value. Depending on the other planets’ theta as a function of time, this need not be true. So here we are making the simplifying assumption that all planets travel at a constant speed.
$endgroup$
– darksky
Mar 13 at 11:20
$begingroup$
@darksky Well, we are assuming the orbits are circles, so we are leaving Kepler out of it! Or, rather, we are leaving Kepler in it - “equal areas during equal intervals of time” in a circle would mean constant speed.
$endgroup$
– Amorydai
Mar 13 at 14:27
3
$begingroup$
@darksky Yep, constant speed and circular orbits. Both are not true, but the corrections would be much less than 7% difference between Mercury and Venus
$endgroup$
– yo'
Mar 13 at 14:27
1
$begingroup$
@Amorydai, how did you calculate the integrals?
$endgroup$
– Dr Xorile
Mar 13 at 16:54
1
$begingroup$
How very dare you use the word "Obviously"?! When did puzzling start allowing witchcraft in answers? That formula is satanic... I mean, what sort of wizard-y demon-scratch is that un-figurable? Let's have a vote on a site-wide ban on pomped up NASA geniuses... (in English: "This made me feel dumb")
$endgroup$
– Brent Hackers
Mar 14 at 11:39
$begingroup$
The key in this calculation is “divide by pi”. Here you are assuming a uniform distribution of theta. That is, you assume the probability of the other planet being in any other theta is equal, and then you take the expected value. Depending on the other planets’ theta as a function of time, this need not be true. So here we are making the simplifying assumption that all planets travel at a constant speed.
$endgroup$
– darksky
Mar 13 at 11:20
$begingroup$
The key in this calculation is “divide by pi”. Here you are assuming a uniform distribution of theta. That is, you assume the probability of the other planet being in any other theta is equal, and then you take the expected value. Depending on the other planets’ theta as a function of time, this need not be true. So here we are making the simplifying assumption that all planets travel at a constant speed.
$endgroup$
– darksky
Mar 13 at 11:20
$begingroup$
@darksky Well, we are assuming the orbits are circles, so we are leaving Kepler out of it! Or, rather, we are leaving Kepler in it - “equal areas during equal intervals of time” in a circle would mean constant speed.
$endgroup$
– Amorydai
Mar 13 at 14:27
$begingroup$
@darksky Well, we are assuming the orbits are circles, so we are leaving Kepler out of it! Or, rather, we are leaving Kepler in it - “equal areas during equal intervals of time” in a circle would mean constant speed.
$endgroup$
– Amorydai
Mar 13 at 14:27
3
3
$begingroup$
@darksky Yep, constant speed and circular orbits. Both are not true, but the corrections would be much less than 7% difference between Mercury and Venus
$endgroup$
– yo'
Mar 13 at 14:27
$begingroup$
@darksky Yep, constant speed and circular orbits. Both are not true, but the corrections would be much less than 7% difference between Mercury and Venus
$endgroup$
– yo'
Mar 13 at 14:27
1
1
$begingroup$
@Amorydai, how did you calculate the integrals?
$endgroup$
– Dr Xorile
Mar 13 at 16:54
$begingroup$
@Amorydai, how did you calculate the integrals?
$endgroup$
– Dr Xorile
Mar 13 at 16:54
1
1
$begingroup$
How very dare you use the word "Obviously"?! When did puzzling start allowing witchcraft in answers? That formula is satanic... I mean, what sort of wizard-y demon-scratch is that un-figurable? Let's have a vote on a site-wide ban on pomped up NASA geniuses... (in English: "This made me feel dumb")
$endgroup$
– Brent Hackers
Mar 14 at 11:39
$begingroup$
How very dare you use the word "Obviously"?! When did puzzling start allowing witchcraft in answers? That formula is satanic... I mean, what sort of wizard-y demon-scratch is that un-figurable? Let's have a vote on a site-wide ban on pomped up NASA geniuses... (in English: "This made me feel dumb")
$endgroup$
– Brent Hackers
Mar 14 at 11:39
|
show 5 more comments
$begingroup$
I must admit, I'm a bit rusty at calculus. So here's an attempt at an answer free of calculations, but with some more visual reasoning.
First, let's draw out the orbits of the planets. Because there is no resonance, let's assume Earth is still and they all rotate at the same speed. Also, for reference we'll draw a circle of distance 1 AU around Earth.
Now, notice that:
I've drawn a few dotted lines here. If we imagine these circles as pie charts, the left part of the charts represent the time spent more than 1 AU away from the Earth.
So:
Let's plot the distance away from the Earth over time, which looks vaguely like this:
Here, you should note:
The arrows along the bottom, telling you the time spent above the green line (1AU still) and the arrows in the middle, telling you that for the Venus and Mercury, (maximum distance - 1AU) is equal to (1AU - minimum distance).
Also, Mars is clearly out of the question. Goodbye Mars.
So:
It's between Mercury and Venus. From calculus or intuition, we know that the average distance of the planet is proportional to the area under the curve. And now it gets a bit non-technical.
The area under the curve is equal to 2π AU·rad + (bit above the curve) - (bits below the curve). But the bit above the curve is approximately the same shape as the bit below the curve (if we shift the right bit under the curve to the left side of the graph), and since they are the same height their area is probably proportional to their width. And since Venus' width of bit above the curve to width of bit below the curve ratio is bigger than that of Mercury, and the fact that those bits are taller than Mercury's bits, I estimate Venus' total area is probably more than that of Mercury's.
So my guess is that:
Mercury is on average closest to the Earth. (I'd love to know how accurate this argument is, but that maths is beyond me.)
NB: Click on images for slightly higher quality if they're a bit fuzzy.
answered Mar 13 at 6:02
boboquackboboquack
15.9k150119
$endgroup$
4
$begingroup$
Heck, I so love this one. If you wanna teach mathematical/physical intuition, this is the way!
$endgroup$
– yo'
Mar 13 at 14:29
$begingroup$
Love this answer. Super helpful for gaining the intuition.
$endgroup$
– Dr Xorile
Mar 13 at 16:48
$begingroup$
One of my favorite answers in this whole site. I would actually recommend reading this before reading the accepted answer so it's easier to digest the technical math.
$endgroup$
– yushi
Mar 15 at 10:07
add a comment |
$begingroup$
I must admit, I'm a bit rusty at calculus. So here's an attempt at an answer free of calculations, but with some more visual reasoning.
First, let's draw out the orbits of the planets. Because there is no resonance, let's assume Earth is still and they all rotate at the same speed. Also, for reference we'll draw a circle of distance 1 AU around Earth.
Now, notice that:
I've drawn a few dotted lines here. If we imagine these circles as pie charts, the left part of the charts represent the time spent more than 1 AU away from the Earth.
So:
Let's plot the distance away from the Earth over time, which looks vaguely like this:
Here, you should note:
The arrows along the bottom, telling you the time spent above the green line (1AU still) and the arrows in the middle, telling you that for the Venus and Mercury, (maximum distance - 1AU) is equal to (1AU - minimum distance).
