Question

A solid sphere of radius $r$ is made of a material having variable density. Its density varies with distance $r$ from centre as $\rho = \frac{kr}{R}$, where $k$ is a constant.

The gravitational field due to this sphere at $r=\frac{R}{2}$ is $g_1$ and at $r=2R$ is $g_2$. Find the ratio $\frac{g_1}{g_2}$.

Answer 1

1

Answer 2

To find the ratio $\frac{g_1}{g_2}$, we need to calculate the gravitational field at $r = \frac{R}{2}$ and $r = 2R$.

1. Mass Enclosed (r ≤ R): The density is given by $\rho(s) = \frac{ks}{R}$. The mass enclosed within a radius $r$ is:

   $$M(r) = \int_0^r 4\pi s^2 \rho(s) \, ds = \int_0^r 4\pi s^2 \left(\frac{ks}{R}\right) \, ds = \frac{4\pi k}{R} \int_0^r s^3 \, ds = \frac{4\pi k}{R} \cdot \frac{r^4}{4} = \frac{\pi k r^4}{R}$$

2. Gravitational Field Inside (at $r = \frac{R}{2}$): Using $g = \frac{GM(r)}{r^2}$,

   $$g_1 = \frac{G\left(\frac{\pi k (\frac{R}{2})^4}{R}\right)}{(\frac{R}{2})^2} = \frac{G\left(\frac{\pi k R^4}{16R}\right)}{\frac{R^2}{4}} = \frac{G\left(\frac{\pi k R^3}{16}\right)}{\frac{R^2}{4}} = G \frac{\pi k R^3}{16} \cdot \frac{4}{R^2} = \frac{\pi G k R}{4}$$

3. Gravitational Field Outside (at $r = 2R$): The total mass of the sphere is:

   $$M(R) = \frac{\pi k R^4}{R} = \pi k R^3$$

   Thus, at $r = 2R$,

   $$g_2 = \frac{GM(R)}{(2R)^2} = \frac{G \pi k R^3}{4R^2} = \frac{\pi G k R}{4}$$

4. Ratio:

   $$\frac{g_1}{g_2} = \frac{\frac{\pi G k R}{4}}{\frac{\pi G k R}{4}} = 1$$

Therefore, the ratio $\frac{g_1}{g_2}$ is 1.