If `g_(1)` and `g_(2)` denote acceleration due to gravity on the surface of the earth and on a planet whose mass and radius is thrice that of earth, then

Options

A. `g_(1)=9g_(2)`
B. `g_(2)=9g_(1)`
C. `g_(1)=3g_(2)`
D. `g_(2)=3g_(1)`

Answer

To solve the problem, we need to find the relationship between the acceleration due to gravity on the surface of the Earth (denoted as \( g_1 \)) and the acceleration due to gravity on a planet whose mass and radius are both three times that of Earth (denoted as \( g_2 \)).

### Step-by-Step Solution:

1. **Understand the formula for acceleration due to gravity**:
The acceleration due to gravity \( g \) on the surface of a celestial body is given by the formula:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the body, and \( R \) is its radius.

2. **Identify the parameters for Earth**:
For Earth, we denote:
- Mass of Earth: \( M_1 \)
- Radius of Earth: \( R_1 \)
- Acceleration due to gravity on Earth: \( g_1 = \frac{GM_1}{R_1^2} \)

3. **Identify the parameters for the planet**:
According to the problem, the planet has:
- Mass: \( M_2 = 3M_1 \) (three times the mass of Earth)
- Radius: \( R_2 = 3R_1 \) (three times the radius of Earth)
- Acceleration due to gravity on the planet: \( g_2 = \frac{GM_2}{R_2^2} \)

4. **Substitute the values for the planet**:
Substitute \( M_2 \) and \( R_2 \) into the formula for \( g_2 \):
\[ g_2 = \frac{G(3M_1)}{(3R_1)^2} \]
Simplifying this gives:
\[ g_2 = \frac{3GM_1}{9R_1^2} = \frac{1}{3} \cdot \frac{GM_1}{R_1^2} \]

5. **Relate \( g_2 \) to \( g_1 \)**:
Since \( g_1 = \frac{GM_1}{R_1^2} \), we can substitute this into the equation for \( g_2 \):
\[ g_2 = \frac{1}{3} g_1 \]

6. **Express \( g_1 \) in terms of \( g_2 \)**:
Rearranging the equation gives:
\[ g_1 = 3g_2 \]

### Final Result:
Thus, the relationship between the acceleration due to gravity on Earth and the planet is:
\[ g_1 = 3g_2 \]