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.

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