Question

A satellite is in insufficiently low orbit so that it encounters air drag and if its orbit changes from r to \[r - \Delta r\]. Find the change in orbital velocity and change in Potential energy.
A) \[\dfrac{{\Delta r}}{2}\sqrt {\dfrac{{GM}}{{{r^3}}}} ,\dfrac{{GMm\Delta r}}{{{r^2}}}\]
B) \[\dfrac{{\Delta r}}{2}\sqrt {\dfrac{{GM}}{{{r^2}}}} ,\dfrac{{GMm}}{{2{r^2}}}\]
C) \[\dfrac{{\Delta r}}{2}\sqrt {\dfrac{{GM}}{{{r^3}}}} ,\dfrac{{GMm\Delta r}}{{2{r^2}}}\]
D) None of these

Answer 1

Hint: Orbital velocity is the minimum velocity that is required by a body in order to move in the given orbit. The satellite’s orbital velocity depends on the gravitational pull on the satellite and the inertia of the motion of the satellite. Due to air drag, its velocity will get affected (decrease). Considering these factors and finding the orbital velocity of the satellite.

Complete step by step solution:
Step I: Let the initial orbital velocity is ‘$V$’.
\[V = \sqrt {\dfrac{{GM}}{r}} \]……(i)
Due to air drag, the orbital velocity is decreased from $r$ to $r-\Delta r$.
Now orbital velocity be
\[V' = \sqrt {\dfrac{{GM}}{{r - \Delta r}}} \]……(ii)
Step II: Change in velocity is
 \[V' - V = \sqrt {\dfrac{{GM}}{{r - \Delta r}}} - \sqrt {\dfrac{{GM}}{r}} \]
Step III :Take out the common terms, the equation becomes
\[V' - V = \sqrt {GM} [{(r - \Delta r)^{ - 1/2}} - {(r)^{ - 1/2}}]\]
Step IV: Also taking \[{r^{ - 1/2}}\] common from the bracket,
= \[\sqrt {GM} {(r)^{ - 1/2}}[{(1 - \dfrac{{\Delta r}}{r})^{ - 1/2}} - 1]\]
Step V: Using Binomial Expansion \[{(1 + x)^n} = 1 + nx\] in the bracket
 \[V' - V = \sqrt {GM} {(r)^{ - 1/2}}[(1 + \dfrac{{\Delta r}}{{2r}}) - 1]\]
 \[V' - V = \sqrt {GM} {(r)^{ - 1/2}}(\dfrac{{\Delta r}}{{2r}})\]
Step VI: Taking \[\Delta r\]and 2 ,out of the bracket and combining like terms of $r$,
\[V' - V = \sqrt {GM} \dfrac{{\Delta r}}{2}(\dfrac{{{r^{ - 1/2}}}}{r})\]
\[V' - V = \sqrt {GM} \dfrac{{\Delta r}}{2}({r^{ - 1/2}} - {r^{ - 1}})\]
\[V' - V = \sqrt {GM} \dfrac{{\Delta r}}{2}{(r)^{ - 3/2}}\]
\[V' - V = \dfrac{{\Delta r}}{2}\sqrt {GM} \dfrac{1}{{{{({r^3})}^{1/2}}}}\]
Step VII: Change in orbital velocity of the satellite is
\[V' - V = \dfrac{{\Delta r}}{2}\sqrt {\dfrac{{GM}}{{{r^3}}}} \]
Step VIII: Now change in potential energy is to find out. Since the satellite is encountering an air drag so its velocity will decrease and some of the energy is lost in the form of heat. So change in potential energy will be twice as much as kinetic energy.
$\text{Potential Energy} = -2 \text{Kinetic Energy}$
\[\Delta U = - 2\Delta K.E.\]……(iii)
Step IX: Formula for kinetic energy is \[\dfrac{1}{2}m{v^2}\]. Substituting this value in equation (iii)
\[\Delta U = - 2(\dfrac{1}{2}m{v^2})\]
Step X: Since there is a change in orbital velocity, so the value of $V$ will also change.
\[\Delta U = - 2(\dfrac{1}{2}m)[V{'^2} - {V^2}]\]
Substituting values of V’ and V
\[ = m[\dfrac{{GM}}{{r - \Delta r}} - \dfrac{{GM}}{r}]\]
Step XI: Taking \[\dfrac{{GM}}{R}\] common from bracket,
\[ = - \dfrac{{GMm}}{r}[1 - \dfrac{{\Delta r}}{r} - 1]\]
\[ = - \dfrac{{GMm}}{r}[ - \dfrac{{\Delta r}}{r}]\]

$\therefore $ The change in orbital velocity and change in Potential energy is $\dfrac{{GMm\Delta r}}{{{r^2}}}$. Hence, option (A) is the correct answer.

Note:
It is to be noted that the orbital velocity of a satellite depends on the distance of the satellite from the center of the Earth. In the equations given above $G$ is the universal gravitational constant and its value is constant. On the other hand, the potential energy of the satellite depends on the height of the satellite. If the satellite experiences air drag then the total energy of the system will decrease hence decreasing potential energy.