A planet has density p, radius R and acceleration due to gravity as g. If the radius of the planet were doubled keeping the density same, the acceleration due to gravity at the surface will be

Options

A. 4g
B. 2g
C. g
D. g/2

Explanation

To solve the problem, we need to determine how the acceleration due to gravity (g') changes when the radius of a planet is doubled while keeping its density (ρ) constant.

### Step-by-Step Solution:

1. **Understand the Formula for Acceleration Due to Gravity**:
The acceleration due to gravity 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 the radius of the planet.

2. **Express Mass in Terms of Density**:
The mass \( M \) of the planet can be expressed in terms of its volume and density. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, the mass \( M \) can be expressed as:
\[ M = \text{Density} \times \text{Volume} = \rho \cdot V = \rho \cdot \left(\frac{4}{3} \pi R^3\right) \]

3. **Substitute Mass into the Gravity Formula**:
Substitute the expression for mass \( M \) into the formula for \( g \):
\[ g = \frac{G \cdot \left(\rho \cdot \frac{4}{3} \pi R^3\right)}{R^2} \]
Simplifying this gives:
\[ g = \frac{4}{3} \pi G \rho R \]

4. **Determine New Acceleration Due to Gravity (g')**:
If the radius of the planet is doubled (i.e., \( R' = 2R \)), and the density remains the same, we can find the new acceleration due to gravity \( g' \):
\[ g' = \frac{4}{3} \pi G \rho (2R) = \frac{4}{3} \pi G \rho \cdot 2R \]
This simplifies to:
\[ g' = 2 \left(\frac{4}{3} \pi G \rho R\right) = 2g \]

5. **Conclusion**:
Therefore, the new acceleration due to gravity \( g' \) when the radius is doubled while keeping the density the same is:
\[ g' = 2g \]

### Final Answer:
The acceleration due to gravity at the surface will be **twice the original acceleration due to gravity**.

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