Question

A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of ${{g}_{0}}$ , the value of acceleration due to gravity at the earth’s surface, is
A. $-\dfrac{2m{{g}_{0}}{{R}^{2}}}{R+h}$
B. $\dfrac{m{{g}_{0}}{{R}^{2}}}{2\left( R+h \right)}$
C. $-\dfrac{m{{g}_{0}}{{R}^{2}}}{2\left( R+h \right)}$
D. $\dfrac{2m{{g}_{0}}{{R}^{2}}}{R+h}$

Answer 1

To solve this question, find the value of the tangential velocity of satellites around earth. Find the total energy of the satellite as the sum of the kinetic energy and potential energy. Then express it in terms of the acceleration due to gravity by transforming the equation.

Complete answer:
A satellite is moving around the earth a t height of h. The total energy of the satellite is given by the addition of the kinetic energy and potential energy. Since, external work done is zero the addition of kinetic energy and potential energy at every point will be the same.
Now, for a satellite moving in orbit around earth at height h, the tangential velocity is given by,
$v=\sqrt{\dfrac{Gm}{R+h}}$
Now, the kinetic energy of the satellite is given by,
$\begin{aligned} & KE=\dfrac{1}{2}m{{v}^{2}} \\\ & KE=\dfrac{1}{2}m{{\left( \sqrt{\dfrac{GM}{R+h}} \right)}^{2}} \\\ & KE=\dfrac{GMm}{2\left( R+h \right)} \\\ \end{aligned}$
Again, potential energy of the satellite at an height of h is given by,
$PE=-\dfrac{GMm}{(R+h)}$
Total energy of the system is given as,
$\begin{aligned} & E=KE+PE=\dfrac{GMm}{2\left( R+h \right)}-\dfrac{GMm}{\left( R+h \right)} \\\ & E=-\dfrac{GMm}{2\left( R+h \right)} \\\ \end{aligned}$
Now we can define the relation between the acceleration due to gravity and the universal gravitational constant as,
$g=\dfrac{GM}{{{r}^{2}}}$
Where, g is the acceleration due to gravity, M is mass of earth, G is the gravitational constant and r is the distance from the centre of earth.
Now, at the surface of earth,
${{g}_{0}}=\dfrac{GM}{{{R}^{2}}}$
Where R is the earth's radius.
Now, in the energy equation in the left-hand side, multiply ${{g}_{0}}$ in both the denominator and numerator.
$E=-\dfrac{GMm{{g}_{0}}}{2\left( R+h \right){{g}_{0}}}$
Now, putting the value of ${{g}_{0}}$ in the denominator, we get that,
$\begin{aligned} & E=-\dfrac{GMm{{g}_{0}}}{2\left( R+h \right)\dfrac{GM}{{{R}^{2}}}} \\\ & E=-\dfrac{m{{g}_{0}}{{R}^{2}}}{2\left( R+h \right)} \\\ \end{aligned}$

So, the correct answer is “Option C”.

Note:
The value of the acceleration due to gravity for a planet changes with height. With the increase in height the acceleration due to gravity will be lower and with the decrease in height the acceleration due to gravity will be higher.