Question

The energy required to take a satellite to a height h above the Earth’s surface (radius of Earth $ = 6.4 \times {10^3}$km) is ${E_1}$ and kinetic energy required for the satellite to be in a circular orbit at this height is ${E_2}$. The value of h for which ${E_1}$ and ${E_2}$ are equal is:
(A) $1.28 \times {10^4}$km
(B) $6.4 \times {10^3}$km
(C) $3.2 \times {10^3}$km
(D) $1.6 \times {10^3}$km

Answer 1

Hint
The energy possessed by an object due to its position is given by the potential energy. At any point above the Earth’s surface, this energy will be the sum of the potential energy at the surface and the kinetic energy used to bring the object to the current position.
Formula used: $U = - \dfrac{{GMm}}{r}$, where U is the potential energy of the object at a distance r. M is the mass of the Earth and m is the mass of the object in consideration. G is the universal gravitational constant.

Complete step by step answer
We have a satellite at some height h above the surface, and we are provided with the energy that is required to take this satellite to the height h in the atmosphere. We are asked to find the value of h for which the kinetic energy of the orbital motion of the satellite would be equal to this energy. The information that is provided to us is:
Radius of the Earth $R = 6.4 \times {10^3}$ km
Energy to take the satellite to height h ${E_1}$
Kinetic energy of orbital motion ${E_2}$
Mass of the satellite is $m$
Now, we know that the potential energy at the height h would be a combination of the potential energy at the surface and ${E_1}$ as:
${U_{surface}} + {E_1} = {U_h}$
Substituting the value for potential; energy would give us:
$ - \dfrac{{GMm}}{R} + {E_1} = - \dfrac{{GMm}}{{(R + h)}}$
$ \Rightarrow {E_1} = - \dfrac{{GMm}}{{(R + h)}} + \dfrac{{GMm}}{R}$
Taking out common terms:
${E_1} = GMm\left( {\dfrac{1}{R} - \dfrac{1}{{R + h}}} \right) = \dfrac{{GMmh}}{{R(R + h)}}$ [Eq. 1]
We also know that the force generated due to the orbital motion is given as:
$F = \dfrac{{m{v^2}}}{{R + h}}$ [Since, the orbit is at height h]
This force is also equal to the force due to gravitational attraction between the Earth and the satellite as:
$F = \dfrac{{GMm}}{{{{(R + h)}^2}}}$
Equating the two gives us”
$\dfrac{{m{v^2}}}{{R + h}} = \dfrac{{GMm}}{{{{(R + h)}^2}}}$
$ \Rightarrow m{v^2} = \dfrac{{GMe}}{{R + h}}$
Also, the kinetic energy is given as:
${E_2} = \dfrac{{m{v^2}}}{2} = \dfrac{{GMm}}{{2(R + h)}}$
According to the question:
${E_1} = {E_2}$
$\dfrac{{GMmh}}{{R(R + h)}} = \dfrac{{GMm}}{{2(R + h)}}$ [From Eq. 1]
Solving for h, we get:
$\dfrac{h}{R} = \dfrac{1}{2}$
Hence, $h = \dfrac{1}{2}R = \dfrac{1}{2}(6.4 \times {10^3}) = 3.2 \times {10^3}$km
Hence, the correct answer is option (C).

Note
We used the force due to gravitational acceleration to calculate the energy. This energy can also be used to calculate using the orbital velocity of the satellite which is given by $v = \sqrt {\dfrac{{GM}}{r}} $. The orbital velocity is an important parameter in satellite launches. The more orbital velocity we provide to a probe, the lesser fuel it requires to reach the orbital speed by taking advantage of the Earth’s spin.