Question

A satellite revolving around a planet in stationary orbit has time period 6 hours. The mass of planet is one-fourth the mass of earth. The radius orbit of planet is: (Given = Radius of geostationary orbit for earth is $4.2 \times 10^4$ km

• A. $1.4 \times 10^4$ km
• B. $8.4 \times 10^1$ km
• C. $1.68 \times 10^5$ km
• D. $1.05 \times 10^4$ km

Answer 1

The Correct Option is D

Provided Data:
- Satellite orbital period around planet: \( T_1 = 6 \, \text{hours} \)
- Geostationary satellite orbital period around Earth: \( T_2 = 24 \, \text{hours} \)
- Radius of geostationary orbit around Earth: \( r_2 = 4.2 \times 10^4 \, \text{km} \)
- Mass of the planet: \( M_1 = \frac{M}{4} \) (where \( M \) is Earth's mass)

Step 1: Applying the Time Period Formula for Circular Orbits
The time period \( T \) of a satellite in orbit is calculated using:

\[ T = 2\pi \sqrt{\frac{r^3}{GM}}. \]

The ratio of the time periods for the satellite and Earth's geostationary satellite is:

\[ \frac{T_1}{T_2} = \left( \frac{r_1}{r_2} \right)^{3/2} \left( \frac{M_2}{M_1} \right)^{1/2}, \]

where:
- \( r_1 \) and \( r_2 \) represent the respective orbital radii,
- \( M_1 \) and \( M_2 \) represent the masses of the respective planets.

Step 2: Inserting Given Values
With the given values substituted:

\[ \frac{6}{24} = \left( \frac{r_1}{4.2 \times 10^4} \right)^{3/2} \left( \frac{M}{M/4} \right)^{1/2}. \]

Simplified expression:

\[ \frac{1}{4} = \left( \frac{r_1}{4.2 \times 10^4} \right)^{3/2} \times 2. \]

Dividing both sides by 2 yields:

\[ \frac{1}{8} = \left( \frac{r_1}{4.2 \times 10^4} \right)^{3/2}. \]

Taking the cube root:

\[ \left( \frac{r_1}{4.2 \times 10^4} \right) = \left( \frac{1}{8} \right)^{2/3} \approx 0.25. \]

Consequently:

\[ r_1 \approx 0.25 \times 4.2 \times 10^4 = 1.05 \times 10^4 \, \text{km}. \]

The calculated orbital radius for the planet is \( 1.05 \times 10^4 \, \text{km} \).