A satellite of mass m is revolving close to surface of a plant of density d with time period T . The value of universal gravitational constant on planet is given by

Options

A. `2a^(2)Tpi`
B. `dT^(2)pi`
C. `1/(d^(2)Tpi)`
D. `(3pi)/(dT^(2))`

Answer

To find the value of the universal gravitational constant \( G \) on a planet where a satellite of mass \( m \) is revolving close to its surface with a time period \( T \) and the planet has a density \( d \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between time period and gravitational force**:
The time period \( T \) of a satellite in orbit is given by the formula:
\[ T^2 = \frac{4\pi^2 R^3}{GM} \]
where \( R \) is the orbital radius (which is equal to the radius of the planet when the satellite is close to the surface), \( G \) is the universal gravitational constant, and \( M \) is the mass of the planet.

2. **Express the mass of the planet in terms of its density**:
The mass \( M \) of the planet can be expressed in terms of its density \( d \) and volume \( V \):
\[ M = d \cdot V \]
For a spherical planet, the volume \( V \) is given by:
\[ V = \frac{4}{3}\pi R^3 \]
Therefore, the mass of the planet becomes:
\[ M = d \cdot \frac{4}{3}\pi R^3 \]

3. **Substitute the mass of the planet into the time period formula**:
Now, substitute \( M \) back into the time period formula:
\[ T^2 = \frac{4\pi^2 R^3}{G \left(d \cdot \frac{4}{3}\pi R^3\right)} \]
Simplifying this gives:
\[ T^2 = \frac{4\pi^2 R^3}{\frac{4}{3}Gd\pi R^3} \]
The \( R^3 \) terms cancel out:
\[ T^2 = \frac{3\pi}{Gd} \]

4. **Rearranging to find \( G \)**:
To find \( G \), rearrange the equation:
\[ G = \frac{3\pi}{dT^2} \]

### Final Result:
Thus, the value of the universal gravitational constant \( G \) on the planet is given by:
\[ G = \frac{3\pi}{dT^2} \]