Also, Mars is clearly out of the question. Goodbye Mars.
So:
It's between Mercury and Venus. From calculus or intuition, we know that the average distance of the planet is proportional to the area under the curve. And now it gets a bit non-technical.
The area under the curve is equal to 2π AU·rad + (bit above the curve) - (bits below the curve). But the bit above the curve is approximately the same shape as the bit below the curve (if we shift the right bit under the curve to the left side of the graph), and since they are the same height their area is probably proportional to their width. And since Venus' width of bit above the curve to width of bit below the curve ratio is bigger than that of Mercury, and the fact that those bits are taller than Mercury's bits, I estimate Venus' total area is probably more than that of Mercury's.
So my guess is that:
Mercury is on average closest to the Earth. (I'd love to know how accurate this argument is, but that maths is beyond me.)
NB: Click on images for slightly higher quality if they're a bit fuzzy.
answered Mar 13 at 6:02
boboquackboboquack
15.9k150119
$endgroup$
4
$begingroup$
Heck, I so love this one. If you wanna teach mathematical/physical intuition, this is the way!
$endgroup$
– yo'
Mar 13 at 14:29
$begingroup$
Love this answer. Super helpful for gaining the intuition.
$endgroup$
– Dr Xorile
Mar 13 at 16:48
$begingroup$
One of my favorite answers in this whole site. I would actually recommend reading this before reading the accepted answer so it's easier to digest the technical math.
$endgroup$
– yushi
Mar 15 at 10:07
add a comment |
$begingroup$
I must admit, I'm a bit rusty at calculus. So here's an attempt at an answer free of calculations, but with some more visual reasoning.
First, let's draw out the orbits of the planets. Because there is no resonance, let's assume Earth is still and they all rotate at the same speed. Also, for reference we'll draw a circle of distance 1 AU around Earth.
Now, notice that:
I've drawn a few dotted lines here. If we imagine these circles as pie charts, the left part of the charts represent the time spent more than 1 AU away from the Earth.
So:
Let's plot the distance away from the Earth over time, which looks vaguely like this:
Here, you should note:
The arrows along the bottom, telling you the time spent above the green line (1AU still) and the arrows in the middle, telling you that for the Venus and Mercury, (maximum distance - 1AU) is equal to (1AU - minimum distance).
Also, Mars is clearly out of the question. Goodbye Mars.
So:
It's between Mercury and Venus. From calculus or intuition, we know that the average distance of the planet is proportional to the area under the curve. And now it gets a bit non-technical.
The area under the curve is equal to 2π AU·rad + (bit above the curve) - (bits below the curve). But the bit above the curve is approximately the same shape as the bit below the curve (if we shift the right bit under the curve to the left side of the graph), and since they are the same height their area is probably proportional to their width. And since Venus' width of bit above the curve to width of bit below the curve ratio is bigger than that of Mercury, and the fact that those bits are taller than Mercury's bits, I estimate Venus' total area is probably more than that of Mercury's.
So my guess is that:
Mercury is on average closest to the Earth. (I'd love to know how accurate this argument is, but that maths is beyond me.)
NB: Click on images for slightly higher quality if they're a bit fuzzy.
answered Mar 13 at 6:02
boboquackboboquack
15.9k150119
$endgroup$
I must admit, I'm a bit rusty at calculus. So here's an attempt at an answer free of calculations, but with some more visual reasoning.
First, let's draw out the orbits of the planets. Because there is no resonance, let's assume Earth is still and they all rotate at the same speed. Also, for reference we'll draw a circle of distance 1 AU around Earth.
Now, notice that:
I've drawn a few dotted lines here. If we imagine these circles as pie charts, the left part of the charts represent the time spent more than 1 AU away from the Earth.
So:
Let's plot the distance away from the Earth over time, which looks vaguely like this:
Here, you should note:
The arrows along the bottom, telling you the time spent above the green line (1AU still) and the arrows in the middle, telling you that for the Venus and Mercury, (maximum distance - 1AU) is equal to (1AU - minimum distance).
Also, Mars is clearly out of the question. Goodbye Mars.
So:
It's between Mercury and Venus. From calculus or intuition, we know that the average distance of the planet is proportional to the area under the curve. And now it gets a bit non-technical.
The area under the curve is equal to 2π AU·rad + (bit above the curve) - (bits below the curve). But the bit above the curve is approximately the same shape as the bit below the curve (if we shift the right bit under the curve to the left side of the graph), and since they are the same height their area is probably proportional to their width. And since Venus' width of bit above the curve to width of bit below the curve ratio is bigger than that of Mercury, and the fact that those bits are taller than Mercury's bits, I estimate Venus' total area is probably more than that of Mercury's.
So my guess is that:
Mercury is on average closest to the Earth. (I'd love to know how accurate this argument is, but that maths is beyond me.)
NB: Click on images for slightly higher quality if they're a bit fuzzy.
answered Mar 13 at 6:02
boboquackboboquack
15.9k150119
edited Mar 13 at 10:03
answered Mar 13 at 6:02
boboquackboboquack
15.9k150119
answered Mar 13 at 6:02
boboquackboboquack
15.9k150119
answered Mar 13 at 6:02
boboquackboboquack
15.9k150119
15.9k150119
4
$begingroup$
Heck, I so love this one. If you wanna teach mathematical/physical intuition, this is the way!
$endgroup$
– yo'
Mar 13 at 14:29
$begingroup$
Love this answer. Super helpful for gaining the intuition.
$endgroup$
– Dr Xorile
Mar 13 at 16:48
$begingroup$
One of my favorite answers in this whole site. I would actually recommend reading this before reading the accepted answer so it's easier to digest the technical math.
$endgroup$
– yushi
Mar 15 at 10:07
add a comment |
4
$begingroup$
Heck, I so love this one. If you wanna teach mathematical/physical intuition, this is the way!
$endgroup$
– yo'
Mar 13 at 14:29
$begingroup$
Love this answer. Super helpful for gaining the intuition.
$endgroup$
– Dr Xorile
Mar 13 at 16:48
$begingroup$
One of my favorite answers in this whole site. I would actually recommend reading this before reading the accepted answer so it's easier to digest the technical math.
$endgroup$
– yushi
Mar 15 at 10:07
4
4
$begingroup$
Heck, I so love this one. If you wanna teach mathematical/physical intuition, this is the way!
$endgroup$
– yo'
Mar 13 at 14:29
$begingroup$
Heck, I so love this one. If you wanna teach mathematical/physical intuition, this is the way!
$endgroup$
– yo'
Mar 13 at 14:29
$begingroup$
Love this answer. Super helpful for gaining the intuition.
$endgroup$
– Dr Xorile
Mar 13 at 16:48
$begingroup$
Love this answer. Super helpful for gaining the intuition.
