The mass of a planet is twice the mass of earth and diameter of the planet is thrie the diameter of the earth, then the acceleration due to gravity on the planet's surface is

Options

A. `g//2`
B. `2g`
C. `2g//9`
D. `3g//sqrt(2)`

Explanation

To find the acceleration due to gravity on the surface of a planet whose mass is twice that of Earth and whose diameter is three times that of Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the known values**:
- Let \( M_E \) be the mass of Earth.
- The mass of the planet \( M_P = 2 M_E \).
- Let \( D_E \) be the diameter of Earth.
- The diameter of the planet \( D_P = 3 D_E \).
- The radius of the planet \( R_P = \frac{D_P}{2} = \frac{3 D_E}{2} = \frac{3}{2} R_E \).

2. **Write the formula for acceleration due to gravity**:
The acceleration due to gravity \( g \) on the surface of a planet is given by the formula:
\[ g_P = \frac{G M_P}{R_P^2} \]
where \( G \) is the universal gravitational constant.

3. **Substitute the values for the planet**:
Substitute \( M_P \) and \( R_P \) into the formula:
\[ g_P = \frac{G (2 M_E)}{(R_P)^2} = \frac{G (2 M_E)}{\left(\frac{3}{2} R_E\right)^2} \]

4. **Simplify the expression**:
Calculate \( R_P^2 \):
\[ R_P^2 = \left(\frac{3}{2} R_E\right)^2 = \frac{9}{4} R_E^2 \]
Now substitute this back into the equation for \( g_P \):
\[ g_P = \frac{G (2 M_E)}{\frac{9}{4} R_E^2} = \frac{2 G M_E}{\frac{9}{4} R_E^2} \]
This simplifies to:
\[ g_P = \frac{2 \cdot 4 G M_E}{9 R_E^2} = \frac{8 G M_E}{9 R_E^2} \]

5. **Relate this to the acceleration due to gravity on Earth**:
The acceleration due to gravity on Earth \( g_E \) is given by:
\[ g_E = \frac{G M_E}{R_E^2} \]
Thus, we can express \( g_P \) in terms of \( g_E \):
\[ g_P = \frac{8}{9} g_E \]

6. **Final Result**:
If we take the value of \( g_E \) to be approximately \( 9.8 \, \text{m/s}^2 \), then:
\[ g_P = \frac{8}{9} \times 9.8 \approx 8.73 \, \text{m/s}^2 \]

### Conclusion:
The acceleration due to gravity on the surface of the planet is approximately \( 8.73 \, \text{m/s}^2 \).