Value of g on the surface of earth is `9.8 m//s^(2)`. Find its value on the surface of a planet whose mass and radius both are two times that of earth.

Explanation

To find the value of gravitational acceleration (g) on the surface of a planet whose mass and radius are both two times that of Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the known values:**
- The gravitational acceleration on Earth, \( g = 9.8 \, \text{m/s}^2 \).
- Let the mass of Earth be \( m \) and the radius of Earth be \( r \).

2. **Determine the mass and radius of the planet:**
- According to the problem, the mass of the planet \( M_p = 2m \) (twice the mass of Earth).
- The radius of the planet \( R_p = 2r \) (twice the radius of Earth).

3. **Use the formula for gravitational acceleration:**
- The formula for gravitational acceleration at the surface of a planet is given by:
\[ g_p = \frac{G M_p}{R_p^2} \]
where \( G \) is the universal gravitational constant.

4. **Substitute the values for the planet:**
- Substitute \( M_p = 2m \) and \( R_p = 2r \) into the formula:
\[ g_p = \frac{G (2m)}{(2r)^2} \]

5. **Simplify the equation:**
- The denominator can be simplified:
\[ (2r)^2 = 4r^2 \]
- Thus, the equation becomes:
\[ g_p = \frac{2Gm}{4r^2} \]

6. **Further simplify:**
- This simplifies to:
\[ g_p = \frac{Gm}{2r^2} \]

7. **Relate it to the gravitational acceleration on Earth:**
- We know that:
\[ g = \frac{Gm}{r^2} \]
- Therefore, we can express \( g_p \) in terms of \( g \):
\[ g_p = \frac{g}{2} \]

8. **Substitute the value of \( g \):**
- Now, substituting \( g = 9.8 \, \text{m/s}^2 \):
\[ g_p = \frac{9.8}{2} = 4.9 \, \text{m/s}^2 \]

### Final Answer:
The value of gravitational acceleration on the surface of the planet is \( g_p = 4.9 \, \text{m/s}^2 \).

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