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

Answer: (D)

$1.05 \times 10^4$ km

Answer 2

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 $1.05 \times 10^4 \, \text{km}$.