Gravitation MCQs for NEET — Physics Questions with Answers

Orbital velocity of satellite v=$\sqrt{\frac{GM}{r}}$

$\Rightarrow \frac{v_A}{v_B}=\sqrt{\frac{r_B}{r_A}}=\sqrt{\frac{R}{4R}}=\frac{1}{2}$

$\therefore \frac{v_A}{v_B}=\frac{3v}{v_B}=\frac{1}{2}$

$\therefore v_B=6v$

$$\begin{gathered} \mathrm{Total} \mathrm{KE}\hspace{0.17em}\mathrm{required} \mathrm{to} \mathrm{change} \mathrm{the} \mathrm{orbit} \mathrm{will} \mathrm{be} \mathrm{equal} \mathrm{to} \mathrm{change} \mathrm{in} \mathrm{total} \mathrm{energy} \mathrm{as} \mathrm{PE} \mathrm{as} \mathrm{well} \mathrm{as} \mathrm{KE}, \mathrm{both} \mathrm{are} \mathrm{changing}. \\ ∆\mathrm{E}=\left(-\frac{GMm}{2R_2}\right)-\left(-\frac{GMm}{2R_1}\right) \\ {\mathrm{KE}}_{\mathrm{extra}}=\frac{GMm}{2}\left[\frac{1}{R_1}-\frac{1}{R_2}\right] \end{gathered}$$

According to the NCERT text, 'The force of attraction between a hollow spherical shell of uniform density and a point mass situated outside is just as if the entire mass of the shell is concentrated at the centre of the shell.' Therefore, the force is $G \frac{Mm}{r^2}$.

As stated in the NCERT text, 'The force of attraction due to a hollow spherical shell of uniform density, on a point mass situated inside it is zero.' This is because the gravitational forces from various regions of the shell cancel each other completely.

The NCERT text states, 'If a particle is inside a homogeneous solid sphere, the force on the particle acts toward the centre of the sphere. This force is exerted by the spherical mass interior to the particle.'

The NCERT text highlights, 'The gravitational for ce on a particle inside a spherical shell is zer o. However, (unlike a metallic shell which shields electrical forces) the shell does not shield other bodies outside it from exerting gravitational forces on a particle inside. Gravitational shielding is not possible.'

As per the NCERT text, 'For a spherically symmetric body however the force on a particle external to the body is as if the mass is concentrated at the centre and this force is therefore central.'

The NCERT text explains, 'The force on m due to the outer shell of thickness d is zero because the result quoted in the previous section. As far as the smaller sphere of radius $(R_E - d)$ is concerned, the point mass is outside it and hence according to the result quoted earlier, the for ce due to this smaller sphere is just as if the entire mass of the smaller sphere is concentrated at the centre.'

The force on mass m at P (distance r from center) inside a homogeneous solid sphere is given by $F = G \frac{M_r m}{r^2}$. Since density is uniform, $M_r = \frac{4}{3}\pi r^3 \rho$. Also, $M_E = \frac{4}{3}\pi R_E^3 \rho$. So, $M_r = M_E \frac{r^3}{R_E^3}$. Substituting this into the force equation gives $F = G \frac{M_E m r}{R_E^3}$. Therefore, F is proportional to r.

The NCERT states, 'We have to add up these forces vectorially for all the point masses in the extended object to get the total force.' Also, 'From the principle of superposition each force acts independently and uninfluenced by the other bodies. The resultant force $F_R$ is then found by vector addition'.