Question

The ratio of energy required to raise a satellite to a height $h$ above the earth to the kinetic energy of the satellite into the orbit is
A. $h:R$
B. $R:2h$
C. $2h:R$
D. $R:h$

Answer 1

The energy possessed by an object due to its position is given by the potential energy. At any point over the Earth’s surface, the energy is the sum of the potential energy at the surface, The kinetic energy gives the objects to the present position.

Complete step by step solution:
We have a satellite at some height h above the surface, and we are giving the energy which is needed to move this satellite to the height h over the atmosphere. The orbital motion of the satellite is equal to its energy
The energy needed to raise the satellite height is as follows,
$E = - GMm\left[ {\dfrac{1}{{R + h}} - \dfrac{1}{R}} \right]$
Now,
$GM = g{R^2}$
Hence,
${E_1} = \dfrac{{gRmh}}{{R(R + h)}}$………….(1)
${E_2} =$ is the energy needed to put a satellite in orbit
${E_2} = \dfrac{1}{2}mv_0^2$
Here,
$v = \sqrt {\dfrac{{g{R^2}}}{{R + h}}}$
Now substitute the equation in ${E_2}$ it becomes,
${E_2} = \dfrac{1}{2} \times m \times \dfrac{{g{R^2}}}{{R + h}}$………..(2)
Now divide the equation (1) by (2) then we get,
$\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{\dfrac{{gRmh}}{{R(R + h)}}}}{{\dfrac{1}{2} \times m \times \dfrac{{g{R^2}}}{{R + h}}}}$
Now simplify the above equation we get,
$\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{2h}}{R}$
Hence the ratio is $2h:R$
So, the correct answer is $2h:R$.

Note:
The force due to the gravitational acceleration helps to calculate the energy.
This energy is calculated using the orbital velocity of the satellite; an important parameter in satellite launches is the orbital velocity. if we give more orbital velocity to a probe, it takes only a smaller fuel to reach the orbital by taking the advantage of earth spin.