$endgroup$
– Dr Xorile
Mar 13 at 16:48
$begingroup$
One of my favorite answers in this whole site. I would actually recommend reading this before reading the accepted answer so it's easier to digest the technical math.
$endgroup$
– yushi
Mar 15 at 10:07
$begingroup$
One of my favorite answers in this whole site. I would actually recommend reading this before reading the accepted answer so it's easier to digest the technical math.
$endgroup$
– yushi
Mar 15 at 10:07
add a comment |
$begingroup$
The answer:
Mercury
Reasoning:
The average position of Earth (and indeed all planets in a circular orbit) is the middle of the Sun. Since Mercury's orbit is closest to the sun, it's the nearest on average to the Earth, and indeed all the other planets.
answered Mar 13 at 6:12
Matthew BarberMatthew Barber
84516
$endgroup$
5
$begingroup$
Following that reasonning, wouldn't the answer be "All planets are equally close to the Earth on average, as all of their average positions fall at the same place (middle of the Sun)" ?
$endgroup$
– Soltius
Mar 13 at 10:54
1
$begingroup$
By this logic, the average closest planet to our moon is also Mercury.
$endgroup$
– BlueRaja - Danny Pflughoeft
Mar 13 at 17:30
$begingroup$
@BlueRaja That's only true if you assume that the orbits of the Moon and the Earth around the sun are independent, which they obviously aren't.
$endgroup$
– Matthew Barber
Mar 13 at 21:32
1
$begingroup$
@Soltius No, that does not follow, as you cannot calculate average distances entirely from average positions. Averaging the position of the Earth won't help you calculate the distance either. It's just something that illustrates why the distance from the sun of the other planet is the only thing that matters in the ordering.
$endgroup$
– Matthew Barber
Mar 13 at 22:03
add a comment |
$begingroup$
The answer:
Mercury
Reasoning:
The average position of Earth (and indeed all planets in a circular orbit) is the middle of the Sun. Since Mercury's orbit is closest to the sun, it's the nearest on average to the Earth, and indeed all the other planets.
answered Mar 13 at 6:12
Matthew BarberMatthew Barber
84516
$endgroup$
5
$begingroup$
Following that reasonning, wouldn't the answer be "All planets are equally close to the Earth on average, as all of their average positions fall at the same place (middle of the Sun)" ?
$endgroup$
– Soltius
Mar 13 at 10:54
1
$begingroup$
By this logic, the average closest planet to our moon is also Mercury.
$endgroup$
– BlueRaja - Danny Pflughoeft
Mar 13 at 17:30
$begingroup$
@BlueRaja That's only true if you assume that the orbits of the Moon and the Earth around the sun are independent, which they obviously aren't.
$endgroup$
– Matthew Barber
Mar 13 at 21:32
1
$begingroup$
@Soltius No, that does not follow, as you cannot calculate average distances entirely from average positions. Averaging the position of the Earth won't help you calculate the distance either. It's just something that illustrates why the distance from the sun of the other planet is the only thing that matters in the ordering.
$endgroup$
– Matthew Barber
Mar 13 at 22:03
add a comment |
$begingroup$
The answer:
Mercury
Reasoning:
The average position of Earth (and indeed all planets in a circular orbit) is the middle of the Sun. Since Mercury's orbit is closest to the sun, it's the nearest on average to the Earth, and indeed all the other planets.
answered Mar 13 at 6:12
Matthew BarberMatthew Barber
84516
$endgroup$
The answer:
Mercury
Reasoning:
The average position of Earth (and indeed all planets in a circular orbit) is the middle of the Sun. Since Mercury's orbit is closest to the sun, it's the nearest on average to the Earth, and indeed all the other planets.
answered Mar 13 at 6:12
Matthew BarberMatthew Barber
84516
answered Mar 13 at 6:12
Matthew BarberMatthew Barber
84516
answered Mar 13 at 6:12
Matthew BarberMatthew Barber
84516
answered Mar 13 at 6:12
Matthew BarberMatthew Barber
84516
84516
5
$begingroup$
Following that reasonning, wouldn't the answer be "All planets are equally close to the Earth on average, as all of their average positions fall at the same place (middle of the Sun)" ?
$endgroup$
– Soltius
Mar 13 at 10:54
1
$begingroup$
By this logic, the average closest planet to our moon is also Mercury.
$endgroup$
– BlueRaja - Danny Pflughoeft
Mar 13 at 17:30
$begingroup$
@BlueRaja That's only true if you assume that the orbits of the Moon and the Earth around the sun are independent, which they obviously aren't.
$endgroup$
– Matthew Barber
Mar 13 at 21:32
1
$begingroup$
@Soltius No, that does not follow, as you cannot calculate average distances entirely from average positions. Averaging the position of the Earth won't help you calculate the distance either. It's just something that illustrates why the distance from the sun of the other planet is the only thing that matters in the ordering.
$endgroup$
– Matthew Barber
Mar 13 at 22:03
add a comment |
5
$begingroup$
Following that reasonning, wouldn't the answer be "All planets are equally close to the Earth on average, as all of their average positions fall at the same place (middle of the Sun)" ?
$endgroup$
– Soltius
Mar 13 at 10:54
1
$begingroup$
By this logic, the average closest planet to our moon is also Mercury.
$endgroup$
– BlueRaja - Danny Pflughoeft
Mar 13 at 17:30
$begingroup$
@BlueRaja That's only true if you assume that the orbits of the Moon and the Earth around the sun are independent, which they obviously aren't.
$endgroup$
– Matthew Barber
Mar 13 at 21:32
1
$begingroup$
@Soltius No, that does not follow, as you cannot calculate average distances entirely from average positions. Averaging the position of the Earth won't help you calculate the distance either. It's just something that illustrates why the distance from the sun of the other planet is the only thing that matters in the ordering.
$endgroup$
– Matthew Barber
Mar 13 at 22:03
5
5
$begingroup$
Following that reasonning, wouldn't the answer be "All planets are equally close to the Earth on average, as all of their average positions fall at the same place (middle of the Sun)" ?
$endgroup$
– Soltius
Mar 13 at 10:54
$begingroup$
Following that reasonning, wouldn't the answer be "All planets are equally close to the Earth on average, as all of their average positions fall at the same place (middle of the Sun)" ?
$endgroup$
– Soltius
Mar 13 at 10:54
1
1
$begingroup$
By this logic, the average closest planet to our moon is also Mercury.
$endgroup$
– BlueRaja - Danny Pflughoeft
Mar 13 at 17:30
$begingroup$
By this logic, the average closest planet to our moon is also Mercury.
$endgroup$
– BlueRaja - Danny Pflughoeft
Mar 13 at 17:30
$begingroup$
@BlueRaja That's only true if you assume that the orbits of the Moon and the Earth around the sun are independent, which they obviously aren't.
