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What is the acceleration due to gravity on the top of Mount Everest? Mount Everest is the highest mountain peak of the world at the height of \[8848\text{ }m\]. The value at sea level is $9.80m/{{s}^{2}}$.

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Last updated date: 03rd Mar 2024
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IVSAT 2024
Answer
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Hint: The acceleration due to gravity depends on the mass of Earth.
The acceleration due to gravity is calculated using Newton’s law of gravity.
To find the acceleration due to gravity we use the formula,
$g=\dfrac{GM}{{{r}^{2}}}$
Where, $M$ is the mass of the body and $r$ is the distance from the center of mass of the body to the point where we need to calculate the acceleration due to gravity. Acceleration due to gravity is defined as the gravitational force of attraction acting per unit mass.

Complete step by step solution:
Using Newton’s law of gravitation,
$g\left( r \right)=\dfrac{GM}{{{r}^{2}}}$
Where,
$g\left( r \right)=$The gravitational field strength
G = The universal gravitational constant
M = The mass of earth
r = The distance from the center of earth
If a point is at an altitude of $h$ from the surface of earth, then value of $r$ can be calculated as
$r={{R}_{e}}+h$, where, ${{R}_{e}}$ is the radius of the earth

Putting $h=0$ we get the gravitational field strength at the surface of earth.
It is given that the acceleration due to gravity on the surface is $9.8m/{{s}^{2}}$
From the formula for acceleration due to gravity,
$ g=\dfrac{GM}{R_{e}^{2}} $
 $ 9.8m/{{s}^{2}}=\dfrac{GM}{R_{e}^{2}}\ldots \ldots \left( i \right)
$
At Mount Everest, let the acceleration due to gravity is ${{g}_{ME}}$
Then,
${{g}_{ME}}=\dfrac{GM}{{{r}^{2}}} $
 $=\dfrac{\left( \dfrac{GM}{R_{e}^{2}} \right)}{\left( \dfrac{{{r}^{2}}}{R_{e}^{2}} \right)}$
$ =\dfrac{g}{{{\left( \dfrac{r}{{{R}_{e}}} \right)}^{2}}}\ldots \ldots \left( \text{from equation }i \right) $
 $ =\dfrac{g}{{{\left( \dfrac{{{R}_{e}}+h}{{{R}_{e}}} \right)}^{2}}}$
 $=g{{\left( 1+\dfrac{h}{{{R}_{e}}} \right)}^{-2}}$

As, $h\ll {{R}_{e}}$
Using approximation we get
${{g}_{ME}}=g\left( 1-\dfrac{2h}{{{R}_{e}}} \right)$
Radius of earth ${{R}_{e}}=6371km=6371000m$
Putting the value in acceleration due to gravity at Mount Everest, we get
$ {{g}_{ME}}=9.8\left( 1-\dfrac{2\times 8848}{6371000} \right)m/{{s}^{2}}$
 $ =9.773m/{{s}^{2}} $

Hence, the acceleration due to gravity at a point on Mount Everest is $9.773m/{{s}^{2}}$.

Note: The magnitude of acceleration due to gravity decreases as we move above the surface of the earth or down the surface of the earth. We should be careful while using the approximation. As if the point is very close to the surface of earth then acceleration due to gravity is approximately the same as that on the sea level.