Question

Calculate the value of the acceleration due to gravity at a place 3,200km above the surface of the earth.

Answer 1

Hint: Acceleration due to gravity varies with height. Here, to solve this, a gravitational formula will be used in which the radius of the earth and its gravity will be taken as two presumed constants and height will be variable due to which the acceleration also varies with respect to height from the surface level.

Formula used:
${{\text{g}}^{\text{ }\!\!'\!\!\text{ }}}\text{=g}\left( \dfrac{{{\text{R}}^{\text{2}}}}{{{\text{(R+h)}}^{\text{2}}}} \right)$

Complete answer:
In the question it is given that the height =3,200km
And we need to find the acceleration due to gravity at that point.
Now, we know that the acceleration due to gravity varies with height,
We have one formula which relates the acceleration due to gravity and height,
${{\text{g}}^{\text{ }\!\!'\!\!\text{ }}}\text{=g}\left( \dfrac{{{\text{R}}^{\text{2}}}}{{{\text{(R+h)}}^{\text{2}}}} \right)$
In the above formula,
$\text{g }\!\!'\!\!\text{ }$ Represents the acceleration due to gravity at specific height h,
h represents the height,
g is the gravity of the earth and R is the Radius of the earth.
Now, we have to simply put the values in the formula to get the required answer,
So let’s first make a list of values to be used in the formula,
$\text{g }\!\!'\!\!\text{ }$= we will find this value
$\text{g=9}\text{.8}\dfrac{\text{m}}{{{\text{s}}^{\text{2}}}}$
$\text{R=6}\text{.4 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{6}}}\text{m}$
$\text{h=3200km=3}\text{.2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{6}}}\text{m}$ (Given)

Now, let’s put the values in our formula to get the answer,
${{\text{g}}^{\text{ }\!\!'\!\!\text{ }}}\text{=9}\text{.8}\left( \dfrac{{{(6.4\times {{10}^{6}})}^{\text{2}}}}{{{\text{(}6.4\times {{10}^{6}}\text{+3}.2\times {{10}^{6}}\text{)}}^{\text{2}}}} \right)$
$\implies{{\text{g}}^{\text{ }\!\!'\!\!\text{ }}}\text{=9}\text{.8}\left( \dfrac{40.96\times {{10}^{12}}}{\text{92}.16\times {{10}^{12}}} \right)$
$\implies {{\text{g}}^{\text{ }\!\!'\!\!\text{ }}}\text{=9}\text{.8}\left( \dfrac{40.96}{\text{92}.16} \right)$
$\implies \text{g=9}\text{.8 }\!\!\times\!\!\text{ 0}\text{.44}$
$\implies \text{g = 4}\text{.35 }\simeq \text{ 4}\text{.4 m}{{\text{s}}^{-2}}$
Hence, we can conclude that the value of the acceleration due to gravity at a place 3,200 km above the surface of the earth will be $\text{4}\text{.4 m}{{\text{s}}^{\text{-2}}}$.

Note:
First of all we need to convert the values of all the quantities which we are supposed to use in the formula, in a particular standard unit, that is meter unit. After this we will put the values in the formula to find the acceleration due to gravity at that height in $\text{m}{{\text{s}}^{\text{-2}}}$ .