Question

Determine the acceleration due to gravity at a height of $h=2R_E$ from the earth's surface ? Take acceleration due to gravity at earth's surface to be $10{\text{ms}}^{-2}$.

Answer 1

Acceleration due to gravity:

1. It is the acceleration experienced by an object under the gravitational attraction of a planet.
2. It depends on the mass and radius of the planet and hence it is different for different planets.

Derivation:

The acceleration due to gravity on the surface of Earth is given by,

$g=\frac{GM}{{R_E}^2}\to \left(1\right)$

G = gravitational constant ($6.67\times {10}^{-11}\raisebox{1ex}{${\text{Nm}}^2$}\!\left/ \!\raisebox{-1ex}{${\text{kg}}^2$}\right.$)

M = Mass of Earth

$R_E$ = Radius of Earth.

Assume acceleration due to gravity at the earth's surface, $g=10{\text{ms}}^{-2}$

Let $g\text{'}$be the acceleration due to gravity at a height $\text{'}h\text{'}$from Earth's surface.

So, $g\text{'}=\frac{GM}{{(R_E+h)}^2}\to \left(2\right)$

In comparing (1) and (2),

$\begin{array}{rcl}\frac{g\text{'}}{g}& =& \frac{{\displaystyle \raisebox{1ex}{$GM$}\!\left/ \!\raisebox{-1ex}{${(R_E+h)}^2$}\right.}}{{\displaystyle \raisebox{1ex}{$GM$}\!\left/ \!\raisebox{-1ex}{${R_E}^2$}\right.}}\\ & =& \frac{{R_E}^2}{{(R_E+h)}^2}\end{array}$

Thus, we can write

$\begin{array}{rcl}g\text{'}& =& g\frac{{R_E}^2}{{(R_E+h)}^2}\\ & =& \frac{g}{{(1+{\displaystyle \raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$R_E$}\right.})}^2}\end{array}$

At $h=2R_E$

,$\begin{array}{rcl}g\text{'}& =& \frac{g}{{(1+{\displaystyle \raisebox{1ex}{$2R_E$}\!\left/ \!\raisebox{-1ex}{$R_E$}\right.})}^2}\\ & =& \frac{g}{3^2}\\ & =& \frac{g}{9}\\ & =& \frac{10}{9}{\text{ms}}^{-2}\end{array}$

Hence, the acceleration due to gravity at a height of $h=2R_E$ from the earth's surface is $\frac{\begin{array}{r}10\end{array}}{9}{\text{ms}}^{-2}$.