A satellite moves in a circular orbit around the earth at a speed of 6.9 km/s. determine the satellite's altitude above the surface of the earth. assume the earth is a homogeneous sphere of radius 6370 km and mass 5.98 * 1024 kg. the value of the universal gravitational constant is 6.67259 * 10-11 n · m2 / kg2. answer in units of km.

Given: Velocity of satellite Vsat = 6.9 Km/s convert to m/s Vsat = 6900 m/s

Radius of Earth re = 6,370 Km; Mass of Earth Me = 5.98 x 10²⁴ Kg

Universal Gravitational constant G = 6.67 x 10⁻¹¹ N. m²/Kg²

Required: Altitude of satellite r = ?

Formulas. F = ma; a = V²/r F = GMeMsat/r²

Equate for "r " from the three equations

GMeMsat/r² = MsatV²/r

r = GMe/V²

r = (6.67 x 10⁻¹¹ N. m²/Kg²) (5.98 x 10²⁴ Kg) / 6,900 m/s) ²

r = 8,379,537.82 m or 8,379.53 Km

Deduct the the radius of our planet to find the altitude of the satellite

r = rsat - re

r = 8, 379.59 Km - 6,370 Km

r = 2009.59 km or

r = 2010 Km altitude above the earth.