Earth’s moon has a mass of approximately 1/81th that of Earth, and orbits at an average altitude (semi-major axis) of 384,399 km.
Ceres, (the largest body in the asteroid belt) has a mass of approximately 1/699th that of Mars—90,000 times the mass of Phobos. If Ceres could be moved into Martian orbit, at what semi-major axis (altitude) would it have a similar tidal effect on Mars as the Moon has on Earth?
Is there a better candidate for a new Martian moon to produce lunar like effects on Mars?
If Mars had a lunar type satellite, would Mars be more Earth-like? Would the tidal forces be sufficient to generate a Van-Allen belt for the planet and/or stabilize the planet’s orbit as well as its rotation?
?2011-12-02T14:34:24Z
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What do you mean by "similar tidal effect"? There are a few things that you could interpret this to be.
One interpretation is the absolute pull given by the moon when at closest approach. Also absolute force/acceleration difference between a point nearest to the moon and a point furthest away, another would be the relative difference. I'll assume the relative difference for now. With Mars's smaller gravity, this would occur at a higher altitude.
Earth (no moon) local acceleration = GM/r^2 M = 5.974x10^24kg r = 6371km g = 9.824 m/s^2
Lunar acceleration on earth toward or away from moon r = lunar distance +/- 6371km r = 384400km +/- 6371km r1 = 378029km r2 = 390771km g1 = 3.432 x 10^-5 m/s^2 g2 = 3.212 x 10^-5 m/s^2 g1-g2 = 2.2 x10^-6 m/s^2
Reversing, mars g is g(mars) = 3.72m/s^2, so the difference for it's moon should be: 2.72m/s^2 * 2.5x10^-7 8.39 x 10^-7m/s^2
g1 - g2 = 8.39x10^-7ms/^2 m(ceres) = 9.43x10^20kg
Rather than solving directly, I eyeballed and found that somewhere south of 100000km should be about right. g1 = G m(ceres) / (100000km - 3390km)^2 g1 = 6.74x10^-6 m/s^2 g2 = G m(ceres) / (100000km + 3390km)^2 g2 = 5.89x10^-6 m/s^2 g1-g2 = 8.55x10^-7m/s^2
If you did want the absolute difference to be the same (2.2x10^-6m/s^2), then that would occur at an altitude somewhere around 70000km.
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