Question

At what height above the Earth’s surface would the Earth’s gravitational field strength be equal to $7.5\,N{\text{ }}k{g^{ - 1}}$ ?

Answer 1

Hint: To solve this question, one who knows the formula can easily solve this question. Here we will simply apply the formula and we will get the required solution.In addition we have to remember the values of G and radius of the earth and then put all the values given in the formula and after solving this we will get the required answer.

Formula used:
$g' = \dfrac{{GM}}{{{{\left( {R + h} \right)}^2}}}$
Where, $g'$ is the acceleration due to gravity above the surface of earth, $G = {6.67310^{ - 11}}$, $R$ is the radius of the earth and $h$ is the height above the earth surface.

Complete step by step answer:
According to the question, acceleration due to gravity above the surface of earth, $g' = 7.5\,N{\text{ }}k{g^{ - 1}}$.
And we know that the radius of the earth is, $R = {6.38106^6}$
And we know that the acceleration due to gravity above the earth's surface is given by,
$g' = \dfrac{{GM}}{{{{\left( {R + h} \right)}^2}}}$
Now, simply substituting all the given data in above formula we get,
\[g' = \dfrac{{GM}}{{{{\left( {R + h} \right)}^2}}} \\
\Rightarrow R + h = \sqrt {\dfrac{{GM}}{{g'}}} \\
\Rightarrow R + h = \dfrac{{6.67 \times {{10}^{ - 11}} \times 5.98 \times {{10}^{24}}}}{{7.5}} \\
\Rightarrow R + h = 7.3 \times {10^6}m \\ \]
And now final answer,
\[R + h = 7.3 \times {10^6}m \\
\Rightarrow h = 7.3 \times {10^6} - 6.4 \times {10^6} \\
\Rightarrow h = 900000\,m \\
\therefore h = 900\,km \\ \]
Hence, the height is $900\,km$ above the earth’s surface at which the gravitational field strength will be $7.5\,N{\text{ k}}{{\text{g}}^{ - 1}}$.

Note: The force of gravity on the equator is smaller than the force an item experiences at the poles, which is an intriguing fact concerning acceleration due to gravity. Gravitational acceleration is greatest at the poles. This occurs because $g$ has an inverse square dependence on $R$, which is the distance from the earth's centre. As $R$ changes, so does the value of $g$.