Question

A body stretches a spring by a particular length at the earth’s surface at the equator. At what height above the South Pole will it stretch the same spring by the same length. Assume the earth to be spherical.

Answer 1

Hint:The acceleration due to gravity is different at different positions on the earth surface and it gets changed if the acceleration due to gravity is checked above the surface of earth or below the surface of earth.

Formula used:The formula of acceleration due to gravity at equator is given by,
$ \Rightarrow {g_e} = g - {\omega ^2} \cdot R$
Where the acceleration on the equator is ${g_e}$ the acceleration due to gravity at normal earth surface is $g$ the angular velocity is $\omega $ and the radius of curvature of the earth is$R$.
The formula of acceleration due to gravity at some height is given by,
$ \Rightarrow {g_h} = g\left( {1 - \dfrac{{2h}}{R}} \right)$
Where the acceleration due to gravity at height h is ${g_h}$ the acceleration at normal earth surface is $g$ the height from the earth surface is h and the radius of curvature is R.

Complete step by step solution:
It is given in the problem that a body stretches a spring by a particular length at the earth’s surface at equator and we need to find the height above the South Pole at which the same spring can be stretched at some length if earth is considered spherical.
The formula of acceleration due to gravity at equator is given by,
$ \Rightarrow {g_e} = g - {\omega ^2} \cdot R$
Where the acceleration on the equator is ${g_e}$ the acceleration due to gravity at normal earth surface is $g$ the angular velocity is $\omega $ and the radius of curvature of the earth is$R$.
The acceleration due to gravity at the equator is equal to,
$ \Rightarrow {g_e} = g - {\omega ^2} \cdot R$………eq. (1)
The formula of acceleration due to gravity at some height is given by,
$ \Rightarrow {g_h} = g\left( {1 - \dfrac{{2h}}{R}} \right)$
Where the acceleration due to gravity at height h is ${g_h}$ the acceleration at normal earth surface is $g$ the height from the earth surface is h and the radius of curvature is R.
The acceleration due to gravity at some height from the earth surface is equal to,
$ \Rightarrow {g_h} = g\left( {1 - \dfrac{{2h}}{R}} \right)$………eq. (2)
The acceleration due to gravity at the equator should be equal to acceleration due to gravity at some height in order to get the same force on the spring.
Equating equation (1) and equation (2).
$ \Rightarrow {g_e} = {g_h}$
$ \Rightarrow g - {\omega ^2} \cdot R = g\left( {1 - \dfrac{{2h}}{R}} \right)$
$ \Rightarrow 1 - \dfrac{{{\omega ^2} \cdot R}}{g} = 1 - \dfrac{{2h}}{R}$
$ \Rightarrow \dfrac{{{\omega ^2} \cdot R}}{g} = \dfrac{{2h}}{R}$
$ \Rightarrow \dfrac{{2h}}{R} = \dfrac{{{\omega ^2} \cdot R}}{g}$
$ \Rightarrow h = \dfrac{{{\omega ^2} \cdot {R^2}}}{{2 \cdot g}}$………eq. (3)
Let us calculate the angular velocity of the earth.
$ \Rightarrow \omega = \dfrac{{2\pi }}{T}$
Where $\omega $ is angular velocity and $T$ is the time taken for one complete revolution.
The time taken for the one complete revolution is equal to,
$ \Rightarrow T = 24 \times 60 \times 60$
$ \Rightarrow T = 86400\sec \cdot $
The angular velocity will be equal to,
$ \Rightarrow \omega = \dfrac{{2\pi }}{T}$
$ \Rightarrow \omega = \dfrac{{2\pi }}{{86400}}$
$ \Rightarrow \omega = 7 \cdot 3 \times {10^{ - 5}}{s^{ - 1}}$………eq. (4)
Replacing the value of equation (4) in equation (3) we get.
$ \Rightarrow h = \dfrac{{{\omega ^2} \cdot {R^2}}}{{2 \cdot g}}$
$ \Rightarrow h = \dfrac{{{{\left( {7 \cdot 3 \times {{10}^{ - 5}}} \right)}^2} \cdot {R^2}}}{{2 \cdot g}}$
The value of the radius of earth surface is $R = 6 \cdot 4 \times {10^6}m$ and the value of$g = 9 \cdot 8m{s^{ - 2}}$.
$ \Rightarrow h = \dfrac{{{{\left( {7 \cdot 3 \times {{10}^{ - 5}}} \right)}^2} \cdot {R^2}}}{{2 \cdot g}}$
$ \Rightarrow h = \dfrac{{{{\left( {7 \cdot 3 \times {{10}^{ - 5}}} \right)}^2} \times {{\left( {6 \cdot 4 \times {{10}^6}m} \right)}^2}}}{{\left( {2 \times 9 \cdot 8} \right)}}$
$ \Rightarrow h = \dfrac{{53 \cdot 29 \times {{10}^{ - 10}} \times 40 \cdot 96 \times {{10}^{12}}}}{{19 \cdot 6}}$
$ \Rightarrow h = \dfrac{{2182 \cdot 76 \times {{10}^2}}}{{19 \cdot 6}}$
$ \Rightarrow h = 11136 \cdot 5m$
$ \Rightarrow h \approx 10km$.

The height from the surface of the earth at which the spring force is equal to the spring force at equator is equal to $h \approx 10km$.

Note:It is advisable for students to understand and remember the formula of the acceleration due to gravity at equator and the formula of acceleration due to gravity at some height from the earth surface. The acceleration due to gravity decreases with increase of height from the earth surface.