A satellite orbits the Earth (mass = 5.98 x 1024 kg) once every = 8.00 hours. At what radius does the satellite orbit?

Answer 1

There is another approach to this problem that give essentially the same result as William W. The force of attraction of the earth to the satellite is given by GMm/R² where G is the gravitational constant, M is the earth's mass, m is the mass of the satellite, and R is the radius of the orbit (assumed to be circular).

This can be equated to the centripetal force on the satellite, so that mV²/ R = GMm/R² . V is the orbital velocity which is (2πR/28800 sec. This can be squared and substituted in the previous equation. m cancels out, and the resulting equation can be simplified and solved for R after substituting the known quantities.

I haven't tried it, but Kepler's third law equation might be derivable from the approach given above.

I obtained a result of R = 20.0 x 10⁶ meters.

Answer 2

Using the variation of Kepler's third law:

you can solver for r:

r = [T√(GM_E)/(2π)]^2/3

But you must make sure you use units of seconds for the period. (8 hrs)(3600 sec/hr) = 28,800 seconds

G is the universal gravitational constant and is 6.674 x 10⁻¹¹ m³/(kg⋅s²)

r = [28800√(6.674 x 10⁻¹¹•5.98 x 10²⁴)/(2π)]^2/3

r = [28800√(3.991 x 10¹⁴)/6.283]^2/3

r = [9.157 x 10¹⁰]^2/3

r = 20.3 x 10⁶ meters