Given M = 5.98 x 1024 kg
m = 103 kg
R = 6.37 x 106 m
Centripetal acceleration of the satellite is provided by the gravitational force exerted by Earth.
\(\frac{mv^2}{r}=\frac{GMm}{r^2}\)
v2 = GM/r
Total energy of the system = K.E. + Gravitational Potential energy
= 1/2 mv2 - GMm/r
= 1/2 m GM/r - GMm/r
= - 1/2 GMm/r
Therefore, energy required to move on Earth satellite radius 2R to orbital radius 3R.
E = GMm/2 (1/2R - 1/3R)
E = GMm/12R
\(E=\frac{6.67\times10^{-11}\times5.98\times10^{24}\times10^3}{12\times6.37\times10^6}\)
\(E=\frac{39.88\times10^{10}}{76.44}\)
⇒ 0.5217 x 1010 J