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 1.889×10⁴ km , measured along the surface of Planet X.

a) How long is the day on Planet X?

b) If a 4.800×10⁴ kg satellite is placed in a circular orbit 4000 km above the surface of Planet X, what will be its orbital period?

Answer 1

m = 946N / 9.8m/s² = 96.53 kg ? mass of astronaut

F = GmM/r²
920 N = 6.674*10^-11N·m²/kg² * 96.53kg * M / (1.889*10⁷m)²
so M = 5.09566*10²⁵ 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*10²⁵kg from above.
So ? = ?(6.674*10^-11N·m²/kg² * 5.09566*10²⁵kg / (1.889*10⁷m + 4*10⁶m)³ ) = 5.325e-4 rad/s
T = 2?/? = 11799.27 s