What is the altitude for synchronous orbit around Jupiter?

Answer 1

To be in a synchronous orbit around a planet, the velocity of the satellite or space craft has to equal to rotational velocity at some point on the planet. For Jupiter, I found that, at the equator, such a velocity is 12 600 m/s (which is really amazingly fast when you think about it, Jupiter has the fastest rotation of all the planets in the solar system). Anyway, for a circular orbit, velocity is given by:

v = sqrt[Gm/R] where G is the gravitational constant, m is the mass of Jupiter (about 1.90x10^27 kg) and R is the radius of Jupiter (in one of physics textbooks it is given as 6.91x10^7, we'll have to assume that this is the radius at the equator, although it is probably different than this). So rearranging this expression we get that:

R = Gm/v^2

Plugging in the numbers from above and then sutracting the given radius of Jupiter (since this equation is measuring from the center of the planet outward, we just want the distance from the surface in this case) you get:

729148929.2 m or 729148.9 km

Of course, you can change this answer to suit whatever data you have, I'm not sure if the numbers I had were completely accurate, so the answer might be off a bit. This number seems rather large, but then again Jupiter is a huge planet, so it might be right.

Answer 2

My guess would be 22370 mi, although that is the standard for earth.

Answer 3

infinity absolute