The value of acceleration due to gravity `.g_p.` on the surface of a planet with radius double that of earth and same mean density as that of the earth is
(`g_e to` acceleration due to gravity on the surface of earth)

Options

A. `g_p=2g_e`
B. `g_p=g_e//2`
C. `r_p=g_e//4`
D. `g_p=4g_e`

Answer

To find the value of acceleration due to gravity \( g_p \) on the surface of a planet with a radius double that of Earth and the same mean density as that of Earth, we can follow these steps:

### Step 1: Understand the relationship between mass, volume, and density
The density \( \rho \) of an object is defined as:
\[ \rho = \frac{M}{V} \]
where \( M \) is the mass and \( V \) is the volume.

### Step 2: Calculate the volume of the planet
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
For the planet with radius \( R_p = 2R_e \) (where \( R_e \) is the radius of Earth), the volume becomes:
\[ V_p = \frac{4}{3} \pi (2R_e)^3 = \frac{4}{3} \pi (8R_e^3) = \frac{32}{3} \pi R_e^3 \]

### Step 3: Calculate the mass of the planet
Since the density is the same as that of Earth, we can express the mass of the planet \( M_p \) as:
\[ M_p = \rho V_p = \rho \left(\frac{32}{3} \pi R_e^3\right) \]
The mass of Earth \( M_e \) is:
\[ M_e = \rho \left(\frac{4}{3} \pi R_e^3\right) \]
Thus, we can relate the mass of the planet to the mass of Earth:
\[ M_p = 8M_e \]

### Step 4: Use the formula for acceleration due to gravity
The acceleration due to gravity \( g \) at the surface of a planet is given by:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

### Step 5: Calculate \( g_p \) for the planet
For the planet, we have:
\[ g_p = \frac{G M_p}{R_p^2} = \frac{G (8M_e)}{(2R_e)^2} = \frac{G (8M_e)}{4R_e^2} = 2 \frac{G M_e}{R_e^2} = 2g_e \]

### Final Answer
Thus, the value of acceleration due to gravity \( g_p \) on the surface of the planet is:
\[ g_p = 2g_e \]