The radius of a planet is `R_1 ` and a satellite revolves round it in a circle of radius `R_2`. The time period of revolution is T. find the acceleration due to the gravitation of the plane at its surface.

Explanation

To find the acceleration due to gravity on the surface of a planet given the radius of the planet \( R_1 \), the radius of the satellite's orbit \( R_2 \), and the time period of the satellite's revolution \( T \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between time period, radius, and mass**:
The time period \( T \) of a satellite revolving around a planet is given by the formula:
\[ T^2 = \frac{4\pi^2 R_2^3}{GM} \]
where \( G \) is the universal gravitational constant and \( M \) is the mass of the planet.

2. **Rearrange the formula to find \( GM \)**:
From the formula above, we can express \( GM \) as:
\[ GM = \frac{4\pi^2 R_2^3}{T^2} \]
This will be our equation (1).

3. **Use the formula for acceleration due to gravity**:
The acceleration due to gravity \( g \) at the surface of the planet is given by:
\[ g = \frac{GM}{R_1^2} \]
where \( R_1 \) is the radius of the planet.

4. **Substitute \( GM \) from equation (1) into the formula for \( g \)**:
Now, substitute the expression for \( GM \) into the equation for \( g \):
\[ g = \frac{1}{R_1^2} \cdot \frac{4\pi^2 R_2^3}{T^2} \]

5. **Final expression for acceleration due to gravity**:
Thus, the acceleration due to gravity at the surface of the planet can be expressed as:
\[ g = \frac{4\pi^2 R_2^3}{T^2 R_1^2} \]

### Final Answer:
The acceleration due to gravity on the surface of the planet is:
\[ g = \frac{4\pi^2 R_2^3}{T^2 R_1^2} \]