Question

Calculate the acceleration due to gravity at a point
a] \[64km\]above and
b] \[32km\]below the surface of earth
given radius of earth as \[R = 6400km\] . Acceleration due to gravity at the surface of earth \[9.8m/{s^2}\].

Answer 1

Hint:As we know the formulas of calculating acceleration due to gravity at height and depth related to earth’s surface are \[{g_h} = \dfrac{{{g_e}}}{{{{\left( {1 + \dfrac{h}{R}} \right)}^2}}}\]and \[{g_d} = \dfrac{{{g_e}}}{{{{\left( {1 - \dfrac{d}{R}} \right)}^{ - 1}}}}\]so substitute the value in the formulas and get your answer.

Complete step-by-step solution:
As in the given question we are given with radius of earth as, \[R = 6400km\]
And we are also given with earth's acceleration due to gravity at earth’s surface as, \[9.8m/{s^2}\]
So to calculate acceleration due to gravity at height \[64km\]above the earth’s surface we will use
\[{g_h} = \dfrac{{{g_e}}}{{{{\left( {1 + \dfrac{h}{R}} \right)}^2}}}\], where height is given in the question.
\[{g_h} = \dfrac{{9.8}}{{{{\left( {1 + \dfrac{{64}}{{6400}}} \right)}^2}}}\], and we know the value of acceleration due to gravity is constant on earth's surface but can vary to some places on earth but in this question value is written above.
\[{g_h} = \dfrac{{9.8}}{{{{\left( {1 + \dfrac{1}{{100}}} \right)}^2}}}\]
\[{g_h} = \dfrac{{9.8}}{{{{\left( {1.01} \right)}^2}}}\]
\[{g_h} = 9.606\], now to calculate the acceleration due to gravity\[32km\] below the surface of earth as,
\[{g_d} = \dfrac{{{g_e}}}{{{{\left( {1 - \dfrac{d}{R}} \right)}^{ - 1}}}}\], where depth is known to us
\[{g_d} = \dfrac{{9.8}}{{{{\left( {1 - \dfrac{{32}}{{6400}}} \right)}^{ - 1}}}}\]
\[{g_d} = \dfrac{{9.8}}{{{{\left( {1 - \dfrac{1}{{200}}} \right)}^{ - 1}}}}\]
\[{g_d} = \dfrac{{9.8}}{{{{\left( {0.995} \right)}^{ - 1}}}}\]
\[{g_d} = 9.8 \times 0.995\]
\[{g_d} = 9.751\]
Therefore we have calculated acceleration due to gravity art height and depth.

Note:- As in the given question we have used value of \[g = 9.8\]which is assumed to be constant to solve questions but more precisely it has different values on different places of earth as earth is not perfect sphere so it has different values at different parts of earth. The value of acceleration due to gravity is least at equator and maximum at the poles.