$endgroup$
– Matthew Barber
Mar 13 at 21:32
$begingroup$
@BlueRaja That's only true if you assume that the orbits of the Moon and the Earth around the sun are independent, which they obviously aren't.
$endgroup$
– Matthew Barber
Mar 13 at 21:32
1
1
$begingroup$
@Soltius No, that does not follow, as you cannot calculate average distances entirely from average positions. Averaging the position of the Earth won't help you calculate the distance either. It's just something that illustrates why the distance from the sun of the other planet is the only thing that matters in the ordering.
$endgroup$
– Matthew Barber
Mar 13 at 22:03
$begingroup$
@Soltius No, that does not follow, as you cannot calculate average distances entirely from average positions. Averaging the position of the Earth won't help you calculate the distance either. It's just something that illustrates why the distance from the sun of the other planet is the only thing that matters in the ordering.
$endgroup$
– Matthew Barber
Mar 13 at 22:03
add a comment |
$begingroup$
Others have done all the necessary calculations, so here's some hairy maths. I assume as per the question that all orbits are circular and that the planets move in such a way that the average distance equals the average over all angular differences. Then it turns out that the average distance from earth to a planet whose orbit has radius $r$ astronomical units (i.e., $r$ times the radius of the earth's orbit) is $frac2pi(1+r)E(frac4r1+r^2)$ astronomical units, where E is the so-called complete elliptic integral of the second kind, what Mathematica calls EllipticE.
So what we'd like to be true is that this is an increasing function of $r$. This does appear to be true, but proving it is not so trivial.
Rather than looking at the average over the whole orbit, let's look at just two antipodal points. So, suppose the angle between earth's position and the other planet's position is $theta$, so that the distance is $sqrt(r-costheta)^2+sin^2theta$; half-way around the orbit the other planet's position is $theta+pi$ and the distance is $sqrt(r+costheta)^2+sin^2theta$. The sum of these is $f(r,theta):=sqrt(r-costheta)^2+sin^2theta+sqrt(r+costheta)^2+sin^2theta$, our average is the average of this over all values of $theta$, and it will be an increasing function of $r$ if $f$ is for every $theta$. This will be true if it's true when we consider instead $g(r,u,v):=sqrt(r-u)^2+v^2+sqrt(r+u)^2+v^2$ and allow $u,v$ to take any value at all. (Which just corresponds to letting the earth's distance from the sun be something other than 1 unit.)
The derivative of this thing is $fracr+usqrt(r+u)^2+v^2+fracr-usqrt(r-u)^2+v^2$. Obviously this is positive when $r>u$. When $r<u$ it's $h(u+r,v)-h(u-r,v)$ where $h(p,q)=fracpsqrtp^2+q^2=costan^-1frac qp$. But this is obviously a decreasing function of $q/p$, hence an increasing function of $p$, which means that $h(u+r,v)>h(u-r,v)$, which means that $fracpartial gpartial r>0$, which means that $fracpartial fpartial r>0$, which means that $fracpartialint fpartial r>0$, which means that indeed the average distance is an increasing function of $r$.
I suspect there may be an easier more purely geometrical way to do this.
answered Mar 13 at 11:05
Gareth McCaughan♦Gareth McCaughan
67.3k3170261
$endgroup$
$begingroup$
@boboquack gets at a nice intuition for why this is true. I think also looking at the pairs of points: in line with the earth they balance out (so 0 difference) and at right angles the further out the orbit the further out the distance. So a continuity argument says that the overall average is monotonic.
$endgroup$
– Dr Xorile
Mar 13 at 16:52
add a comment |
$begingroup$
Others have done all the necessary calculations, so here's some hairy maths. I assume as per the question that all orbits are circular and that the planets move in such a way that the average distance equals the average over all angular differences. Then it turns out that the average distance from earth to a planet whose orbit has radius $r$ astronomical units (i.e., $r$ times the radius of the earth's orbit) is $frac2pi(1+r)E(frac4r1+r^2)$ astronomical units, where E is the so-called complete elliptic integral of the second kind, what Mathematica calls EllipticE.
So what we'd like to be true is that this is an increasing function of $r$. This does appear to be true, but proving it is not so trivial.
Rather than looking at the average over the whole orbit, let's look at just two antipodal points. So, suppose the angle between earth's position and the other planet's position is $theta$, so that the distance is $sqrt(r-costheta)^2+sin^2theta$; half-way around the orbit the other planet's position is $theta+pi$ and the distance is $sqrt(r+costheta)^2+sin^2theta$. The sum of these is $f(r,theta):=sqrt(r-costheta)^2+sin^2theta+sqrt(r+costheta)^2+sin^2theta$, our average is the average of this over all values of $theta$, and it will be an increasing function of $r$ if $f$ is for every $theta$. This will be true if it's true when we consider instead $g(r,u,v):=sqrt(r-u)^2+v^2+sqrt(r+u)^2+v^2$ and allow $u,v$ to take any value at all. (Which just corresponds to letting the earth's distance from the sun be something other than 1 unit.)
The derivative of this thing is $fracr+usqrt(r+u)^2+v^2+fracr-usqrt(r-u)^2+v^2$. Obviously this is positive when $r>u$. When $r<u$ it's $h(u+r,v)-h(u-r,v)$ where $h(p,q)=fracpsqrtp^2+q^2=costan^-1frac qp$. But this is obviously a decreasing function of $q/p$, hence an increasing function of $p$, which means that $h(u+r,v)>h(u-r,v)$, which means that $fracpartial gpartial r>0$, which means that $fracpartial fpartial r>0$, which means that $fracpartialint fpartial r>0$, which means that indeed the average distance is an increasing function of $r$.
I suspect there may be an easier more purely geometrical way to do this.
answered Mar 13 at 11:05
Gareth McCaughan♦Gareth McCaughan
67.3k3170261
$endgroup$
$begingroup$
@boboquack gets at a nice intuition for why this is true. I think also looking at the pairs of points: in line with the earth they balance out (so 0 difference) and at right angles the further out the orbit the further out the distance. So a continuity argument says that the overall average is monotonic.
$endgroup$
– Dr Xorile
Mar 13 at 16:52
add a comment |
$begingroup$
Others have done all the necessary calculations, so here's some hairy maths. I assume as per the question that all orbits are circular and that the planets move in such a way that the average distance equals the average over all angular differences. Then it turns out that the average distance from earth to a planet whose orbit has radius $r$ astronomical units (i.e., $r$ times the radius of the earth's orbit) is $frac2pi(1+r)E(frac4r1+r^2)$ astronomical units, where E is the so-called complete elliptic integral of the second kind, what Mathematica calls EllipticE.
So what we'd like to be true is that this is an increasing function of $r$. This does appear to be true, but proving it is not so trivial.
