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
- $1.4 \times 10^4$ km
- $8.4 \times 10^1$ km
- $1.68 \times 10^5$ km
- $1.05 \times 10^4$ km
The Correct Option is D
Approach Solution - 1
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 \( 1.05 \times 10^4 \, \text{km} \).
Approach Solution -2
Time period of satellite around the planet, T = 6 hours
Mass of planet, M = (1/4) × Me (where Me is the mass of Earth)
Radius of geostationary orbit of Earth, Re = 4.2 × 10⁴ km
Time period for Earth's geostationary satellite, Te = 24 hours
Step 2: Formula for time period of a satellite.
The time period is related to radius and mass of the planet as:
\[ T^2 = \frac{4 \pi^2 R^3}{G M} \] Hence, \[ T^2 \propto \frac{R^3}{M} \]
Step 3: Write ratio for two satellites (Earth and the given planet).
\[ \frac{T^2}{T_e^2} = \frac{R^3 / M}{R_e^3 / M_e} \] \[ \frac{T^2}{T_e^2} = \frac{R^3 M_e}{R_e^3 M} \] Substitute M = (1/4) Me:
\[ \frac{T^2}{T_e^2} = \frac{R^3 M_e}{R_e^3 (M_e / 4)} = 4 \frac{R^3}{R_e^3} \]
\[ \frac{R^3}{R_e^3} = \frac{1}{4} \left(\frac{T}{T_e}\right)^2 \]
Step 4: Substitute the given values.
T = 6 hours, Te = 24 hours, Re = 4.2 × 10⁴ km
\[ \frac{R^3}{(4.2 \times 10^4)^3} = \frac{1}{4} \left(\frac{6}{24}\right)^2 = \frac{1}{4} \times \frac{1}{16} = \frac{1}{64} \]
\[ R^3 = \frac{(4.2 \times 10^4)^3}{64} \] \[ R = \frac{4.2 \times 10^4}{4} = 1.05 \times 10^4 \, \text{km} \]
Step 5: Final Answer.
The radius of the stationary orbit of the planet is:
\[ \boxed{1.05 \times 10^4 \, \text{km}} \]
Final Answer: 1.05 × 10⁴ km
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