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 with a mass twice that of Earth and a diameter three times that of Earth, we can follow these steps:

### Step 1: Understand the formula for acceleration due to gravity
The acceleration due to gravity \( g \) at the surface of a planet is given by the formula:
\[ g = \frac{G \cdot M}{R^2} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the planet,
- \( R \) is the radius of the planet.

### Step 2: Define the mass and radius of the planet
Let:
- \( M_E \) be the mass of Earth,
- \( R_E \) be the radius of Earth.

According to the problem:
- The mass of the planet \( M = 2M_E \),
- The diameter of the planet is three times that of Earth, hence the radius of the planet \( R = \frac{3D_E}{2} = \frac{3 \cdot 2R_E}{2} = 3R_E \).

### Step 3: Substitute the values into the formula
Now we can substitute the values of \( M \) and \( R \) into the formula for \( g \):
\[ g_{planet} = \frac{G \cdot (2M_E)}{(3R_E)^2} \]

### Step 4: Simplify the equation
Now we simplify the equation:
\[ g_{planet} = \frac{G \cdot (2M_E)}{9R_E^2} \]

### Step 5: Relate it to Earth's gravity
We know the acceleration due to gravity on Earth is:
\[ g_E = \frac{G \cdot M_E}{R_E^2} \]

Now, we can express \( g_{planet} \) in terms of \( g_E \):
\[ g_{planet} = \frac{2G \cdot M_E}{9R_E^2} = \frac{2}{9} \cdot \frac{G \cdot M_E}{R_E^2} = \frac{2}{9} g_E \]

### Step 6: Calculate \( g_{planet} \)
Assuming \( g_E \approx 9.8 \, \text{m/s}^2 \):
\[ g_{planet} = \frac{2}{9} \cdot 9.8 \approx 2.18 \, \text{m/s}^2 \]

### Final Answer
Thus, the acceleration due to gravity on the planet's surface is approximately \( 2.18 \, \text{m/s}^2 \).

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