Rather than looking at the average over the whole orbit, let's look at just two antipodal points. So, suppose the angle between earth's position and the other planet's position is $theta$, so that the distance is $sqrt(r-costheta)^2+sin^2theta$; half-way around the orbit the other planet's position is $theta+pi$ and the distance is $sqrt(r+costheta)^2+sin^2theta$. The sum of these is $f(r,theta):=sqrt(r-costheta)^2+sin^2theta+sqrt(r+costheta)^2+sin^2theta$, our average is the average of this over all values of $theta$, and it will be an increasing function of $r$ if $f$ is for every $theta$. This will be true if it's true when we consider instead $g(r,u,v):=sqrt(r-u)^2+v^2+sqrt(r+u)^2+v^2$ and allow $u,v$ to take any value at all. (Which just corresponds to letting the earth's distance from the sun be something other than 1 unit.)
The derivative of this thing is $fracr+usqrt(r+u)^2+v^2+fracr-usqrt(r-u)^2+v^2$. Obviously this is positive when $r>u$. When $r<u$ it's $h(u+r,v)-h(u-r,v)$ where $h(p,q)=fracpsqrtp^2+q^2=costan^-1frac qp$. But this is obviously a decreasing function of $q/p$, hence an increasing function of $p$, which means that $h(u+r,v)>h(u-r,v)$, which means that $fracpartial gpartial r>0$, which means that $fracpartial fpartial r>0$, which means that $fracpartialint fpartial r>0$, which means that indeed the average distance is an increasing function of $r$.
I suspect there may be an easier more purely geometrical way to do this.
answered Mar 13 at 11:05
Gareth McCaughan♦Gareth McCaughan
67.3k3170261
$endgroup$
Others have done all the necessary calculations, so here's some hairy maths. I assume as per the question that all orbits are circular and that the planets move in such a way that the average distance equals the average over all angular differences. Then it turns out that the average distance from earth to a planet whose orbit has radius $r$ astronomical units (i.e., $r$ times the radius of the earth's orbit) is $frac2pi(1+r)E(frac4r1+r^2)$ astronomical units, where E is the so-called complete elliptic integral of the second kind, what Mathematica calls EllipticE.
So what we'd like to be true is that this is an increasing function of $r$. This does appear to be true, but proving it is not so trivial.
Rather than looking at the average over the whole orbit, let's look at just two antipodal points. So, suppose the angle between earth's position and the other planet's position is $theta$, so that the distance is $sqrt(r-costheta)^2+sin^2theta$; half-way around the orbit the other planet's position is $theta+pi$ and the distance is $sqrt(r+costheta)^2+sin^2theta$. The sum of these is $f(r,theta):=sqrt(r-costheta)^2+sin^2theta+sqrt(r+costheta)^2+sin^2theta$, our average is the average of this over all values of $theta$, and it will be an increasing function of $r$ if $f$ is for every $theta$. This will be true if it's true when we consider instead $g(r,u,v):=sqrt(r-u)^2+v^2+sqrt(r+u)^2+v^2$ and allow $u,v$ to take any value at all. (Which just corresponds to letting the earth's distance from the sun be something other than 1 unit.)
The derivative of this thing is $fracr+usqrt(r+u)^2+v^2+fracr-usqrt(r-u)^2+v^2$. Obviously this is positive when $r>u$. When $r<u$ it's $h(u+r,v)-h(u-r,v)$ where $h(p,q)=fracpsqrtp^2+q^2=costan^-1frac qp$. But this is obviously a decreasing function of $q/p$, hence an increasing function of $p$, which means that $h(u+r,v)>h(u-r,v)$, which means that $fracpartial gpartial r>0$, which means that $fracpartial fpartial r>0$, which means that $fracpartialint fpartial r>0$, which means that indeed the average distance is an increasing function of $r$.
I suspect there may be an easier more purely geometrical way to do this.
answered Mar 13 at 11:05
Gareth McCaughan♦Gareth McCaughan
67.3k3170261
answered Mar 13 at 11:05
Gareth McCaughan♦Gareth McCaughan
67.3k3170261
answered Mar 13 at 11:05
Gareth McCaughan♦Gareth McCaughan
67.3k3170261
answered Mar 13 at 11:05
Gareth McCaughan♦Gareth McCaughan
67.3k3170261
67.3k3170261
$begingroup$
@boboquack gets at a nice intuition for why this is true. I think also looking at the pairs of points: in line with the earth they balance out (so 0 difference) and at right angles the further out the orbit the further out the distance. So a continuity argument says that the overall average is monotonic.
$endgroup$
– Dr Xorile
Mar 13 at 16:52
add a comment |
$begingroup$
@boboquack gets at a nice intuition for why this is true. I think also looking at the pairs of points: in line with the earth they balance out (so 0 difference) and at right angles the further out the orbit the further out the distance. So a continuity argument says that the overall average is monotonic.
$endgroup$
– Dr Xorile
Mar 13 at 16:52
$begingroup$
@boboquack gets at a nice intuition for why this is true. I think also looking at the pairs of points: in line with the earth they balance out (so 0 difference) and at right angles the further out the orbit the further out the distance. So a continuity argument says that the overall average is monotonic.
$endgroup$
– Dr Xorile
Mar 13 at 16:52
$begingroup$
@boboquack gets at a nice intuition for why this is true. I think also looking at the pairs of points: in line with the earth they balance out (so 0 difference) and at right angles the further out the orbit the further out the distance. So a continuity argument says that the overall average is monotonic.
$endgroup$
– Dr Xorile
Mar 13 at 16:52
add a comment |
$begingroup$
It's
Mercury.
Because:
Fix the position of the Earth, and let the planets move in orbit. We want the average distance. If the Sun was an object to consider, the radius R would be the average. If there was another planet on Earth's orbit, it's average would be greater than R, as most of the orbit is at a further distance than R from the Earth (draw a circle radius R from the Earth - it cuts the other orbit before the halfway points).
As this is a continuous and monotonic increasing function, the planet closest on average is Mercury.
answered Mar 13 at 6:03
JonMark PerryJonMark Perry
20.7k64099
$endgroup$
$begingroup$
Why is it continuous and monotonic? And how did you deal with Mars?
$endgroup$
– boboquack
Mar 13 at 6:52
$begingroup$
@boboquack; the orbit moves continuously and so therefore does the average function, which is a quadratic, and therefore has a monotonic differential. Larger orbits just get bigger (monotonic increasing remember!). Also see en.wikipedia.org/wiki/Orbit.
$endgroup$
– JonMark Perry
Mar 13 at 7:20
$begingroup$
Average distance to earth is not a quadratic function of orbit radius because for very large orbits it's approximately equal to the radius.
$endgroup$
– Gareth McCaughan♦
Mar 13 at 8:49
$begingroup$
@GarethMcCaughan; this doesn't change my argument much though. the average function depends on R and only approximates R locally.
$endgroup$
– JonMark Perry
Mar 13 at 9:07
$begingroup$
The function in question is monotone increasing, though. At least, I think it is, though I haven't tried to prove it; it's a pretty ugly function involving elliptic integrals.
