A planet of mass 4 times earth spins about itself and completes one rotation is 96 hours . The radius of a secondary stationary satellite about this planet in comparisons to the radius of the geostationary orbit around the earth is

Options

A. 4 times
B. `(1/4)` times
C. 2 times
D. `(1/2)` times

Answer

To solve the problem, we need to find the radius of a secondary stationary satellite around a planet with a mass 4 times that of Earth, which completes one rotation in 96 hours. We will compare this radius with the radius of a geostationary orbit around Earth.

### Step 1: Understand the relationship between gravitational force and centripetal force

The gravitational force acting on a satellite provides the necessary centripetal force for its circular motion. The equation can be expressed as:

\[ \frac{G M m}{r^2} = m \omega^2 r \]

Where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the planet,
- \( m \) is the mass of the satellite,
- \( r \) is the radius of the orbit,
- \( \omega \) is the angular velocity.

### Step 2: Simplify the equation

We can cancel \( m \) from both sides of the equation:

\[ \frac{G M}{r^2} = \omega^2 r \]

Rearranging gives:

\[ \frac{G M}{\omega^2} = r^3 \]

### Step 3: Relate angular velocity to the time period

The angular velocity \( \omega \) is related to the time period \( T \) by:

\[ \omega = \frac{2\pi}{T} \]

Substituting this into our equation gives:

\[ r^3 = \frac{G M T^2}{(2\pi)^2} \]

### Step 4: Calculate for the planet

For the planet with mass \( M' = 4M \) and a time period \( T' = 96 \text{ hours} = 4 \times 24 \text{ hours} \):

\[ r'^3 = \frac{G (4M) (96 \text{ hours})^2}{(2\pi)^2} \]

### Step 5: Calculate for Earth

For Earth, with mass \( M \) and time period \( T = 24 \text{ hours} \):

\[ r^3 = \frac{G M (24 \text{ hours})^2}{(2\pi)^2} \]

### Step 6: Set up the ratio

Now, we can set up the ratio of the two radii:

\[ \frac{r'^3}{r^3} = \frac{4M \cdot (96)^2}{M \cdot (24)^2} \]

### Step 7: Simplify the ratio

This simplifies to:

\[ \frac{r'^3}{r^3} = 4 \cdot \frac{(96)^2}{(24)^2} = 4 \cdot \frac{(4 \cdot 24)^2}{(24)^2} = 4 \cdot 16 = 64 \]

Taking the cube root:

\[ \frac{r'}{r} = \sqrt[3]{64} = 4 \]

### Conclusion

Thus, the radius of the secondary stationary satellite around the planet is 4 times the radius of the geostationary orbit around the Earth.

### Final Answer
The radius of the secondary stationary satellite about this planet in comparison to the radius of the geostationary orbit around the Earth is **4**.

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