Asatellite 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.

The sample with the lowest specific heat capacity will have the greatest temperature increase. in this case, it would be platinum.