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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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- 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 \[661.5N\]

- 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.