The magnitude of force of gravitational attraction between two point masses $m_1$ and $m_2$ separated by a distance $r$ is given by the equation

$F=\frac{G{m}_1{m}_2}{r^2}$

Here $G$ is the universal gravitational constant.

$G=6.67\times 10^{-11}\frac{\mathrm{N}\cdot {\mathrm{m}}^2}{{\mathrm{kg}}^2}$

Acceleration due to gravity ( $g$ ) near the surface of a planet of mass $M$ and radius $R$ is given as :

$g_{\mathrm{p}}=\frac{GM}{R^2}\text{. }$

As the height increases from the surface of the planet to $h$, it is given as :

$g^{\prime }\approx g_{\mathrm{p}}\left(1-\frac{2h}{R}\right),$

provided h << R$.

However, with the increase in depth from the surface of the planet to $h$, it is given as :

$g^{\prime }=g_p\left(1-\frac{h}{R}\right)$

Gravitational potential (V) due to a point mass M, at a distance R is given as :

$V=-\frac{GM}{R}$

Outside the shell

$\mathrm{V}=-\frac{\mathrm{GM}}{\mathrm{r}},\mathrm{r}>\mathrm{R}$

Inside/on the surface the shell

$\mathrm{V}=-\frac{\mathrm{GM}}{\mathrm{R}},\mathrm{r}\le \mathrm{R}$

Outside the sphere, $V=-\frac{GM}{r},r>R$

On the surface, $\mathrm{V}=-\frac{\mathrm{GM}}{\mathrm{R}},\mathrm{r}=\mathrm{R}$

Inside the sphere, $\mathrm{V}=-\frac{\mathrm{GM}\left(3{\mathrm{R}}^2-{\mathrm{r}}^2\right)}{2{\mathrm{R}}^3},\mathrm{r}<\mathrm{R}$

Potential energy due mutual gravitational interaction between particles of masses $m_1$ and $m_2$ separated by a distance $r$ is given by the following equation.

$U_r=-\frac{G{m}_1{m}_2}{r}$

When separation between the particles is infinitely large, the potential energy is arbitrarily assumed zero.

E_g $=\frac{Gm}{r^2}$ [Radially inwards]

${\mathrm{E}}_{\mathrm{g}}=\frac{\mathrm{Gm}}{{\left({\mathrm{R}}^2+{\mathrm{x}}^2\right)}^{3/2}}\mathrm{x}$ [Directed towards centre]

E_g is max at $\mathrm{x}=\pm \frac{\mathrm{R}}{\sqrt{2}}$

Outside the shell $E_g=\frac{GM}{r^2}$, where $r>R$

On the surface ${\mathrm{E}}_g=\frac{\mathrm{GM}}{{\mathrm{r}}^2}$, where $\mathrm{r}=\mathrm{R}$

Inside the shell ${\mathrm{E}}_{\mathrm{g}}=0$, where $\mathrm{r}<\mathrm{R}$

[Note : Direction always towards the centre of the sphere]

Outside the sphere $E_g=\frac{GM}{r^2}$, where $r>R$

On the surface $E_g=\frac{GM}{r^2}$, where $r=R$

Inside the sphere ${\mathrm{E}}_{\mathrm{g}}=\frac{\mathrm{GMr}}{{\mathrm{R}}^3}$, where $\mathrm{r}<\mathrm{R}$

${\mathrm{v}}_e=\sqrt{\frac{2\mathrm{GM}}{\mathrm{R}}}$

$v_0=\sqrt{\frac{GM}{r}}=\sqrt{\frac{GM}{(R+h)}}$

For nearby satellite ${\mathrm{v}}_0=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}=\frac{{\mathrm{v}}_e}{\sqrt{2}}$

Here $v_e=$ escape velocity on earth surface.

Potential energy : $U=-\frac{GMm}{r}$

Kinetic energy : $\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2=\frac{\mathrm{GMm}}{2\mathrm{r}}$

Mechanical energy : $E=U+K=-\frac{GMm}{2r}$

Binding energy : $\mathrm{BE}=-\mathrm{E}=\frac{\mathrm{GMm}}{2\mathrm{r}}$