4.1 Gravitational potential due to a point mass

1.1 Characteristics of gravitational force
1.2 Universal gravitational constant
1.3 Principle of superposition of gravitation
1.4 Gravity
1.5 Acceleration due to gravity
1.6 Relation between $g$ and $G$

2.1 Variation of acceleration due to gravity $(g)$ due to shape of the earth
2.2 Variation of acceleration due to gravity $(g)$ due to altitude
2.3 Variation of acceleration due to gravity $(g)$ due to depth
2.4 Variation of acceleration due to gravity $(g)$ due to rotation of earth

3.1 Gravitational field due to a point mass
3.2 Gravitational field due to a uniform solid sphere
3.3 Gravitational field due to a uniform spherical shell
3.4 Gravitatioal field due to a uniform circular ring at a point on its axis

4.1 Gravitational potential due to a point mass
4.2 Gravitational potential due to a uniform solid sphere
4.3 Gravitational potential due to a uniform thin spherical shell
4.4 Gravitational potential due to a uniform ring at a point on its centre
4.5 Relation between gravitational field and gravitational potential

5.1 Gravitational potential energy for a system of particles
5.2 Gravitational potential energy of a body on earth's surface

6.1 Orbital speed of satellite
6.2 Time period of a satellite
6.3 Angular momentum of a satellite
6.4 Energy of a satellite
6.5 Types of satellite
6.6 Binding energy
6.7 Escape velocity
6.8 Weightlessness

7.1 Kepler's first law
7.2 Kepler's second law
7.3 Kepler's third law

Suppose a point mass $M$ is situated at a point $O$ , then the gravitational force on mass $m$ at point $P$ which is at a distance $r$ is,

$$F = \frac{{GMm}}{{{r^2}}}$$

Work done for displacement $dr$ is, $$dW = \overrightarrow F .\,d\overrightarrow r $$$$dW = \frac{{GMm}}{{{r^2}}}dr\,\cos \pi $$$$dW = - \frac{{GMm}}{{{r^2}}}dr$$

As we know, $$dV = - \frac{{dW}}{m}$$

Gravitational potential $(V)$ is the amount of work done in bringing a unit mass from infinity to point $P$, $$dV = \frac{{GM}}{{{r^2}}}dr$$

Integrating with proper limits we get,

$$\int\limits_{{V_\infty }}^V {dV} = \int\limits_\infty ^r {\frac{{GM}}{{{r^2}}}dr} $$$$\left[ V \right]_{{V_\infty }}^V = GM\left[ { - \frac{1}{r}} \right]_\infty ^r$$$$V - {V_\infty } = - GM\left[ {\frac{1}{r} - \frac{1}{\infty }} \right]$$$$V = - \frac{{GM}}{r}$$

As we cannot define the absolute potential at any point. So, we take reference at infinity stating ${V_\infty } = 0$.