Question

A 38.0 kg satellite has a circular orbit with a period of 1.30 h and a radius of 7.90 × 106 m around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is 30.0 m/s2, what is the radius of the planet

Answer 1

Answer:

5.44×10⁶ m

Explanation:

For a satellite with period t and orbital radius r, the velocity is:

v = 2πr/t

So the centripetal acceleration is:

a = v² / r

a = (2πr/t)² / r

a = (2π/t)² r

This is equal to the acceleration due to gravity at that elevation:

g = MG / r²

(2π/t)² r = MG / r²

M = (2π/t)² r³ / G

At the surface of the planet, the acceleration due to gravity is:

g = MG / R²

Substituting our expression for the mass of the planet M:

g = [(2π/t)² r³ / G] G / R²

g = (2π/t)² r³ / R²

R² = (2π/t)² r³ / g

R = (2π/t) √(r³ / g)

Given that t = 1.30 h = 4680 s, r = 7.90×10⁶ m, and g = 30.0 m/s²:

R = (2π / 4680 s) √((7.90×10⁶ m)³ / 30.0 m/s²)

R = 5.44×10⁶ m

Notice we didn't need to know the mass of the satellite.