Question

A satellite moves eastwards very near the surface of the earth in an equatorial plane with speed $({v_0})$ . Another satellite moves at the same height with the same speed in the equatorial plane but westwards. If $R = $ radius of the earth and $\omega $ be its angular speed of the earth about its own axis. Then find the approximate difference in the two-time period as observed on the earth.
A. $\dfrac{{4\pi \omega {R^2}}}{{{v_0}^2 + {R^2}{\omega ^2}}}$
B. $\dfrac{{2\pi \omega {R^2}}}{{{v_0}^2 - {R^2}{\omega ^2}}}$
C. $\dfrac{{4\pi \omega {R^2}}}{{{v_0}^2 - {R^2}{\omega ^2}}}$
D. $\dfrac{{2\pi \omega {R^2}}}{{{v_0}^2 + {R^2}{\omega ^2}}}$

Answer 1

Hint: In this case, when a problem is based on Gravitation, we know that relative velocity in two different directions will play a significant role in finding out the relative time period hence, the relative velocity of satellites with respect to earth must be evaluated in both east and west direction first in order to get an accurate solution.

Formula used:
The expression of velocity of earth from east to west is,
${v_{eastwards}} = R\omega $
where $R = $ radius of the earth
$\omega = $ angular speed of the earth

Complete step by step solution:
As we know that the Earth moves in a direction from east to west therefore the velocity of the earth will be: -
${v_{eastwards}} = R\omega $
We know the Velocity of the satellite moving eastwards with respect to earth will be: -
${v_{satellite}} = {v_0} - R\omega $......(eastwards)
And, Velocity of the satellite moving westwards with respect to earth will be: -
${v_{satellite}} = {v_0} + R\omega $.....(westwards)

Now, the Time period in an eastward direction,
${T_{east}} = \dfrac{{2\pi R}}{{{v_0} - R\omega }}$
Time period in westward direction,
${T_{west}} = \dfrac{{2\pi R}}{{{v_0} + R\omega }}$
Therefore, the approximate difference in two-time period i.e., $\Delta T$ can be calculated as: -
$\Delta T = {T_{east}} - {T_{west}} = \dfrac{{2\pi R}}{{{v_0} - R\omega }} - \dfrac{{2\pi R}}{{{v_0} + R\omega }}$

Taking $2\pi R$ common, we get
$\Delta T = 2\pi R\left( {\dfrac{{{v_0} + R\omega - ({v_0} - R\omega )}}{{({v_0} - R\omega )({v_0} + R\omega )}}} \right) \\ $
$\Rightarrow \Delta T = 2\pi R\left( {\dfrac{{2R\omega }}{{{v_0}^2 - {R^2}{\omega ^2}}}} \right) \\
\therefore \Delta T = \dfrac{{4\pi \omega {R^2}}}{{{v_0}^2 - {R^2}{\omega ^2}}}$
Thus, the approximate difference in the two-time period as observed on the earth is $\dfrac{{4\pi \omega {R^2}}}{{{v_0}^2 - {R^2}{\omega ^2}}}$.

Hence, the correct option is C.

Note: Since this is a problem related to Binding Energy and Satellites in the gravitation concept hence, given conditions must be analyzed carefully and quantities that are required to calculate the difference in the time period, must be identified on a prior basis as it gives a better understanding of the problem.