Physics - Gravitation Question with Solution | TestHub

If$M$is the total mass of the system of particles, then similar to a satellite revolving around the earth, the orbital velocity of the test mass is given by

$v=\sqrt{\frac{GM}{r}}$

For$r\le R,\mathrm{ }\mathrm{ }v=\sqrt{\frac{G\cdot \frac{4}{3}\pi r^3{\rho }_0}{r}}$

i.e.$v\propto \sqrt{r^2}$or$v\propto r$

i.e.$v$increases linearly with$r$up to$r=R$

Thus$\left(1\right)$and$\left(4\right)$are wrong but$\left(2\right)$and$\left(3\right)$satisfy this.

$\therefore \text{For }r>R,v=\sqrt{\frac{GM}{r}}\text{or }v\propto \frac{1}{\sqrt{r}}$

Thus is correctly shown in figure$\left(2\right)$. Figure$\left(3\right)$is wrong because for$r>R,v$is shown as constant.