$endgroup$
– Gareth McCaughan♦
Mar 13 at 9:58
|
show 2 more comments
$begingroup$
It's
Mercury.
Because:
Fix the position of the Earth, and let the planets move in orbit. We want the average distance. If the Sun was an object to consider, the radius R would be the average. If there was another planet on Earth's orbit, it's average would be greater than R, as most of the orbit is at a further distance than R from the Earth (draw a circle radius R from the Earth - it cuts the other orbit before the halfway points).
As this is a continuous and monotonic increasing function, the planet closest on average is Mercury.
answered Mar 13 at 6:03
JonMark PerryJonMark Perry
20.7k64099
$endgroup$
$begingroup$
Why is it continuous and monotonic? And how did you deal with Mars?
$endgroup$
– boboquack
Mar 13 at 6:52
$begingroup$
@boboquack; the orbit moves continuously and so therefore does the average function, which is a quadratic, and therefore has a monotonic differential. Larger orbits just get bigger (monotonic increasing remember!). Also see en.wikipedia.org/wiki/Orbit.
$endgroup$
– JonMark Perry
Mar 13 at 7:20
$begingroup$
Average distance to earth is not a quadratic function of orbit radius because for very large orbits it's approximately equal to the radius.
$endgroup$
– Gareth McCaughan♦
Mar 13 at 8:49
$begingroup$
@GarethMcCaughan; this doesn't change my argument much though. the average function depends on R and only approximates R locally.
$endgroup$
– JonMark Perry
Mar 13 at 9:07
$begingroup$
The function in question is monotone increasing, though. At least, I think it is, though I haven't tried to prove it; it's a pretty ugly function involving elliptic integrals.
$endgroup$
– Gareth McCaughan♦
Mar 13 at 9:58
|
show 2 more comments
$begingroup$
It's
Mercury.
Because:
Fix the position of the Earth, and let the planets move in orbit. We want the average distance. If the Sun was an object to consider, the radius R would be the average. If there was another planet on Earth's orbit, it's average would be greater than R, as most of the orbit is at a further distance than R from the Earth (draw a circle radius R from the Earth - it cuts the other orbit before the halfway points).
As this is a continuous and monotonic increasing function, the planet closest on average is Mercury.
answered Mar 13 at 6:03
JonMark PerryJonMark Perry
20.7k64099
$endgroup$
It's
Mercury.
Because:
Fix the position of the Earth, and let the planets move in orbit. We want the average distance. If the Sun was an object to consider, the radius R would be the average. If there was another planet on Earth's orbit, it's average would be greater than R, as most of the orbit is at a further distance than R from the Earth (draw a circle radius R from the Earth - it cuts the other orbit before the halfway points).
As this is a continuous and monotonic increasing function, the planet closest on average is Mercury.
answered Mar 13 at 6:03
JonMark PerryJonMark Perry
20.7k64099
edited Mar 13 at 6:09
answered Mar 13 at 6:03
JonMark PerryJonMark Perry
20.7k64099
answered Mar 13 at 6:03
JonMark PerryJonMark Perry
20.7k64099
answered Mar 13 at 6:03
JonMark PerryJonMark Perry
20.7k64099
20.7k64099
$begingroup$
Why is it continuous and monotonic? And how did you deal with Mars?
$endgroup$
– boboquack
Mar 13 at 6:52
$begingroup$
@boboquack; the orbit moves continuously and so therefore does the average function, which is a quadratic, and therefore has a monotonic differential. Larger orbits just get bigger (monotonic increasing remember!). Also see en.wikipedia.org/wiki/Orbit.
$endgroup$
– JonMark Perry
Mar 13 at 7:20
$begingroup$
Average distance to earth is not a quadratic function of orbit radius because for very large orbits it's approximately equal to the radius.
$endgroup$
– Gareth McCaughan♦
Mar 13 at 8:49
$begingroup$
@GarethMcCaughan; this doesn't change my argument much though. the average function depends on R and only approximates R locally.
$endgroup$
– JonMark Perry
Mar 13 at 9:07
$begingroup$
The function in question is monotone increasing, though. At least, I think it is, though I haven't tried to prove it; it's a pretty ugly function involving elliptic integrals.
$endgroup$
– Gareth McCaughan♦
Mar 13 at 9:58
|
show 2 more comments
$begingroup$
Why is it continuous and monotonic? And how did you deal with Mars?
$endgroup$
– boboquack
Mar 13 at 6:52
$begingroup$
@boboquack; the orbit moves continuously and so therefore does the average function, which is a quadratic, and therefore has a monotonic differential. Larger orbits just get bigger (monotonic increasing remember!). Also see en.wikipedia.org/wiki/Orbit.
$endgroup$
– JonMark Perry
Mar 13 at 7:20
$begingroup$
Average distance to earth is not a quadratic function of orbit radius because for very large orbits it's approximately equal to the radius.
$endgroup$
– Gareth McCaughan♦
Mar 13 at 8:49
$begingroup$
@GarethMcCaughan; this doesn't change my argument much though. the average function depends on R and only approximates R locally.
$endgroup$
– JonMark Perry
Mar 13 at 9:07
$begingroup$
The function in question is monotone increasing, though. At least, I think it is, though I haven't tried to prove it; it's a pretty ugly function involving elliptic integrals.
$endgroup$
– Gareth McCaughan♦
Mar 13 at 9:58
$begingroup$
Why is it continuous and monotonic? And how did you deal with Mars?
$endgroup$
– boboquack
Mar 13 at 6:52
$begingroup$
Why is it continuous and monotonic? And how did you deal with Mars?
$endgroup$
– boboquack
Mar 13 at 6:52
$begingroup$
@boboquack; the orbit moves continuously and so therefore does the average function, which is a quadratic, and therefore has a monotonic differential. Larger orbits just get bigger (monotonic increasing remember!). Also see en.wikipedia.org/wiki/Orbit.
$endgroup$
– JonMark Perry
Mar 13 at 7:20
$begingroup$
@boboquack; the orbit moves continuously and so therefore does the average function, which is a quadratic, and therefore has a monotonic differential. Larger orbits just get bigger (monotonic increasing remember!). Also see en.wikipedia.org/wiki/Orbit.
$endgroup$
– JonMark Perry
Mar 13 at 7:20
$begingroup$
Average distance to earth is not a quadratic function of orbit radius because for very large orbits it's approximately equal to the radius.
$endgroup$
– Gareth McCaughan♦
Mar 13 at 8:49
$begingroup$
Average distance to earth is not a quadratic function of orbit radius because for very large orbits it's approximately equal to the radius.
$endgroup$
– Gareth McCaughan♦
Mar 13 at 8:49
$begingroup$
@GarethMcCaughan; this doesn't change my argument much though. the average function depends on R and only approximates R locally.
