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

Given:
- Time period of the satellite around the planet: \( T_1 = 6 \, \text{hours} \)
- Time period of a geo-stationary satellite around Earth: \( T_2 = 24 \, \text{hours} \)
- Radius of geo-stationary orbit around Earth: \( r_2 = 4.2 \times 10^4 \, \text{km} \)
- Mass of the planet: \( M_1 = \frac{M}{4} \) (where \( M \) is the mass of the Earth)

Step 1: Using the Time Period Relation for Circular Orbits
The formula for the time period of a satellite in orbit is given by:

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

Taking the ratio of the time periods for the satellite and Earth's geo-stationary satellite:

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

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

Step 2: Substituting the Given Values
Substituting the given values:

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

Simplifying:

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

Dividing both sides by 2:

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

Taking the cube root:

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

Thus:

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

Therefore, the radius of the orbit of the planet is \( 1.05 \times 10^4 \, \text{km} \).