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) $\dfrac{r}{2}$
(b) $\dfrac{r}{{2\sqrt 2 }}$
(c) $\dfrac{r}{{{4^{1/3}}}}$
(d) $\dfrac{r}{{{2^{1/3}}}}$

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:
$T = \dfrac{{2\pi }}{\omega }$ …… (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 ${\omega _e}^{'}$be the new angular velocity of Earth. It is given that:
${\omega _e}^{'} = 2{\omega _e}$ …… (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:
${\omega _e}^{'} = {\omega _2}$ …… (5)

Comparing eq (3), (4) and (5):
${\omega _2} = 2{\omega _1}$
$\dfrac{{{\omega _2}}}{{{\omega _1}}} = 2$ …… (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:
${T_1} = \dfrac{{2\pi }}{{{\omega _1}}}$ …… (7)
\[{T_2} = \dfrac{{2\pi }}{{{\omega _2}}}\] ……. (8)

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

Substitute eq (6) in eq (9):
$\dfrac{{{T_1}}}{{{T_2}}} = 2$ …… (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):
$\dfrac{{{T_1}}}{{{T_2}}} = {(\dfrac{{{r_1}}}{{{r_2}}})^{3/2}}$

Substituting eq (10) in eq (12):
$
  2 = {(\dfrac{{{r_1}}}{{{r_2}}})^{3/2}} \\
  {2^{2/3}} = \dfrac{{{r_1}}}{{{r_2}}} \\
 $

Rearranging:
$
  {r_2} = \dfrac{{{r_1}}}{{{2^{2/3}}}} \\
  {r_2} = \dfrac{{{r_1}}}{{{4^{1/3}}}} \\
 $

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

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.