View Full Version : our nearest planet

profloater

2019-Jan-11, 06:11 PM

a surprising result on the radio today. I would have thought that either Venus or Mars would be nearest to us depending on their orbits and indeed they are but on average Mercury is closer than them. Once heard it is kind of obvious that when Venus and Mars are out the other side of the sun, Mercury is closer to us. They also pointed out that despite being closer mor often, a trip to Mercury is difficult because of the need to slow down.

Swift

2019-Jan-11, 06:30 PM

The closest planet to me is Earth. ;)

I also suspect it depends on what you mean by closest; do you mean at a single point in time, or do you mean on average. I suspect it is Venus on average.

Found this from UT (https://www.universetoday.com/14447/what-is-the-closest-planet-to-earth/) (which disagrees with Mercury as the answer:

When Venus lies between Earth and the Sun, it experiences what is known as an inferior conjunction. It is at this point that it makes its closest approach to Earth (and that of any planet) with an average distance of 41 million km (25,476,219 mi). On average, Venus achieves an inferior conjunction with Earth every 584 days.

profloater

2019-Jan-11, 06:38 PM

The closest planet to me is Earth. ;)

I also suspect it depends on what you mean by closest; do you mean at a single point in time, or do you mean on average. I suspect it is Venus on average.

Found this from UT (https://www.universetoday.com/14447/what-is-the-closest-planet-to-earth/) (which disagrees with Mercury as the answer:

Indeed, Mars is closest more of the time than Venus if you take a long average but over that long time like ten years, Mercury is closest the mostest. Of course it never gets as close as Venus does, that's a different statistic.

George

2019-Jan-15, 05:12 PM

What does "on average" actually mean? If you take the average of the closest and farthest distances using their semimajor axis then Mercury and Venus have the same average -- 1AU, of course. The Geocentric model puts Mercury closer but they were guessing. Ptolemy added his epicycles and equants, so it gets complicated though the goal is to match today's "appearances".

So over enough time to allow precession and our system's barycenter wiggle to do its thing, then are they not on average the same distance?

Delvo

2019-Jan-15, 11:17 PM

It sounds like your thinking can be summed up with the simple diagram I'll describe...

Picture the sun in the middle of your mental page, and Earth straight out to the right. A straight vertical line in the middle through the sun would divide the Earth-sun system into two halves, right & left, so, when we add Venus and Mercury, there are three possible arrangements:

1. Both on the left: Mercury is closer

2. Both on the right: Venus is closer

3. One on each side: the one on the right is closer

Is that the idea here?

But there's a catch. Picture both of them straight above or below the sun, exactly on that straight vertical line. Their distances from Earth are not the same. The Earth-sun-Mercury and Earth-sun-Venus right triangles share a base and the right angle, but Venus's triangle is taller, which gives it a longer hypotenuse, and it's the hypotenuses that are their distances from Earth. (Both have a base of 1 AU but a hypotenuse greater than 1 AU, and Venus's is greater by a wider margin.) To make the two planets' distances from earth the same, you would need to move Venus right or Mercury left (or both). That means a line that truly divided the diagram into two parts yielding the three simple rules as described above could not be a straight vertical line. It would need to cross Mercury's orbit slightly left of center and cross Venus's orbit slightly right of center. Then we get a diagram in which Mercury spends slightly more time in the right section and Venus spends slightly more time in the left section.

Jens

2019-Jan-16, 05:39 AM

The closest planet to me is Earth. ;)

I also suspect it depends on what you mean by closest; do you mean at a single point in time, or do you mean on average. I suspect it is Venus on average.

I think it means on average, and that Mercury is on average closer to the earth than Venus. I can understand why that would happen, because if you think about the situation where the earth is at 12 o'clock and both Venus and Mercury are at 3 or 9 o'clock, Mercury will be closer to the earth than Venus. When all three are at 12 o'clock, Venus will be closer, and when the earth is at 12 o'clock and both Venus and Mercury are at 6 o'clock, Mercury will be closer. So it makes sense intuitively that on average, Mercury will be closer.

