Question

A satellite must move in the equatorial plane of earth close to its surface either in the earth’s rotation direction or against it. Find how many times the kinetic energy of the satellite in the latter case exceeds that in the former case (in the reference frame fixed to the earth).
A. 1.27
B. 1
C. 2
D. None of these

Answer 1

Hint: Here, we have two cases of motion of the satellite. Find the relative velocity of the satellite with respect to earth. The obtain the kinetic energy of the satellite for both cases and take their ratio. Express the terms in terms of the acceleration due to gravity and then we can find our answer.

Complete step by step answer:
Let, the orbital velocity of the satellite in the equatorial plane of earth is ${{v}_{0}}$
Let us consider the case where the satellite is moving in the earth’s rotation direction.
Let, the relative velocity of the satellite in the rest frame of earth is, ${{v}_{1}}$
So, we can write,
${{v}_{1}}={{v}_{0}}-\omega R$
Where $\omega $ is the angular velocity of rotation of earth around its own axis and R is the radius of orbit of the satellite around earth.
Now,
$\omega $ can be written as,
$\omega =\dfrac{2\pi }{t}$
So,
${{v}_{1}}={{v}_{0}}-\dfrac{2\pi R}{T}$
So, the kinetic energy of the satellite will be,
${{T}_{1}}=\dfrac{1}{2}m{{\left( {{v}_{0}}-\dfrac{2\pi R}{T} \right)}^{2}}$
Where, m is the mass of the satellite.
Now, considering the second case where the satellite moves in the opposite direction to the earth’s rotational motion, let, the relative velocity of the satellite is ${{v}_{2}}$
So,
${{v}_{2}}={{v}_{0}}+\dfrac{2\pi R}{T}$
So, the kinetic energy of the satellite will be,
${{T}_{2}}=\dfrac{1}{2}m{{\left( {{v}_{0}}+\dfrac{2\pi R}{T} \right)}^{2}}$
Taking the ratio of the two kinetic energies, we get that,
$\begin{align}
  & \dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{\dfrac{1}{2}m{{\left( {{v}_{0}}+\dfrac{2\pi R}{T} \right)}^{2}}}{\dfrac{1}{2}m{{\left( {{v}_{0}}-\dfrac{2\pi R}{T} \right)}^{2}}} \\
 & \dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{{{\left( {{v}_{0}}+\dfrac{2\pi R}{T} \right)}^{2}}}{{{\left( {{v}_{0}}-\dfrac{2\pi R}{T} \right)}^{2}}} \\
\end{align}$
Now, the satellite moves around the earth due to gravitational pull of earth and the required centripetal force on the satellite is equal to this gravitational pull.
So, we can write that,
$\dfrac{GMm}{{{R}^{2}}}=\dfrac{m{{v}_{0}}^{2}}{R}$
Where, M is the Earth's mass and G is the gravitational constant.
${{v}_{0}}=\sqrt{\dfrac{GM}{{{R}^{2}}}R}$
Again, acceleration due to gravity can be given by,
$g=\dfrac{GM}{{{R}^{2}}}$
So,
${{v}_{0}}=\sqrt{gR}$
Putting this value on the kinetic energy ratio we get,
$\dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{{{\left( \sqrt{gR}+\dfrac{2\pi R}{T} \right)}^{2}}}{{{\left( \sqrt{gR}-\dfrac{2\pi R}{T} \right)}^{2}}}$
Now, the satellite is moving near the surface of earth. So, R will be almost equal to the radius of earth.
$R=6.4\times {{10}^{6}}m$
$g=10m{{s}^{-2}}$
$T=24h=24\times 60\times 60s$
Putting these values on the above equation, we get that,
$\begin{align}
  & \dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{{{\left( \sqrt{10\times 6.4\times {{10}^{6}}}+\dfrac{2\pi \times 6.4\times {{10}^{6}}}{24\times 60\times 60} \right)}^{2}}}{{{\left( \sqrt{10\times 6.4\times {{10}^{6}}}-\dfrac{2\pi \times 6.4\times {{10}^{6}}}{24\times 60\times 60} \right)}^{2}}} \\
 & \dfrac{{{T}_{2}}}{{{T}_{1}}}=1.27 \\
 & {{T}_{2}}=1.27{{T}_{1}} \\
\end{align}$
The kinetic energy of the satellite in the second case exceeds the first case by 1.27 times.
The correct option is (A).

Note: A satellite moves around the earth due to its gravitational attraction. The gravitational attraction provides the required centripetal force to move the satellite in a circular orbit. Depending on the direction of motion and the height of the satellite the velocity of the satellite will be different and kinetic energy will also be different.