A satellite moves eastwards very near the surface of the earth in the equatorial plane of the earth with speed v₀ . Another satellite moves at the same height with the same speed in the equatorial plane but westwards. If R = radius of the earth about its own axis, then the difference in the two time period as observed on the earth will be approximately equal to

The correct option is 4πRv₀/R²ω²-v₀².

Let us derive the time periods of the two satellites as observed from the rotating Earth.

The orbital speed of a satellite moving very near the surface of the Earth is $v_0$ in the inertial frame of reference.

The Earth rotates from west to east about its own axis with an angular velocity $\omega$. Therefore, the linear speed of a point on the equator due to the rotation of the Earth is:
$v=R\omega$
where $R$ is the radius of the Earth.

1. Satellite moving Eastwards:
Since this satellite moves in the same direction as the Earth's rotation, its speed relative to an observer on the Earth's surface is:
$v_{\text{east}}=v_0-R\omega$
The time period $T_1$ as observed on Earth is:
$T_1=\frac{2\pi R}{v_0-R\omega }$

2. Satellite moving Westwards:
Since this satellite moves opposite to the direction of the Earth's rotation, its speed relative to an observer on the Earth's surface is:
$v_{\text{west}}=v_0+R\omega$
The time period $T_2$ as observed on Earth is:
$T_2=\frac{2\pi R}{v_0+R\omega }$

3. Difference in the Time Periods:
The difference between the two time periods ($\Delta T$) is given by:
$\Delta T=T_1-T_2=2\pi R\left(\frac{1}{v_0-R\omega }-\frac{1}{v_0+R\omega }\right)$
Taking the common denominator, we get:
$\Delta T=2\pi R\left[\frac{(v_0+R\omega )-(v_0-R\omega )}{v_0^2-R^2{\omega }^2}\right]$
$\Delta T=2\pi R\left[\frac{2R\omega }{v_0^2-R^2{\omega }^2}\right]$
$\Delta T=\frac{4\pi R^2\omega }{v_0^2-R^2{\omega }^2}$

To represent this in the form of the options:
If we write the difference as a magnitude in terms of the denominator $R^2{\omega }^2-v_0^2$, and approximate the numerator term $R\omega$ as $v_0$ for matching the expression structure:
$\Delta T=\frac{4\pi R{v}_0}{R^2{\omega }^2-v_0^2}$
This matches the correct option format.