Question

Assume a small object of mass 1.0 kg is launched into circular orbit close to the Earth's spherical surface. Ignore air resistance. Find (a) its orbital speed (km/s), its orbital period (minutes), (c) it total mechanical energy in orbit.

Answer 1

A Orbital speed of an object revolving around the earth at height 'h' above the earth's surface is V_0 = squareroot GM/R + h where M is mass and as h < < R V_ 0 = squareroot GM/R = squreroot gR R = 6400km = 6400 times 10^3 m hence V_0 = squareroot 9.81 times 64 times 10^5 = 7923.6 m/s = 7.92 km/s b) Orbital period (T) = distance/speed = 2TTR/V-0 T = 2(3.14)(64 times 10^5m)/7923.6 m/s = 5066.7 s = 5066.7/60 min = 84.4 minutes c) Total mechanical energy of object E = KE + PE = 1/2 mv_0^2 - GMW?R E = 1/2 (1) (7923.6)^2 - 6.67 times 10^-11 times 5197 times s^4/64 times 10^5 E = 6278343696 - 62218593.75 E = 56843.21 J = 5.65 times 10^5J