$endgroup$
– JonMark Perry
Mar 13 at 9:07
$begingroup$
@GarethMcCaughan; this doesn't change my argument much though. the average function depends on R and only approximates R locally.
$endgroup$
– JonMark Perry
Mar 13 at 9:07
$begingroup$
The function in question is monotone increasing, though. At least, I think it is, though I haven't tried to prove it; it's a pretty ugly function involving elliptic integrals.
$endgroup$
– Gareth McCaughan♦
Mar 13 at 9:58
$begingroup$
The function in question is monotone increasing, though. At least, I think it is, though I haven't tried to prove it; it's a pretty ugly function involving elliptic integrals.
$endgroup$
– Gareth McCaughan♦
Mar 13 at 9:58
|
show 2 more comments
$begingroup$
Here is another way to deduce the same result without any calculations.
First the inner planets.
Imagine that instead of the Earth, there is a wall at 1 AU from the sun. The average distance of any inner planet's orbit to the wall is easily seen to be exactly 1 AU.
This is because you can pair up points on the left and right sides of the orbit. The average distance of those two points to the wall is the same as the distance of their midpoint to the wall.
If you now go back to measuring the distance to the Earth itself, the distances get larger.
Crucially, the larger the orbit, the higher the slopes of the line segments we are measuring, and the further away the average distance is from 1 AU.
This shows that amongst the inner planets, the innermost planet (Mercury) has the smallest average distance to Earth.
What about the outer planets?
Let's go back to the wall replacing the Earth. If an planet's orbit crosses the wall, then when it comes to measuring its distance, we might as well mirror the planet's position in the wall and measure the distance to its mirror image.
When we then do the same trick of pairing points in the two halves of the orbit, it is clear that the average distance to the wall becomes greater than 1 AU to start with. Combine that with the fact that when measuring the distance to Earth itself the slopes of the line segments are even larger than before, it is clear that the average distance to the planet is even greater compared to the inner planets.
answered Mar 14 at 10:32
Jaap ScherphuisJaap Scherphuis
16.6k12872
$endgroup$
add a comment |
$begingroup$
Here is another way to deduce the same result without any calculations.
First the inner planets.
Imagine that instead of the Earth, there is a wall at 1 AU from the sun. The average distance of any inner planet's orbit to the wall is easily seen to be exactly 1 AU.
This is because you can pair up points on the left and right sides of the orbit. The average distance of those two points to the wall is the same as the distance of their midpoint to the wall.
If you now go back to measuring the distance to the Earth itself, the distances get larger.
Crucially, the larger the orbit, the higher the slopes of the line segments we are measuring, and the further away the average distance is from 1 AU.
This shows that amongst the inner planets, the innermost planet (Mercury) has the smallest average distance to Earth.
What about the outer planets?
Let's go back to the wall replacing the Earth. If an planet's orbit crosses the wall, then when it comes to measuring its distance, we might as well mirror the planet's position in the wall and measure the distance to its mirror image.
When we then do the same trick of pairing points in the two halves of the orbit, it is clear that the average distance to the wall becomes greater than 1 AU to start with. Combine that with the fact that when measuring the distance to Earth itself the slopes of the line segments are even larger than before, it is clear that the average distance to the planet is even greater compared to the inner planets.
answered Mar 14 at 10:32
Jaap ScherphuisJaap Scherphuis
16.6k12872
$endgroup$
add a comment |
$begingroup$
Here is another way to deduce the same result without any calculations.
First the inner planets.
Imagine that instead of the Earth, there is a wall at 1 AU from the sun. The average distance of any inner planet's orbit to the wall is easily seen to be exactly 1 AU.
This is because you can pair up points on the left and right sides of the orbit. The average distance of those two points to the wall is the same as the distance of their midpoint to the wall.
If you now go back to measuring the distance to the Earth itself, the distances get larger.
Crucially, the larger the orbit, the higher the slopes of the line segments we are measuring, and the further away the average distance is from 1 AU.
This shows that amongst the inner planets, the innermost planet (Mercury) has the smallest average distance to Earth.
What about the outer planets?
Let's go back to the wall replacing the Earth. If an planet's orbit crosses the wall, then when it comes to measuring its distance, we might as well mirror the planet's position in the wall and measure the distance to its mirror image.
When we then do the same trick of pairing points in the two halves of the orbit, it is clear that the average distance to the wall becomes greater than 1 AU to start with. Combine that with the fact that when measuring the distance to Earth itself the slopes of the line segments are even larger than before, it is clear that the average distance to the planet is even greater compared to the inner planets.
answered Mar 14 at 10:32
Jaap ScherphuisJaap Scherphuis
16.6k12872
$endgroup$
Here is another way to deduce the same result without any calculations.
First the inner planets.
Imagine that instead of the Earth, there is a wall at 1 AU from the sun. The average distance of any inner planet's orbit to the wall is easily seen to be exactly 1 AU.
This is because you can pair up points on the left and right sides of the orbit. The average distance of those two points to the wall is the same as the distance of their midpoint to the wall.
If you now go back to measuring the distance to the Earth itself, the distances get larger.
Crucially, the larger the orbit, the higher the slopes of the line segments we are measuring, and the further away the average distance is from 1 AU.
This shows that amongst the inner planets, the innermost planet (Mercury) has the smallest average distance to Earth.
What about the outer planets?
Let's go back to the wall replacing the Earth. If an planet's orbit crosses the wall, then when it comes to measuring its distance, we might as well mirror the planet's position in the wall and measure the distance to its mirror image.
When we then do the same trick of pairing points in the two halves of the orbit, it is clear that the average distance to the wall becomes greater than 1 AU to start with. Combine that with the fact that when measuring the distance to Earth itself the slopes of the line segments are even larger than before, it is clear that the average distance to the planet is even greater compared to the inner planets.
answered Mar 14 at 10:32
Jaap ScherphuisJaap Scherphuis
16.6k12872
answered Mar 14 at 10:32
Jaap ScherphuisJaap Scherphuis
16.6k12872
answered Mar 14 at 10:32
Jaap ScherphuisJaap Scherphuis
16.6k12872
answered Mar 14 at 10:32
Jaap ScherphuisJaap Scherphuis
16.6k12872
16.6k12872
add a comment |
add a comment |
$begingroup$
I'd say the closest planet to Earth is Earth with an average distance of 0
edited Mar 13 at 10:20
Ahmed Ashour
9741313
answered Mar 13 at 9:21
user57862user57862
471
$endgroup$
4
$begingroup$
Well, unfortunately for you, a neighbour is a well defined notion in geometry, and excludes you yourself...
$endgroup$
– yo'
Mar 13 at 14:31
$begingroup$
@yo' I disagree. In graph theory this might be true, but if you’re talking about metric spaces (which we are), then an epsilon neighborhood around a point always contains that point.
$endgroup$
– Santana Afton
Mar 13 at 22:10
$begingroup$
Well, in clustering theory, statistics etc. it's all pretty clear, and that's the context in which I see this puzzle.
