Question

The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and Radius R in a circular orbit at an altitude of 2R is 

.

Answer 1

Correct option is: \(\frac{5\,GMm}{6R}\)

Given mass of satellite = M

mass of the surface = M

Radius = R

Altitude h = 2R

Gravitational potential energy  = \(-\frac{GMm}{r}\)

Gravitational potential energy at Altitude = \(-\frac{GMm}{r+h}\)

= \(-\frac{GMm}{R+2R}\)

= \(-\frac{GMm}{3R}\)

Orbital velocity V_o² = \(\frac{GM}{R+h}\)

V_o² = \(\frac{GM}{3R}\)

total potential energy E = kinetic energy + potential energy

E_f = \(\frac{1}{2}\)mv_o² + \((-\frac{GMm}{3R})\)

∵ V_o² = \(\frac{GM}{3R}\)

E_f = \(\frac{1}{2}\) \(\frac{GMm}{3R}-\frac{GMm}{3R}\)

E_f = \(\frac{GMm-2\,GMm}{6R}\)

E_f = \(-\frac{GMm}{6R}\)

Ei = Ef

the minium energy required

= \(\frac{GMm}{R}-\frac{GMm}{6R}\)

= \(\frac{6\,GMm-GMm}{6R}\)

Minimum energy required = \(\frac{5\,GMm}{6R}\)