The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet is

Options

A. `9.8 m s^(-2)`
B. `4.9 m s^(-2)`
C. `980 m s^(-2)`
D. `19.6 m s^(-2)`

Answer

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

### Step-by-Step Solution:

1. **Identify the parameters:**
- Let the mass of Earth be \( m_e \).
- The mass of the planet \( m_p = 2 m_e \) (twice the mass of Earth).
- Let the radius of Earth be \( r_e \).
- The diameter of the planet is twice that of Earth, so the radius of the planet \( r_p = 2r_e \).

2. **Use 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, and \( r \) is its radius.

3. **Set up the ratio of gravitational accelerations:**
The acceleration due to gravity on the planet \( g_p \) can be expressed as:
\[ g_p = \frac{G \cdot m_p}{r_p^2} \]
For Earth, it is:
\[ g_e = \frac{G \cdot m_e}{r_e^2} \]

4. **Calculate the ratio of \( g_p \) to \( g_e \):**
\[ \frac{g_p}{g_e} = \frac{G \cdot m_p / r_p^2}{G \cdot m_e / r_e^2} = \frac{m_p}{m_e} \cdot \frac{r_e^2}{r_p^2} \]

5. **Substitute the values:**
Substitute \( m_p = 2 m_e \) and \( r_p = 2 r_e \):
\[ \frac{g_p}{g_e} = \frac{2 m_e}{m_e} \cdot \frac{r_e^2}{(2 r_e)^2} = 2 \cdot \frac{r_e^2}{4 r_e^2} = 2 \cdot \frac{1}{4} = \frac{1}{2} \]

6. **Find \( g_p \):**
Since \( g_e \) (acceleration due to gravity on Earth) is approximately \( 9.8 \, \text{m/s}^2 \):
\[ g_p = \frac{1}{2} g_e = \frac{1}{2} \cdot 9.8 \, \text{m/s}^2 = 4.9 \, \text{m/s}^2 \]

### Final Answer:
The acceleration due to gravity on the surface of the planet is \( 4.9 \, \text{m/s}^2 \).

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