What is the minimum, maximum and average distance of the Venus and Jupiter?
I only found 670,130,000 km and 670,198,000 km as an average distance (here and here).
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1$\begingroup$ You cannot have made any effort to research this (e.g. web search). $\endgroup$– StephenG - Help UkraineNov 12, 2017 at 10:05
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$\begingroup$ I only found 670,130,000 km and 670,198,000 km as an average distance: planetsedu.com/distances-between-the-planets and theplanets.org/distances-between-planets $\endgroup$– hunyadymNov 12, 2017 at 10:14
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$\begingroup$ helpful edit pending acceptance $\endgroup$– uhohNov 12, 2017 at 14:04
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$\begingroup$ Look up each planet's perihelion and aphelion. Also consider that they can be on opposite sides of the sun and quite a bit further apart than 670 million km. Your two numbers are simple approximations of when the planets line up, not average distances but average distances when Venus passes Jupiter relative to the sun. If you want to get even more accurate, you can factor in inclination. Such calculations don't serve much purpose though, unless you care about how bright each planet might appear to the other. $\endgroup$– userLTKNov 12, 2017 at 14:06
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1 Answer
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The furthest distance Venus is from the Sun is 0.73 AU
The furthest distance Jupiter is from the Sun is 5.46 AU
The closest distance Jupiter is fro the Sun is 4.95 AU
So if we assume the orbits are concentric circles (they're not !) then :
- Minimum distance = 4.95-0.73 = 4.22 AU
- Maximum distance = 5.46+0.73 = 6.19 AU
And a simple average of these gets you an average distance of about 5.2 AU.
The problems with that simple estimate are :
- The orbits are not circles, but ellipses
- The orbits are at an inclination
- A more detailed calculation for circular concentric orbits produces a tricky double integral with no easy solution
- Over time (the only way to do a true average) the orbits would change from even "simple" ellipses.
A number of online sources quote the average distance as 4.48 AU. This appears to be nothing more than the semi-major axis of Venus's orbit subtracted from the semi-major axis of Jupiter, which is probably an under-estimate of an average separation, as it neglects that they can be at opposite sides of the Sun sometimes and the same side at others.
Minor update :
For those interested in seeing the effect of a more detailed calculation I tried treating the orbits as concentric circles and doing the full integral on WolframAlpha.com .
The required integral in this case is (units are AU) :
$$L = \frac {5.2044}{4\pi^2} \int_0^{2\pi}\int_0^{2\pi} \left[ ( sin(v)-0.139 cos(u) )^2 + ( cos(v)-0.139 cos(v) )^2 \right]^{1/2}du\, dv$$
where $0.139$ is the ratio of Venus's orbital radius to Jupiter's and $5.2044$ AU is the mean radius of Jupiter's orbit.
And the result of this tedious integral $L=5.2293$ AU, which is not exactly much of an improvement on the much simpler $5.2$ AU given earlier. But it at least dispels then notion on some online pages that it could be $4.48$ AU.
Last (?) minor update :
The above calculation just did a geometric average over the two orbit shapes, but made no attempt to consider the motion. Well it occurred to me to revise that and allow for two bodies in a circular motion in the same plane. Guess what ? Same answer !
Well, almost. In this case it's $5.2297$, which is an insignificant difference as we're already ignoring inclination and using circular orbits. I'm inclined to put down the difference to a numerical calculation margin of error (the integrals do not have simple closed forms and would easily propagate errors, I think).
The integral boils to an average over the synodic period and is a simpler single integral which reduces to :
$$L = \sqrt{a^2+b^2} \frac 1 {2\pi} \int_0^{2\pi} \sqrt{1-2\frac {ab}{a^2+b^2} cos\theta}\, d\theta$$
I do prefer this result myself as it's taking account of orbital motion and over time, rather than purely geometric.
answered Nov 12, 2017 at 15:03