3.1 Gravitational field 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 placed at point $O$, then the gravitational field $\left( {\overrightarrow E } \right)$ at a distance $\overrightarrow r $ from point $O$ is,

$$\overrightarrow E = - \frac{{GM}}{{{r^2}}}\widehat r$$ or $$\overrightarrow E = - \frac{{GM}}{{{r^3}}}\overrightarrow r $$ or $$E = \frac{{GM}}{{{r^2}}}$$

The direction of the force $\overrightarrow F $ is from $P$ to $O$. Therefore, the direction of gravitational field $\overrightarrow E $ is also from $P$ to $O$.

Let the force on mass $m$ by point mass $M$ is, $$F = \frac{{GMm}}{{{r^2}}}$$ or $$\overrightarrow F = - \frac{{GMm}}{{{r^2}}}\widehat r$$

So, the gravitational field is given by, $$\overrightarrow E = \frac{{\overrightarrow F }}{m}$$ or $$\overrightarrow E = - \frac{{GM}}{{{r^2}}}\widehat r$$