If the density of the planet is double that of the earth and the radius 1.5 times that of the earth, the acceleration due to gravity on the planet is

Options

A. `3/4` times that on the surface of the earth
B. 3 times that on the surface of the earth
C. `4/3` times that on the surface of the earth
D. 6 times that on the surface of the earth

Answer

To find the acceleration due to gravity on a planet with a density double that of Earth and a radius 1.5 times that of Earth, we can follow these steps:

### Step 1: Understand the formula for acceleration due to gravity
The acceleration due to gravity \( g \) on the surface of a planet is given by the formula:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

### Step 2: Relate the mass of the planet to its density
The mass \( M \) of a planet can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho V \]
For a sphere, the volume \( V \) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the mass of the planet can be written as:
\[ M = \rho \left(\frac{4}{3} \pi R^3\right) \]

### Step 3: Substitute the mass into the gravity formula
Substituting the expression for mass into the gravity formula gives:
\[ g = \frac{G \left(\rho \frac{4}{3} \pi R^3\right)}{R^2} \]
This simplifies to:
\[ g = \frac{4}{3} \pi G \rho R \]

### Step 4: Determine the density and radius of the new planet
Given:
- The density of the new planet \( \rho_p = 2\rho_e \) (where \( \rho_e \) is the density of Earth)
- The radius of the new planet \( R_p = 1.5 R_e \) (where \( R_e \) is the radius of Earth)

### Step 5: Substitute these values into the gravity formula
Now substituting \( \rho_p \) and \( R_p \) into the gravity formula:
\[ g_p = \frac{4}{3} \pi G (2\rho_e) (1.5 R_e) \]
This simplifies to:
\[ g_p = \frac{4}{3} \pi G \cdot 2\rho_e \cdot 1.5 R_e \]
\[ g_p = 2 \cdot 1.5 \cdot \frac{4}{3} \pi G \rho_e R_e \]
\[ g_p = 3 \cdot \frac{4}{3} \pi G \rho_e R_e \]

### Step 6: Relate this to Earth's gravity
Since \( g_e = \frac{4}{3} \pi G \rho_e R_e \), we can express \( g_p \) in terms of \( g_e \):
\[ g_p = 3g_e \]

### Conclusion
Thus, the acceleration due to gravity on the planet is:
\[ g_p = 3g_e \]

### Final Answer
The acceleration due to gravity on the planet is three times that on the surface of the Earth.

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