A planet has a mass of eight time the mass of earth and denisity is also equal to eight times a the average density of the earth. If g be the acceleration due to earth's gravity on its surface, then acceleration due to gravity on planet's surface will be

Options

A. 2g
B. 4g
C. 8g
D. 16g

Answer

To find the acceleration due to gravity on the surface of a planet that has a mass and density both eight times that of Earth, we can follow these steps:

### Step 1: Define Given Values
Let:
- \( M_e \) = Mass of Earth
- \( D_e \) = Density of Earth
- \( M_p = 8M_e \) = Mass of the planet
- \( D_p = 8D_e \) = Density of the planet

### Step 2: Relate Density to Mass and Volume
The density \( D \) is defined as:
\[ D = \frac{M}{V} \]
For a spherical object, the volume \( V \) is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Thus, we can express the densities as:
\[ D_p = \frac{M_p}{V_p} \quad \text{and} \quad D_e = \frac{M_e}{V_e} \]

### Step 3: Write the Volume in Terms of Radius
For the planet and Earth, we can write:
\[ D_p = \frac{M_p}{\frac{4}{3} \pi r_p^3} \quad \text{and} \quad D_e = \frac{M_e}{\frac{4}{3} \pi r_e^3} \]

### Step 4: Substitute Known Values
Substituting the values for \( M_p \) and \( D_p \):
\[ 8D_e = \frac{8M_e}{\frac{4}{3} \pi r_p^3} \]
And for Earth:
\[ D_e = \frac{M_e}{\frac{4}{3} \pi r_e^3} \]

### Step 5: Set Up the Equation
Equating the two expressions for density:
\[ \frac{8M_e}{\frac{4}{3} \pi r_p^3} = 8 \cdot \frac{M_e}{\frac{4}{3} \pi r_e^3} \]

### Step 6: Simplify the Equation
Cancel out common terms:
\[ \frac{8M_e}{r_p^3} = \frac{8M_e}{r_e^3} \]
This implies:
\[ r_p^3 = r_e^3 \implies r_p = r_e \]

### Step 7: Use the Formula for Acceleration due to Gravity
The formula for acceleration due to gravity \( g \) is given by:
\[ g = \frac{GM}{r^2} \]
For Earth:
\[ g_e = \frac{GM_e}{r_e^2} \]
For the planet:
\[ g_p = \frac{GM_p}{r_p^2} \]

### Step 8: Substitute Known Values
Substituting \( M_p = 8M_e \) and \( r_p = r_e \):
\[ g_p = \frac{G(8M_e)}{(r_e)^2} = 8 \cdot \frac{GM_e}{r_e^2} = 8g_e \]

### Conclusion
Thus, the acceleration due to gravity on the surface of the planet is:
\[ g_p = 8g \]