Question

Planet X rotates in the same manner as the earth, around an axis
through its north and south poles, and is perfectly spherical. An
astronaut who weighs 946.0 N on the earth weighs 920.0 N at the
north pole of Planet X and only 848.0 N at its equator. The
distance from the north pole to the equator is 1.889×10^{4}
km , measured along the surface of Planet X.

a) How long is the day on Planet X?

b) If a 4.800×10^{4} kg satellite is placed in a
circular orbit 4000 km above the surface of Planet X, what will be
its orbital period?

Answer #1

F = GmM/r²

920 N = 6.674*10^{-11}N·m²/kg² * 96.53kg * M /
(1.889*10^{7}m)²

so M = 5.09566*10^{25} kg

The centripetal acceleration at the equator is

a = (920N - 848N) / 96.53kg = 0.7458 m/s² = ?²r = ?² *
(1.889e7m)

? = 1.987e-4 rad/s

T = 2?/? = 34 610 s = 9 h 36 min 50 s

The mass of the satellite is immaterial.

For "orbit", centripetal acceleration = gravitational acceleration,
or

?²r = GM/r² rearranges to ? = ?(GM/r³)

where G = Newton's gravitational constant = 6.674e?11
N·m²/kg²

and M = mass of X = 5.09566*10^{25}kg from above.

So ? = ?(6.674*10^{-11}N·m²/kg² * 5.09566*10^{25}kg
/ (1.889*10^{7}m + 4*10^{6}m)³ ) = 5.325e-4
rad/s

T = 2?/? = 11799.27 s

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