Two planets A and B of radii R and `1.5 R` have densities `rho` and `rho/2` respectively .The radio of acceleration due to gravity at the surface of B to A is :

Options

A. `2:1`
B. `4:3`
C. `2:3`
D. `3:4`

Answer

To find the ratio of acceleration due to gravity at the surface of planets B and A, we can use the formula for gravitational acceleration:

\[ g = \frac{GM}{R^2} \]

where \( g \) is the acceleration due to gravity, \( G \) is the universal gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

### Step 1: Calculate the mass of each planet

The mass \( M \) of a planet can be expressed in terms of its density \( \rho \) and volume \( V \):

\[ M = \rho V \]

The volume \( V \) of a sphere is given by:

\[ V = \frac{4}{3} \pi R^3 \]

Thus, the mass of planet A (\( M_A \)) is:

\[ M_A = \rho \cdot \frac{4}{3} \pi R^3 \]

And the mass of planet B (\( M_B \)), with density \( \frac{\rho}{2} \) and radius \( 1.5R \), is:

\[ M_B = \left(\frac{\rho}{2}\right) \cdot \frac{4}{3} \pi (1.5R)^3 \]

### Step 2: Simplify the mass of planet B

Calculating the volume for planet B:

\[ M_B = \left(\frac{\rho}{2}\right) \cdot \frac{4}{3} \pi (1.5R)^3 = \left(\frac{\rho}{2}\right) \cdot \frac{4}{3} \pi \cdot (3.375R^3) = \frac{4 \cdot 3.375 \pi \rho R^3}{6} = \frac{13.5 \pi \rho R^3}{6} = \frac{2.25 \pi \rho R^3}{1} \]

### Step 3: Calculate the acceleration due to gravity for both planets

Now we can calculate \( g_A \) and \( g_B \):

\[ g_A = \frac{G M_A}{R^2} = \frac{G \left(\rho \cdot \frac{4}{3} \pi R^3\right)}{R^2} = \frac{4 \pi G \rho R}{3} \]

\[ g_B = \frac{G M_B}{(1.5R)^2} = \frac{G \left(\frac{\rho}{2} \cdot \frac{4}{3} \pi (1.5R)^3\right)}{(1.5R)^2} \]

Calculating \( g_B \):

\[ g_B = \frac{G \left(\frac{\rho}{2} \cdot \frac{4}{3} \pi \cdot 3.375 R^3\right)}{2.25 R^2} = \frac{G \cdot \frac{4 \cdot 3.375 \pi \rho R^3}{6}}{2.25 R^2} = \frac{4 \cdot 3.375 \pi G \rho R}{13.5} \]

### Step 4: Find the ratio \( \frac{g_B}{g_A} \)

Now we find the ratio of \( g_B \) to \( g_A \):

\[ \frac{g_B}{g_A} = \frac{\frac{4 \cdot 3.375 \pi G \rho R}{13.5}}{\frac{4 \pi G \rho R}{3}} = \frac{3.375}{13.5} \cdot \frac{3}{1} = \frac{3.375 \cdot 3}{13.5} = \frac{10.125}{13.5} \]

Simplifying this gives:

\[ \frac{g_B}{g_A} = \frac{3}{4} \]

### Final Answer

Thus, the ratio of acceleration due to gravity at the surface of planet B to that at planet A is:

\[ \frac{g_B}{g_A} = \frac{3}{4} \]