# Gravitational Potential and Field

## Problems from IIT JEE

Problem (IIT JEE 1997): A particle is projected vertically upwards from the surface of earth (radius $R$) with a kinetic energy equal to half of the minimum value needed for it to escape. The height to which it rises above the surface of earth is _______.

Solution: The expressions for escape velocity and minimum kinetic energy for a particle to escape from a planet of radius $R$ and mass $M$ are, \begin{align} &v_e=\sqrt{2GM/R},&K_\text{min}=\tfrac{1}{2}mv_e^2={GMm}/{R}.\nonumber \end{align} Initial potential energy and kinetic energy of the projectile are, \begin{align} & U_i=-\frac{GMm}{R}, &K_i=\frac{1}{2}K_\text{min}=\frac{GMm}{2R}.\nonumber \end{align} At the maximum height $h$, the velocity of the projectile is zero. The potential energy and kinetic energy of the projectile at maximum height are, \begin{align} U_f=-\frac{GMm}{R+h}, &&K_f=0.&&\nonumber \end{align} Use conservation of energy, $U_i+K_i=U_f+K_f$, to get $h=R$.

Problem (IIT JEE 1994): The magnitudes of the gravitational field at distance $r_1$ and $r_2$ from the centre of a uniform sphere of radius $R$ and mass $M$ are $F_1$ and $F_2$ respectively, then,

1. $\frac{F_1}{F_2}=\frac{r_1}{r_2}$ if $r_1 < R$ and $r_2 < R$
2. $\frac{F_1}{F_2}=\frac{r_{2}^2}{r_{1}^2}$ if $r_1 > R$ and $r_2 > R$
3. $\frac{F_1}{F_2}=\frac{r_{1}^3}{r_{2}^3}$ if $r_1 < R$ and $r_2 < R$
4. $\frac{F_1}{F_2}=\frac{r_{1}^2}{r_{2}^2}$ if $r_1 < R$ and $r_2 < R$

Solution: The gravitational field outside and inside a uniform solid sphere of radius $R$ and mass $M$ is given by, \begin{align} F=\begin{cases} \frac{GM}{R^3}r, & \text{if $r\leq R$;}\\ \frac{GM}{r^2}, & \text{if $r>R$.} \end{cases}\nonumber \end{align}