Question

A satellite is seen after each $8\text{hours}$ over the equator at a place on the earth when its sense of rotation is opposite to the earth. The time interval after which it can be seen at the same place when the sense of rotation of earth and satellite is same will be:

• A. $8\text{hours}$
• B. $128\text{hours}$
• C. $24\text{hours}$
• D. $6\text{hours}$

Answer 1

The correct option is C $24\text{hours}$

Let ${\omega }_1$ and ${\omega }_2$ be the angular velocities of the earth and the satellite.

Since, ${\omega }_1$ and ${\omega }_2$ are opposite in sense.

$\therefore$ Relative angular velocity,
${\omega }_r=({\omega }_1$+${\omega }_2)$

So, time period $T$ is given by

$T=\frac{2\pi }{{\omega }_r}=\frac{2\pi }{{\omega }_1+{\omega }_1}$

Given, $T=8\text{hours}$

$\Rightarrow 8=\frac{2\pi }{\frac{2\pi }{T_1}+\frac{2\pi }{T_2}}$

$[T_1$ is the time period the earth $T_2$ is the time of the satellite $]$

$\Rightarrow \frac{1}{24}+\frac{1}{T_2}=\frac{1}{8}$

$\Rightarrow \frac{1}{T_2}=\frac{1}{8}-\frac{1}{24}=\frac{2}{24}$

$\Rightarrow T_2=12\text{hours}$

When ${\omega }_1$ and ${\omega }_2$ are in the same sense.

Since, ${\omega }_r^{\prime }={\omega }_1-{\omega }_2$

$\therefore T^{\prime }=\frac{2\pi }{{\omega }_r^{\prime }}=\frac{2\pi }{\frac{2\pi }{T_1}-\frac{2\pi }{T_2}}=\frac{2\pi }{\frac{2\pi }{12}-\frac{2\pi }{24}}$

$\Rightarrow T^{\prime }=24\text{hours}$

Hence, option $\left(c\right)$ is correct.