Found this from UT (https://www.universetoday.com/14447/what-is-the-closest-planet-to-earth/) (which disagrees with Mercury as the answer:

The link you have only says that Mercury never gets as close to the earth as Venus does. The OP clearly states "on average."

Jens

2019-Jan-16, 05:42 AM

What does "on average" actually mean? If you take the average of the closest and farthest distances using their semimajor axis then Mercury and Venus have the same average -- 1AU, of course. The Geocentric model puts Mercury closer but they were guessing. Ptolemy added his epicycles and equants, so it gets complicated though the goal is to match today's "appearances".

I may be misunderstanding you but my understanding is that the exercise is to measure the distance between the earth and Mars and the earth and Venus every day for a year, add the values, and divide by 365, and see which is smaller. I don't think you'll get 1 AU.

profloater

2019-Jan-16, 08:06 AM

The way i heard it on the BBC, on any given day there is one of those three closest to us. If you table all the days for a long period Mercury is closest for the most number of days, then Mars, then Venus. No averaging required.q

grant hutchison

2019-Jan-16, 09:37 AM

Interesting usage of the word "closest". Under those rules, a planet Vulcan, orbiting inside Mercury, would be "closer" than Mercury, for the same geometrical reasons that make Mercury "closer" than Venus. And so on down.

Grant Hutchison

profloater

2019-Jan-16, 01:13 PM

Interesting usage of the word "closest". Under those rules, a planet Vulcan, orbiting inside Mercury, would be "closer" than Mercury, for the same geometrical reasons that make Mercury "closer" than Venus. And so on down.

Grant Hutchison

Yes its an odd fact, it was covered in “more or less” the fact checking programme after some astronomy program debated mars and venus forgetting Mercury for obvious reasons but within those rules it makes sense.

21st Century Schizoid Man

2019-Jan-16, 03:31 PM

If we assume the cow is round and all the planets follow circular orbits in the same plane, I am indeed getting that the one with the smallest orbit is the one closest to us.

Assumptions - planets follow circular orbits, at uniform speeds, in the same plane, but the two planets have different orbital periods. Doesn't matter if they orbit in the same direction or opposite directions.

Then the average distance - that is, the integral of the distance between the two planets from one conjunction to the next, divided by the amount of time passed - appears to be a strictly increasing function of the radius of one orbit, holding the other fixed.

I say "appears to be" because the exact expression contains elliptical integrals, and it's hard to work out the properties of the average distance analytically. But just by plotting it, it certainly looks like it is strictly increasing in one radius, holding the other fixed.

The analysis would appear to hold even if the two radii are the same, provided the planets have different orbital periods, even if this assumption violates Kepler's laws.

So in this simplified model - Mercury is indeed closest to Earth on average. Using some "distance from the sun" numbers I found with a Google search, I get the average distance for Mercury is 155.259 million kilometres, for Venus it is 169.911 million kilometres, for Mars it is 253.198 million kilometres, and for Jupiter, it is 785.704 million kilometres.

Deviations from this analysis could occur because the orbits are elliptical, perturbed from the elliptical, or in different planes. However, my gut feel is that these effects will be relatively minor, and since Venus has an average distance about 9.4% greater than Mercury's, I think the conclusion will survive correction of these idealised assumptions.

So I think the conclusion is correct - Mercury has the lowest average distance to Earth out of all the planets. At least until we discover Vulcan.

21st Century Schizoid Man

2019-Jan-16, 04:31 PM

Details - sun is at the origin, one planet has orbit with radius r_{1}, the other with radius r_{2}. \theta is the angle between the two planets, from the perspective of the sun. Since the orbits are circular, \theta increases (or decreases) at a constant rate.

