Given: Altitude of satellite r = 13,300 Km convert to meter
r = 1.33 x 10⁷ m
Universal Gravitational constant G = 6.67 x 10⁻¹¹ N.m²/Kg²
Mass of the earth Me = 5.98 x 10²⁴ Kg
Required: Period of satellite T = ?
Formula: F = ma; Centripetal acceleration ac = V²/r F = GMeMsat/r²
Velocity of satellite V = 2πr/T
equate T from all given equation.
F = ma
GMeMsat/r² = MsatV²/r cancel Msat and insert V = 2πr/T
GMe/r² = (2πr)²/rT² Equate T or period of satellite
T² = 4π²r³/GMe
T² = 39.48(1.33 X 10⁷ m)³/(6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)
T² = 9.29 x 10²² m³/3.99 x 10¹⁴ m³/s²
T² = 232,832,080.2 s²
T = 15,258.84 seconds or (it can be said around 4.24 Hr)
