Question

Discuss variation of $ g $ with
(A) Altitude
(B) Depth

Answer 1

Hint : Acceleration due to gravity varies differently with altitude and depth. For altitude, we simply put the height $ h $ in the formula and get the required result. But for its accurate value, we need to calculate the mass of the earth in terms of its density and then change the radius according to the depth.
Formula used:
Acceleration due to gravity is given as,
 $\Rightarrow g = \dfrac{{GM}}{{{R^2}}} $
where, $ G $ Universal gravitational constant, $ M $ is the mass of the earth, $ R $ is the radius of the earth.

Complete step by step answer
(a)
Let us calculate $ g $ at a height of $ h $ .
We know that,
 $\Rightarrow g = \dfrac{{GM}}{{{R^2}}} $ $ - - - - (1) $
At the height $ h $ , $ R $ changes to $ R + h $
 $\Rightarrow R \to R + h $
 $\Rightarrow {g_h} = \dfrac{{GM}}{{{{(R + h)}^2}}} $ $ - - - - (2) $
Where, $ {g_h} $ is the value of $ g $ at height $ h $ .
Dividing equation $ (2) $ by $ (1) $ , we get
 $\Rightarrow \dfrac{{{g_h}}}{g} = \dfrac{{{R^2}}}{{{{(R + h)}^2}}} $
 $\Rightarrow \dfrac{{{g_h}}}{g} = \dfrac{{{R^2}}}{{{R^2}{{(1 + \dfrac{h}{R})}^2}}} $
Cancelling out $ {R^2} $ , we get
 $\Rightarrow \dfrac{{{g_h}}}{g} = \dfrac{1}{{{{(1 + \dfrac{h}{R})}^2}}} $
 $\Rightarrow \dfrac{{{g_h}}}{g} = {(1 + \dfrac{h}{R})^{ - 2}} $
Using the binomial theorem, we can write is as
 $\Rightarrow \dfrac{{{g_h}}}{g} = 1 - \dfrac{{2h}}{R} $
 $\Rightarrow {g_h} = \left( {1 - \dfrac{{2h}}{R}} \right)g $
This is the required variation of $ g $ with an altitude $ h $ .
We can vary the value of height i.e. $ h $ and we will get a different value of $ {g_h} $ .
(b)
The density of earth is given as,
 $\Rightarrow \rho = \dfrac{M}{V} $
 $\Rightarrow M = \rho V $ $ - - - - (3) $
For earth, $ V $ can be written as
 $\Rightarrow V = \dfrac{4}{3}\pi {R^3} $
Using this in equation $ (3) $ , we get
 $\Rightarrow M = \rho \times \dfrac{4}{3}\pi {R^3} $
We use this value of $ M $ in equation $ (1) $
 $\Rightarrow g = \dfrac{{G \times \rho \times \dfrac{4}{3}\pi {R^3}}}{{{R^2}}} $
 $\Rightarrow g = G \times \rho \times \dfrac{4}{3}\pi R $ $ - - - - (5) $
Now, the value of $ g $ at a depth $ d $ is given as
 $\Rightarrow {g_d} = G \times \rho \times \dfrac{4}{3}\pi \left( {R - d} \right) $ $ - - - - (6) $
Dividing equation $ (6) $ by $ (5) $ , we get
 $\Rightarrow \dfrac{{{g_d}}}{g} = \dfrac{{G \times \rho \times \dfrac{4}{3}\pi \left( {R - d} \right)}}{{G \times \rho \times \dfrac{4}{3}\pi R}} $
 $\Rightarrow \dfrac{{{g_d}}}{g} = \dfrac{{\left( {R - d} \right)}}{R} $
We can write this as
 $\Rightarrow \dfrac{{{g_d}}}{g} = \left( {1 - \dfrac{d}{R}} \right) $
 $\Rightarrow {g_d} = \left( {1 - \dfrac{d}{R}} \right)g $
This is the required variation of $ g $ with depth.
In this case of measuring $ g $ at a depth $ d $ , we can vary the value of $ d $ and we will get different values of $ {g_d} $ .

Note
In the case of measuring $ g $ at an altitude of $ h $ , while using binomial theorem, we make an assumption that $ h < < R $ . Thus this formula holds true only for those cases where the height at which we are measuring $ g $ is negligible with respect to $ R $ .