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If the radius of earth is $6000km$, what will be the weight of $120kg$ body if taken to a height of $2000km$ above sea level $\left( {i.e.,{\text{ }}surface{\text{ }}of{\text{ }}earth} \right).$

Last updated date: 13th Jun 2024
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Hint:-
- Weight of an object is the product of mass and acceleration due to gravity.
- Recall the formula for variation of acceleration due to gravity.
- The height from the sea level in the question is comparable to the radius of earth.
- Force between earth and an object $f = \dfrac{{GMm}}{{{r^2}}}$

Complete step by step solution:-
According to the Newton’s Law of gravitation
Force between earth and an object $f = \dfrac{{GMm}}{{{r^2}}}$
$G$ is the Gravitational constant.
$r$ is the distance from the centre of earth.
$M$ is the mass of earth.
$m$is the mass of the body. $m = 120kg$
Here we can compare the distance from the earth,
$R$ is the radius of earth.
Already given that,$R = 6000km = 6 \times {10^6}m$
So at surface $r = R$ then,
Force $f = \dfrac{{GMm}}{{{R^2}}}$
This force is equivalent to the weight of the body at the surface of the earth.
Weight is given by $f = mg$
$g$ is the acceleration due to gravity having a value $g = 9.8m/{s^2}$at earth surface.
Equate both forces $mg = \dfrac{{GMm}}{{{R^2}}}$
Cancels the mass $m$from both sides.
$g = \dfrac{{GM}}{{{R^2}}}$ This is the acceleration due to gravity at the surface of earth.
Now we are looking to the problem the body is at a height $h$
Given that, $h = 2000km = 2 \times {10^3}m$
So the gravitational force equation changed to
$f = \dfrac{{GMm}}{{{{(R + h)}^2}}}$
The weight is also changed
$f = mg'$
$g'$is the new acceleration due to gravity.
Compare both equations,
$mg' = \dfrac{{GMm}}{{{{(R + h)}^2}}}$
Cancel $m$from each side.
$g' = \dfrac{{GM}}{{{{(R + h)}^2}}}$
We can divide $g'$ with $g$
$\dfrac{{g'}}{g} = \dfrac{{GM}}{{{{(R + h)}^2}}}/(\dfrac{{GM}}{{{R^2}}})$
Cancel same terms
$\dfrac{{g'}}{g} = \dfrac{1}{{{{(R + h)}^2}}}/(\dfrac{1}{{{R^2}}}) = \dfrac{{{R^2}}}{{{{(R + h)}^2}}}$
$g' = g\dfrac{{{R^2}}}{{{{(R + h)}^2}}}$
Now at height $h$,
Weight $w = mg'$
Substitute $g'$
$w = mg' = \dfrac{{mg{R^2}}}{{{{(R + h)}^2}}}$
Now substitute all values
$w = \dfrac{{120 \times 9.8 \times {{(6 \times {{10}^6})}^2}}}{{{{(6 \times {{10}^6} + 2 \times {{10}^6})}^2}}} = \dfrac{{4.23 \times {{10}^{16}}}}{{6.4 \times {{10}^{13}}}} = 661.5N$

So the answer is weight of the body at a height is $661.5N$

Note:-
- Weight has the same unit of force Newton $N$ .
- Use the conversion$1km = {10^3}m$.
- Value of $G = 6.674 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}$
- From sea level means the surface of earth.
- The heights from the earth are generally referred to from the sea level.
- Mass of the earth is $M = 5.972 \times {10^{24}}kg$
- Different planets have different acceleration due to gravity value.
$g = 10m/{s^2}$at earth poles.