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 given the density \( d \) of the planet and the time period \( T \) of a satellite revolving close to its surface, we can follow these steps:

### Step 1: Write the formula for the time period of a satellite
The time period \( T \) of a satellite in circular orbit is given by the formula:
\[ T^2 = \frac{4 \pi^2 r^3}{G M} \]
where \( r \) is the orbital radius (which is equal to the radius of the planet \( R \) in this case), and \( M \) is the mass of the planet.

### Step 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 sphere, the volume \( V \) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the mass of the planet becomes:
\[ M = d \cdot \frac{4}{3} \pi R^3 \]

### Step 3: Substitute the mass of the planet into the time period formula
Substituting \( M \) into the time period formula gives:
\[ T^2 = \frac{4 \pi^2 R^3}{G \left( d \cdot \frac{4}{3} \pi R^3 \right)} \]
This simplifies to:
\[ T^2 = \frac{4 \pi^2 R^3}{G \cdot \frac{4}{3} \pi R^3} \]

### Step 4: Simplify the equation
Cancel \( R^3 \) and \( 4 \pi \) from both sides:
\[ T^2 = \frac{3 \pi}{G} \]

### Step 5: Solve for \( G \)
Rearranging the equation to solve for \( G \) gives:
\[ G = \frac{3 \pi}{T^2} \]

### Step 6: Substitute the density into the equation
Since we have the relationship between \( G \), \( d \), and \( T \), we can express \( G \) as:
\[ G = \frac{3 \pi}{d T^2} \]

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