Question

An earth satellite remains in orbit at a distance of 13594 km from the center of the earth. what is its period? the universal gravitational constant is 6.67 × 10−11 n · m2 /kg2 and the mass of the earth is 5.98 × 1024 kg.

Answer 1

Given: Mass of earth Me = 5.98 x 10²⁴ Kg
           Radius of satellite - radius of the earth

           r = 13,594 Km - 6,380 Km = 7,214 Km    convert to meter "m"

           r = 7,214,000 m
            G = 6.67 x 10⁻¹¹ N.m²/Kg²
Required: What is the period T = ?
Formula: F = ma;  F = GMeMsat/r²     Centripetal acceleration ac = V²/r
               but V = 2πr/T
equate T from all equation.
F = ma
GMeMsat/r² = Msat4π²/rT²    
GMe = 4π²r³/T²
T² = 4π²r³/GMe  

T² = 39.48(7,214,000 m)³/(6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)

T² = 1.48 x 10²² m³/3.99 x 10¹⁴ m³/s²

T² = 37,092,731.83 s²

T = 6,090.38 seconds    or  1.7 Hr