**Gravitational potential** at a point P is the amount of work required to move a unit mass from infinity to the point P in a gravitational field. The gravitational potential of a particle of mass $M$ at a distance $r$ from it is given by
\begin{align}
V=-\frac{GM}{r}.
\end{align}
The unit of gravitational potential is J/kg. It is a scalar function.

The gravitational potential energy of a mass $m$ placed at a point P having potential $V$ is given by \begin{align} U=mV=-\frac{GMm}{r} \end{align}

A **gravitational field** of a point mass $M$ at a distance $r$ from it is given by
\begin{align}
E=-\frac{GM}{r^2}\hat{r}
\end{align}

**Problem (JEE Mains 2014):**
The gravitational field in a region is given by $\vec{g}=5\,\hat\imath+ 12\,\hat\jmath$ N/kg. The change in the gravitational potential energy of a particle of mass 2 kg when it is taken from the origin to a point (7, -3) metre is

- 71 J
- $13\sqrt{58}$ J
- -71 J
- 1 J

**Solution:**
Suppose a particle moves from a point A to another point B along some path in a gravitational field $\vec{g}$. The gravitational potential at B is given by,
\begin{align}
V_B&=V_{A}-\int_\text{path} \vec{g}\cdot\mathrm{d}\vec{l},\nonumber
\end{align}
where integration is carried out over the path taken by the particle. If $\vec{g}$ is constant then above equation can be written as,
\begin{align}
V_{B}=V_{A}-\vec{g}\cdot(\vec{r}_B-\vec{r}_A),\nonumber
\end{align}
where $\vec{r}_{A}$ and $\vec{r}_B$ are the position vectors of A and B.

The decrease in gravitational potential energy of a particle of mass $m=\unit{2}{\kilogram}$ when it moves from A(0,0) to B(7,-3) in gravitational field $\vec{g}=5\,\hat\imath+12\,\hat\jmath$ is, \begin{align} \Delta U & =m(V_A-V_B) \\ &=m\vec{g}\cdot(\vec{r}_B-\vec{r}_A)\nonumber\\ &=2(5\,\hat\imath+12\,\hat\jmath)\cdot(7\,\hat\imath-3\,\hat\jmath)\nonumber\\ &=2\;\mathrm{J}.\nonumber \end{align}

**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,

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

**Solution:**
The gravitational field due to a uniform solid sphere of radius $R$ and mass $M$ at a distance $r$ from its centre 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}

**Problem (JEE Mains 2020): **
Consider two solid spheres of radii $r_1 =1$ m, $r_2 = 2$ m and masses $m_1$ and $m_2$, respectively. The gravitational field due to sphere 1 and 2 are shown. The value of $m_1/m_2$ is

- 2/3
- 1/6
- 1/2
- 1/3

**Solution:**
The gravitational field on the surface of sphere 1 of radius $r_1=1$ m is,
\begin{align}
g_1&=\frac{Gm_1}{r_1^2} \\
&=\frac{Gm_1}{(1)^2}\\
&=Gm_1\\
&=2.\nonumber
\end{align}
The gravitational field on the surface of sphere 2 of radius $r_2=2$ m is,
\begin{align}
g_2 &=\frac{Gm_2}{r_2^2} \\
&=\frac{Gm_2}{(2)^2} \\
&=\frac{Gm_2}{4}\\
&=3.\nonumber
\end{align}
Divide above equations to get $m_1/m_2=1/6$.

**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.
\end{align}
Use conservation of energy, $U_i+K_i=U_f+K_f$, to get $h=R$.