An earth satellite remains in orbit at a distance of 13300 km from the center of the earth. what is its period? the universal gravitational constant is 6.67 à 10â11 n · m2 / kg2 and the mass of the earth is 5.98 à 1024 kg. answer in units of s.

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)