Then the average is

\frac{1}{2\pi}\int_{0}^{2\pi}\sqrt{\left(r_{2}-r_{1}\cos\theta\right)^{2}+r_{1}^{2}\sin^{2}\theta }d\theta

The integral can be evaluated in terms of elliptical integrals. (Alternate interpretation - we can't evaluate it, so we give the integral a name, so we can say, "yea, look, we can do it!".) But we can also rewrite it as

\frac{1}{2\pi}\int_{0}^{2\pi}\sqrt{r_{1}^{2}+r_{2} ^{2}-2r_{1}r_{2}\cos\theta}d\theta

Taking r_{1} outside, we get

\frac{r_{1}}{2\pi}\int_{0}^{2\pi}\sqrt{1+\left(\fr ac{r_{2}}{r_{1}}\right)^{2}-2\frac{r_{2}}{r_{1}}\cos\theta}d\theta

New notation, r\equiv r_{1} and \alpha\equiv\frac{r_{2}}{r_{1}}. Then the average distance is

\frac{r}{2\pi}\int_{0}^{2\pi}\sqrt{1+\alpha^{2}-2\alpha\cos\theta}d\theta

Evaluating the integral in Mathematica produces a result with elliptical integrals, and plotting the result as a function of \alpha shows that it is increasing in \alpha.

Roger E. Moore

2019-Jan-16, 07:48 PM

If we assume the cow is round and all the planets follow circular orbits in the same plane....

My train of thought keeps derailing at this word. It provokes unusual mental images.

George

2019-Jan-16, 11:06 PM

I may be misunderstanding you but my understanding is that the exercise is to measure the distance between the earth and Mars and the earth and Venus every day for a year, add the values, and divide by 365, and see which is smaller. I don't think you'll get 1 AU. Yeah, I muffed it. I went super simple; the average of the conjunctions is 1.0 AU.

The average for the model shown below gives Mercury the closeness edge by 7% (for the average distances of the points shown).

23918

21st Century Schizoid Man

2019-Jan-17, 12:50 AM

My train of thought keeps derailing at this word. It provokes unusual mental images.

https://en.wikipedia.org/wiki/Spherical_cow

grant hutchison

2019-Jan-17, 01:14 AM

The way I came at it was simple geometry. Assume coplanar circular orbits. The average distance from Earth to a "minimal" orbit (radius zero) is 1 A.U. Now give the orbit some small radius, and draw the two lines of length exactly 1 A.U. that connect Earth to orbit. These will form an equilateral isoceles triangle, apex at Earth, and base spanning a chord of the orbit. Simple geometry demonstrates that the chord is always on the Earth side of the sun, splitting the orbit circle into a short arc closer than 1 A.U., and a long arc farther away than 1 A.U. The average distance of the planet therefore must be greater than 1 A.U.

Enlarge the orbit. Simple geometry demonstrates that the chord will cut the circular orbit into more uneven arcs, so the average distance will now be more biased towards the far side of the orbit. And the orbit itself is bigger, so the proportional effect will be to magnify that bias.

The smaller the orbit, the closer the chord comes to being a diameter, the more evenly the two arcs are matched, and the smaller the effect of the mismatch.

Grant Hutchison

21st Century Schizoid Man

2019-Jan-20, 06:20 PM

The way I came at it was simple geometry. Assume coplanar circular orbits. The average distance from Earth to a "minimal" orbit (radius zero) is 1 A.U. Now give the orbit some small radius, and draw the two lines of length exactly 1 A.U. that connect Earth to orbit. These will form an equilateral isoceles triangle, apex at Earth, and base spanning a chord of the orbit. Simple geometry demonstrates that the chord is always on the Earth side of the sun, splitting the orbit circle into a short arc closer than 1 A.U., and a long arc farther away than 1 A.U. The average distance of the planet therefore must be greater than 1 A.U.

Enlarge the orbit. Simple geometry demonstrates that the chord will cut the circular orbit into more uneven arcs, so the average distance will now be more biased towards the far side of the orbit. And the orbit itself is bigger, so the proportional effect will be to magnify that bias.

The smaller the orbit, the closer the chord comes to being a diameter, the more evenly the two arcs are matched, and the smaller the effect of the mismatch.

Grant Hutchison

That seems reasonable enough.

I have been trying to get an exact result when the orbits are not in the same plane. Even with circular orbits, I'm finding this brutally difficult. Maybe there is an easy way to do it, but if so, it has thus far eluded me.

I believe (but haven't actually finished yet) I can derive a result when there is orbital resonance (any integer-to-integer ratio will do). The general result - ugh.

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