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

Explanation

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 \( g \) at the surface of a planet is given by the formula:
\[ g = \frac{G M}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

### Step 2: Define the variables for Earth and the planet
Let:
- \( R_E \) = radius of Earth
- \( M_E \) = mass of Earth
- \( \rho_E \) = density of Earth

For the new planet:
- The radius \( R_P = 1.5 R_E \)
- The density \( \rho_P = 2 \rho_E \)

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

### Step 4: Substitute the values for density and radius
Substituting the values for the planet:
\[ M_P = (2 \rho_E) \cdot \frac{4}{3} \pi (1.5 R_E)^3 \]
Calculating \( (1.5 R_E)^3 \):
\[ (1.5 R_E)^3 = 3.375 R_E^3 \]
Thus,
\[ M_P = (2 \rho_E) \cdot \frac{4}{3} \pi (3.375 R_E^3) = 9 \rho_E \cdot \frac{4}{3} \pi R_E^3 \]
Since \( M_E = \rho_E \cdot \frac{4}{3} \pi R_E^3 \), we have:
\[ M_P = 9 M_E \]

### Step 5: Substitute mass and radius into the gravity formula
Now substituting \( M_P \) and \( R_P \) into the gravity formula:
\[ g_P = \frac{G M_P}{R_P^2} = \frac{G (9 M_E)}{(1.5 R_E)^2} \]
Calculating \( (1.5 R_E)^2 \):
\[ (1.5 R_E)^2 = 2.25 R_E^2 \]
Thus,
\[ g_P = \frac{G (9 M_E)}{2.25 R_E^2} = \frac{9}{2.25} \cdot \frac{G M_E}{R_E^2} \]
Since \( \frac{G M_E}{R_E^2} = g_E \) (acceleration due to gravity on Earth):
\[ g_P = \frac{9}{2.25} g_E = 4 g_E \]

### Conclusion
The acceleration due to gravity on the planet is:
\[ g_P = 4 g_E \]