A sky laboratory of mass 2 ×103 kg has to be lifted from one circular orbit of radius 2R to another circular orbit of radius 3R, where R is radius of the earth equal to 6.37×106m. Acceleration due to gravity =9.8m/s2 lf the minimum energy required is n×1010J, find n.
Let velocity of laboratory in a circular orbit of radius r be v. Then,
mv2r=GMmr2
Thus, kinetic energy is
K=12mv2=GMm2r
Thus, total energy in orbit is
E=12mv2−GMmr=−GMm2r
Hence, energy required to change orbit is
ΔE=GMm3R−GMm6R=GMm4R=gmR12
ΔE=9.8(2000)(6.37X106)12≈1X1010J
Answer is 1.