$endgroup$
– yo'
Mar 14 at 0:59
1
$begingroup$
@SantanaAfton: a neighbourhood of a point in a topological/metric space contains the point itself, but you don’t speak of points being neighbours of each other in that setting. (That said, I think everyday usage is more to the point here than any of the technical mathematical senses.)
$endgroup$
– Peter LeFanu Lumsdaine
Mar 15 at 10:37
add a comment |
$begingroup$
I'd say the closest planet to Earth is Earth with an average distance of 0
edited Mar 13 at 10:20
Ahmed Ashour
9741313
answered Mar 13 at 9:21
user57862user57862
471
$endgroup$
4
$begingroup$
Well, unfortunately for you, a neighbour is a well defined notion in geometry, and excludes you yourself...
$endgroup$
– yo'
Mar 13 at 14:31
$begingroup$
@yo' I disagree. In graph theory this might be true, but if you’re talking about metric spaces (which we are), then an epsilon neighborhood around a point always contains that point.
$endgroup$
– Santana Afton
Mar 13 at 22:10
$begingroup$
Well, in clustering theory, statistics etc. it's all pretty clear, and that's the context in which I see this puzzle.
$endgroup$
– yo'
Mar 14 at 0:59
1
$begingroup$
@SantanaAfton: a neighbourhood of a point in a topological/metric space contains the point itself, but you don’t speak of points being neighbours of each other in that setting. (That said, I think everyday usage is more to the point here than any of the technical mathematical senses.)
$endgroup$
– Peter LeFanu Lumsdaine
Mar 15 at 10:37
add a comment |
$begingroup$
I'd say the closest planet to Earth is Earth with an average distance of 0
edited Mar 13 at 10:20
Ahmed Ashour
9741313
answered Mar 13 at 9:21
user57862user57862
471
$endgroup$
I'd say the closest planet to Earth is Earth with an average distance of 0
edited Mar 13 at 10:20
Ahmed Ashour
9741313
answered Mar 13 at 9:21
user57862user57862
471
edited Mar 13 at 10:20
Ahmed Ashour
9741313
edited Mar 13 at 10:20
Ahmed Ashour
9741313
edited Mar 13 at 10:20
Ahmed Ashour
9741313
9741313
answered Mar 13 at 9:21
user57862user57862
471
answered Mar 13 at 9:21
user57862user57862
471
answered Mar 13 at 9:21
user57862user57862
471
471
4
$begingroup$
Well, unfortunately for you, a neighbour is a well defined notion in geometry, and excludes you yourself...
$endgroup$
– yo'
Mar 13 at 14:31
$begingroup$
@yo' I disagree. In graph theory this might be true, but if you’re talking about metric spaces (which we are), then an epsilon neighborhood around a point always contains that point.
$endgroup$
– Santana Afton
Mar 13 at 22:10
$begingroup$
Well, in clustering theory, statistics etc. it's all pretty clear, and that's the context in which I see this puzzle.
$endgroup$
– yo'
Mar 14 at 0:59
1
$begingroup$
@SantanaAfton: a neighbourhood of a point in a topological/metric space contains the point itself, but you don’t speak of points being neighbours of each other in that setting. (That said, I think everyday usage is more to the point here than any of the technical mathematical senses.)
$endgroup$
– Peter LeFanu Lumsdaine
Mar 15 at 10:37
add a comment |
4
$begingroup$
Well, unfortunately for you, a neighbour is a well defined notion in geometry, and excludes you yourself...
$endgroup$
– yo'
Mar 13 at 14:31
$begingroup$
@yo' I disagree. In graph theory this might be true, but if you’re talking about metric spaces (which we are), then an epsilon neighborhood around a point always contains that point.
$endgroup$
– Santana Afton
Mar 13 at 22:10
$begingroup$
Well, in clustering theory, statistics etc. it's all pretty clear, and that's the context in which I see this puzzle.
$endgroup$
– yo'
Mar 14 at 0:59
1
$begingroup$
@SantanaAfton: a neighbourhood of a point in a topological/metric space contains the point itself, but you don’t speak of points being neighbours of each other in that setting. (That said, I think everyday usage is more to the point here than any of the technical mathematical senses.)
$endgroup$
– Peter LeFanu Lumsdaine
Mar 15 at 10:37
4
4
$begingroup$
Well, unfortunately for you, a neighbour is a well defined notion in geometry, and excludes you yourself...
$endgroup$
– yo'
Mar 13 at 14:31
$begingroup$
Well, unfortunately for you, a neighbour is a well defined notion in geometry, and excludes you yourself...
$endgroup$
– yo'
Mar 13 at 14:31
$begingroup$
@yo' I disagree. In graph theory this might be true, but if you’re talking about metric spaces (which we are), then an epsilon neighborhood around a point always contains that point.
$endgroup$
– Santana Afton
Mar 13 at 22:10
$begingroup$
@yo' I disagree. In graph theory this might be true, but if you’re talking about metric spaces (which we are), then an epsilon neighborhood around a point always contains that point.
$endgroup$
– Santana Afton
Mar 13 at 22:10
$begingroup$
Well, in clustering theory, statistics etc. it's all pretty clear, and that's the context in which I see this puzzle.
$endgroup$
– yo'
Mar 14 at 0:59
$begingroup$
Well, in clustering theory, statistics etc. it's all pretty clear, and that's the context in which I see this puzzle.
$endgroup$
– yo'
Mar 14 at 0:59
1
1
$begingroup$
@SantanaAfton: a neighbourhood of a point in a topological/metric space contains the point itself, but you don’t speak of points being neighbours of each other in that setting. (That said, I think everyday usage is more to the point here than any of the technical mathematical senses.)
$endgroup$
– Peter LeFanu Lumsdaine
Mar 15 at 10:37
$begingroup$
@SantanaAfton: a neighbourhood of a point in a topological/metric space contains the point itself, but you don’t speak of points being neighbours of each other in that setting. (That said, I think everyday usage is more to the point here than any of the technical mathematical senses.)
$endgroup$
– Peter LeFanu Lumsdaine
Mar 15 at 10:37
add a comment |
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$begingroup$
Though this is a somewhat mathematical puzzle, the answer is a little surprising...
$endgroup$
– Dr Xorile
Mar 13 at 5:10
6
$begingroup$
I assume you mean the closest planet besides Earth?
$endgroup$
– Deusovi♦
Mar 13 at 5:31
3
$begingroup$
Didn't I see someone promoting their paper on this today?
$endgroup$
– Jay
Mar 13 at 5:38
3
$begingroup$
@Deusovi It's not in the body of the question, but "neighbour" in the title clearly excludes the Earth.
$endgroup$
– yo'
Mar 13 at 14:31
1
$begingroup$
A great puzzle for pi day
$endgroup$
– Strawberry
Mar 14 at 12:02