Two plants A and B have the same average density . Their radii `RA and RB` are such that `R_A: R_B = 3 : 1` . If `gA and g_B` are the acceleration due to gravity at the surface of the planets , the `g_A : g_B` equals

Options

A. `3:1`
B. `1:3`
C. `1:9`
D. `sqrt3:1`

Answer

To solve the problem step by step, we need to find the ratio of the acceleration due to gravity on two planets, A and B, given that they have the same average density and their radii are in the ratio of \( R_A : R_B = 3 : 1 \).

### Step 1: Understanding the relationship between mass, density, and volume
The mass \( m \) of a planet can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ m = \rho \cdot V \]
For a spherical planet, the volume \( V \) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the mass of planet A can be expressed as:
\[ m_A = \rho \cdot \frac{4}{3} \pi R_A^3 \]
And the mass of planet B as:
\[ m_B = \rho \cdot \frac{4}{3} \pi R_B^3 \]

### Step 2: Expressing 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 universal gravitational constant.

For planet A:
\[ g_A = \frac{G m_A}{R_A^2} = \frac{G \left(\rho \cdot \frac{4}{3} \pi R_A^3\right)}{R_A^2} \]
This simplifies to:
\[ g_A = \frac{G \rho \cdot \frac{4}{3} \pi R_A}{1} \]

For planet B:
\[ g_B = \frac{G m_B}{R_B^2} = \frac{G \left(\rho \cdot \frac{4}{3} \pi R_B^3\right)}{R_B^2} \]
This simplifies to:
\[ g_B = \frac{G \rho \cdot \frac{4}{3} \pi R_B}{1} \]

### Step 3: Finding the ratio of \( g_A \) to \( g_B \)
Now we can find the ratio of the accelerations due to gravity:
\[ \frac{g_A}{g_B} = \frac{G \rho \cdot \frac{4}{3} \pi R_A}{G \rho \cdot \frac{4}{3} \pi R_B} = \frac{R_A}{R_B} \]
Since \( R_A : R_B = 3 : 1 \), we have:
\[ \frac{g_A}{g_B} = \frac{3}{1} \]

### Step 4: Conclusion
Thus, the ratio of the acceleration due to gravity on planets A and B is:
\[ g_A : g_B = 3 : 1 \]