Question

A satellite can be in a geostationary orbit around earth at a distance r from the center. If the angular velocity of earth about its axis doubles, a satellite can now be in a geostationary orbit around the earth if its distance from the center is:
(a) ${\displaystyle \frac{r}{2}}$
(b) ${\displaystyle \frac{r}{2\sqrt{2}}}$
(c) ${\displaystyle \frac{r}{4^{1/3}}}$
(d) ${\displaystyle \frac{r}{2^{1/3}}}$

Answer 1

Hint: Geostationary satellites appear to be stationary as seen from the Earth. This is because its angular velocity matches with that of the Earth.
Formula used:
Time period:
$T={\displaystyle \frac{2\pi }{\omega }}$ …… (1)
where,
T is the time period of revolution of the body.
$\omega$ is the angular velocity of the body.

Kepler’s 3rd law of motion:
$T^2\propto r^3$
$T\propto r^{3/2}$ …… (2)
where,
T is the time period of revolution of the body.
r is the mean distance of the body from its axis of revolution.

Complete step-by-step answer:
Given:
1. Radius of revolution of satellite = r
2. New angular velocity of Earth = 2 times of old angular velocity.

To find: New radius of revolution of satellite.
Step 1 of 3:
Let ${\omega }_e$be the angular velocity of Earth and ${\omega }_1$ be the angular velocity of the satellite, initially. A geostationary satellite revolves with the same angular velocity as earth:
${\omega }_e={\omega }_1$ …… (3)

Let ${{\omega }_e}^{{}^{\prime }}$be the new angular velocity of Earth. It is given that:
${{\omega }_e}^{{}^{\prime }}=2{\omega }_e$ …… (4)

Let ${\omega }_2$ be the new angular velocity of the satellite. Again, the angular velocity of earth will be same as that of the satellite:
${{\omega }_e}^{{}^{\prime }}={\omega }_2$ …… (5)

Comparing eq (3), (4) and (5):
${\omega }_2=2{\omega }_1$
${\displaystyle \frac{{\omega }_2}{{\omega }_1}}=2$ …… (6)

Step 2 of 3:
Let the old Time period of the satellite be $T_1$ and the new time period be $T_2$. Let the old angular velocity of the satellite be ${\omega }_1$ and the new angular velocity be ${\omega }_2$. Use eq (1) to find the relation between time period and angular velocity in both cases:
$T_1={\displaystyle \frac{2\pi }{{\omega }_1}}$ …… (7)
$$T_2={\displaystyle \frac{2\pi }{{\omega }_2}}$$ ……. (8)

Divide eq (7) and (8) to derive the relation between old time period and new time period:
${\displaystyle \frac{T_1}{T_2}}={\displaystyle \frac{{\omega }_2}{{\omega }_1}}$ …… (9)

Substitute eq (6) in eq (9):
${\displaystyle \frac{T_1}{T_2}}=2$ …… (10)

Step 3 of 3:
Let the old radius be $r_1$ and the new radius be $r_2$. Using eq (2):
$T_1\propto (r_1{)}^{3/2}$ …… (11)
$T_2\propto (r_2{)}^{3/2}$ …… (12)

Divide eq (11) and (12):
${\displaystyle \frac{T_1}{T_2}}=({\displaystyle \frac{r_1}{r_2}}{)}^{3/2}$

Substituting eq (10) in eq (12):
$$\begin{gathered} 2=({\displaystyle \frac{r_1}{r_2}}{)}^{3/2} \\ {2}^{2/3}={\displaystyle \frac{r_1}{r_2}} \end{gathered}$$

Rearranging:
$$\begin{gathered} r_2={\displaystyle \frac{r_1}{2^{2/3}}} \\ {r}_2={\displaystyle \frac{r_1}{4^{1/3}}} \end{gathered}$$

Correct Answer:
New radius of revolution of the satellite is (c) ${\displaystyle \frac{r}{4^{1/3}}}$.

Note: In questions like these, use the definition of geostationary satellite. Apply Kepler’s 3rd law and compare the 2 cases to find the answer.