Question

The ratio of the energy required to raise a satellite upto a height $h$ above the earth of radius $R$ to that the kinetic energy of the satellite into that orbit is

• A. $R : h$
• B. $h : R$
• C. $R : 2h$
• D. $2h : R$

Answer 1

Answer: (D)

$2h : R$

Answer 2

Energy required to raise the satellite to a height h from surface of earth is given by,
$U = \frac{GMm}{\left(R +h\right)}- \left(-\frac{GMm}{R}\right)= \frac{GMmh}{R\left(R+h\right)}$
Kinetic energy of satellite is given by,
$K = \frac{1}{2}m\upsilon^{2}_{0} = \frac{1}{2}m \frac{GM}{\left(R +h\right)} \:\:\: \therefore \frac{U}{K}=\frac{2h}{R}$