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.
