What's the orbital period of the satellite?

Question

A satellite circles planet X in an orbit having a radius of 10^8 m. If the radius of planet X is 10^7 m and the free fall acceleration on the surface of X is 20 m/s^2, what's the orbital period of the satellite?

Answer 1

g = v²/r = ω² r

So, at the surface, ω = √(g_s/r_s)

T_s = 2pi/ω = 2pi√(r_s/g_s) = 4442.9 sec

By Kepler's law: T/T_s = (r/rs)^3/2

Solving for T,

T = 1.41 * 10⁵ sec

Answer 2

First you need to compute the acceleration of gravity at the orbital radius. Since it is 10 times bigger than the planet's radius, the acceleration will be 100 times less due to Newton's inverse square law.

Thus g_orbit = 0.2 m/s². Now you need to use the formula for centripetal acceleration, a = ω²r.

Letting a = g_orbit, and solving for ω gives ω = √(g_orbit / r).

Plugging in the values gives f = ω/2π = √[(0.2 m/s²) / 10⁸ m] / 2π = 7 × 10^-6 orbits/second.

Thus the orbital period is T = 1/f = 1.4 × 10⁵ s.