The radii of two planets A and B are R and 4R and their densities are $$\rho$$ and $$\frac{\rho}{3}$$ respectively. The ratio of acceleration due to gravity at their surfaces $$g_A : g_B$$ will be

The acceleration due to gravity at the surface of a planet is given by:

$$g = \frac{GM}{R^2}$$

Since $$M = \frac{4}{3}\pi R^3 \rho$$, we get:

$$g = \frac{G \times \frac{4}{3}\pi R^3 \rho}{R^2} = \frac{4}{3}\pi G R \rho$$

So $$g \propto R\rho$$.

For Planet A: Radius = $$R$$, Density = $$\rho$$

$$g_A = \frac{4}{3}\pi G R \rho$$

For Planet B: Radius = $$4R$$, Density = $$\frac{\rho}{3}$$

$$g_B = \frac{4}{3}\pi G (4R) \left(\frac{\rho}{3}\right) = \frac{4}{3}\pi G \times \frac{4R\rho}{3}$$

The ratio:

$$\frac{g_A}{g_B} = \frac{R\rho}{\frac{4R\rho}{3}} = \frac{R\rho \times 3}{4R\rho} = \frac{3}{4}$$

Therefore, $$g_A : g_B = 3 : 4$$.