Question

If due to air drag, the orbital radius of the earth decreases from $R$ to $R - \Delta R$, $\Delta R < < R$, the change in K.E, $\Delta K$ is
A. $- \dfrac{{GM}}{{{R^2}}} \times \Delta R$
B. $\dfrac{{GM}}{{{R^2}}} \times \Delta R$
C. $m\dfrac{{GM}}{{2{R^2}}} \times \Delta R$
D.. $- \dfrac{{GM}}{{2{R^2}}} \times \Delta R$

Answer 1

First, find the value of orbital velocity of a body moving around the earth. Write the expression for initial and final orbital velocity separately. Then recall the formula for kinetic energy of a body and use it to find the initial and final kinetic energy. Use these values to find the change in kinetic energy of a body.

Complete step by step answer:
Given, initial orbital radius of the earth, $R$.
Final orbital radius of the earth, $R - \Delta R$.
Also, $\Delta R < < R$
The formula for orbital velocity of a body is,
$v = \sqrt {\dfrac{{GM}}{R}}$ (i)
where $G$ is the gravitational constant, $M$ is the mass of the attracting body and $R$ is the radius of the attracting body.
The formula for kinetic energy is,
$K.E = \dfrac{1}{2}m{v^2}$ (ii)
where $m$ is the mass of the body and $v$ is the velocity of the body.

The initial orbital velocity of the body using the formula from equation (i) is
${v_{in}} = \sqrt {\dfrac{{GM}}{R}}$ (iii)
The initial kinetic energy of the body using the formula from equation (ii) is,
${\left( {K.E} \right)_{in}} = \dfrac{1}{2}m{v_{in}}^2$ (iv)
The final orbital velocity of the body using the formula from equation (i) is
${v_{fi}} = \sqrt {\dfrac{{GM}}{{R - \Delta R}}}$ (v)
The final kinetic energy of the body using the formula from equation (ii) is,
${\left( {K.E} \right)_{fi}} = \dfrac{1}{2}m{v_{fi}}^2$ (vi)
Change in kinetic energy is,
$\Delta K = {\left( {K.E} \right)_{fi}} - {\left( {K.E} \right)_{in}}$
Putting the values of ${\left( {K.E} \right)_{fi}}$ and ${\left( {K.E} \right)_{in}}$ in the above equation we have,
$\Delta K = \dfrac{1}{2}m{v_{fi}}^2 - \dfrac{1}{2}m{v_{in}}^2$
Putting the values of ${v_{fi}}$ and ${v_{in}}$ in the above equation we have,
$\Delta K = \dfrac{1}{2}m{\left( {\sqrt {\dfrac{{GM}}{{R - \Delta R}}} } \right)^2} - \dfrac{1}{2}m{\left( {\sqrt {\dfrac{{GM}}{R}} } \right)^2}$
$\Rightarrow \Delta K = \dfrac{1}{2}m\left( {\dfrac{{GM}}{{R - \Delta R}}} \right) - \dfrac{1}{2}m\left( {\dfrac{{GM}}{R}} \right)$
$\Rightarrow \Delta K = \dfrac{1}{2}mGM\left( {{{\left( {R - \Delta R} \right)}^{ - 1}} - {R^{ - 1}}} \right)$
$\Rightarrow \Delta K = \dfrac{1}{2}mGM{R^{ - 1}}\left( {{{\left( {1 - \dfrac{{\Delta R}}{R}} \right)}^{ - 1}} - 1} \right)$
We will now use the binomial expansion, we says ${\left( {1 + x} \right)^n} = 1 + nx$ for $x < < 1$
$\Delta K = \dfrac{1}{2}mGM{R^{ - 1}}\left( {\left( {1 + \dfrac{{\Delta R}}{R}} \right) - 1} \right)$
$\Rightarrow \Delta K = \dfrac{1}{2}mGM\dfrac{{\Delta R}}{{{R^2}}}$
$\therefore \Delta K = m\dfrac{{GM}}{{2{R^2}}} \times \Delta R$
Therefore, the change in kinetic energy is, $m\dfrac{{GM}}{{2{R^2}}} \times \Delta R$.

Hence, option C is the correct answer.

Note: Orbital velocity is the minimum velocity required by a body to stay in orbit of the attracting body. Orbital velocity is independent of the mass of the body and depends only on the mass and radius of the attracting body. The concept of orbital velocity is used in natural and artificial satellites.