Answer

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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.

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