Question

Above the Earth’s surface, the variation of $g$ w.r.t the height $\left( r \right)$ is correctly represented by which of the following proportionalities
(A) $g \propto \dfrac{1}{{{r^2}}}$
(B) $g \propto r$
(C) $g \propto {r^2}$
(D) $g \propto {r^0}$

Answer 1

Hint
We need to first establish a relationship between the acceleration due to gravity and the universal gravitational constant to understand the variation of $g$with height from the surface of the earth $r$.
$\Rightarrow \dfrac{{{g'}}}{g} = \dfrac{{{R^2}}}{{{{\left( {R + r} \right)}^2}}}$ where $g$ and $g'$ are the accelerations due to gravity on the earth’s surface and at height $r$ respectively.
$\Rightarrow g = \dfrac{{GM}}{{{R^2}}}$ where $g$ is the acceleration due to gravity, $G$ is the universal gravitational constant, $R$ is the radius of the earth and $M$ is the mass of the earth.

Complete step by step answer
The acceleration due to gravity or simply g is the acceleration produced in a body due to the attractive forces of the earth, also known as gravitational force. As the earth rotates about its own axis, different bodies at different latitudes move with different velocities and so, bodies near the equator that are moving with greater linear velocities will experience greater centrifugal force as compared to the ones near the poles. Thus g at different latitudes is affected.
The acceleration due to gravity is measured by the acceleration due to the earth’s attractive force minus the centrifugal acceleration due to the earth’s rotation. Its value is also affected by other factors such as altitude, depth and latitude.
Here, the question asks us to express the relation between g and height from the earth’s surface, also known as altitude $\left( r \right)$
So, let us consider $g$ and ${g'}$ to be the acceleration due to gravity on the earth’s surface and at height $r$ respectively. Then, we have
$\Rightarrow \dfrac{g}{{{g'}}} = \dfrac{{1/{R^2}}}{{1/{{\left( {R + r} \right)}^2}}}$
$\Rightarrow \dfrac{g}{{{g'}}} = \dfrac{{{R^2} + 2Rr + {r^2}}}{{{R^2}}} = \left( {1 + \dfrac{{2r}}{R}} \right)$
Where $R$ is the radius of the earth
$\Rightarrow \dfrac{{{g'}}}{g} = {\left( {1 + \dfrac{{2r}}{R}} \right)^{ - 1}} = \left( {1 - \dfrac{{2r}}{R}} \right)$ (approx.)
$\Rightarrow {g'} = g\left( {1 - \dfrac{{2r}}{R}} \right)$
So, from this expression we can see that as the height from the earth’s surface increases, the acceleration due to gravity decreases.
Now we can also write $\dfrac{{{g'}}}{g} = \dfrac{{{R^2}}}{{{{\left( {R + r} \right)}^2}}}$----equation $\left( 1 \right)$
Now, we know $gR = \dfrac{{GM}}{R}$ where $G$ is called the universal gravitational constant and $M$ is the mass of earth.
Therefore, $g = \dfrac{{GM}}{{{R^2}}}$
Substituting this value in the equation $\left( 1 \right)$ we get,
$\Rightarrow g' = \dfrac{{GM}}{{{{\left( {R + r} \right)}^2}}}$
So, from this equation we can show that $g' \propto \dfrac{1}{{{r^2}}}$
Therefore, the correct option is (A).

Note
As we see from the equation $g = \dfrac{{GM}}{{{R^2}}}$, the acceleration due to gravity depends upon the mass and radius of earth. This means that the value of $g$ will vary for different planets having different mass and radius. In the equation $g' = \dfrac{{GM}}{{{{\left( {R + r} \right)}^2}}}$ the value of $G$, $M$and $R$ are so overpowering that the value of $r$ hardly makes any difference. That is why the value of $g$ for all calculations is taken to be $9.8m{s^{ - 2}}$.