What is the radius of a communications satellite's orbital path that is in a uniform circular orbit around Earth and has a period of exactly 24.0 hours (86,400 seconds). (Measurement is from the center of Earth.) G = 6.67 * 10-11 N. m2/kg2 ME = 5.98 * 1024 kg 1 hr = 3,600 s

If the period of a satellite is T=24 h = 86400 s that means it is in geostationary orbit around Earth. That means that the force of gravity Fg and the centripetal force Fcp are equal:

Fg=Fcp

m*g=m * (v²/R),

where m is mass, v is the velocity of the satelite and R is the height of the satellite and g=G * (M/r²), where G=6.67*10^-11 m³ kg⁻¹ s⁻², M is the mass of the Earth and r is the distance from the satellite.

Masses cancel out and we have:

G * (M/r²) = v²/R, R=r so:

G * (M/r) = v²

r=G * (M/v²), since v=ωr it means v²=ω²r² and we plug it in,

r=G * (M/ω²r²),

r³=G * (M/ω²), ω=2π/T, it means ω²=4π²/T² and we plug that in:

r³=G * (M / (4π²/T²)), and finally we take the third root to get r:

r=∛{ (G*M*T²) / (4π²) }=4.226*10^7 m = 42 260 km which is the height of a geostationary satellite.