Question

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 $$\left(i.e.,surfaceofearth\right).$$

Answer 1

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 $$f={\displaystyle \frac{GMm}{r^2}}$$

Complete step by step solution:-
According to the Newton’s Law of gravitation
Force between earth and an object $$f={\displaystyle \frac{GMm}{r^2}}$$
$$G$$ 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 $$f={\displaystyle \frac{GMm}{R^2}}$$
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 $$mg={\displaystyle \frac{GMm}{R^2}}$$
Cancels the mass $$m$$from both sides.
 $$g={\displaystyle \frac{GM}{R^2}}$$ 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
$$f={\displaystyle \frac{GMm}{{(R+h)}^2}}$$
The weight is also changed
$$f=m{g}^{\prime }$$
$$g^{\prime }$$is the new acceleration due to gravity.
Compare both equations,
$$m{g}^{\prime }={\displaystyle \frac{GMm}{{(R+h)}^2}}$$
Cancel $$m$$from each side.
$$g^{\prime }={\displaystyle \frac{GM}{{(R+h)}^2}}$$
We can divide $$g^{\prime }$$ with $$g$$
$${\displaystyle \frac{g^{\prime }}{g}}={\displaystyle \frac{GM}{{(R+h)}^2}}/({\displaystyle \frac{GM}{R^2}})$$
Cancel same terms
$${\displaystyle \frac{g^{\prime }}{g}}={\displaystyle \frac{1}{{(R+h)}^2}}/({\displaystyle \frac{1}{R^2}})={\displaystyle \frac{R^2}{{(R+h)}^2}}$$
$$g^{\prime }=g{\displaystyle \frac{R^2}{{(R+h)}^2}}$$
Now at height $$h$$,
Weight $$w=m{g}^{\prime }$$
Substitute $$g^{\prime }$$
$$w=m{g}^{\prime }={\displaystyle \frac{mg{R}^2}{{(R+h)}^2}}$$
Now substitute all values
$$w={\displaystyle \frac{120\times 9.8\times {(6\times {10}^6)}^2}{{(6\times {10}^6+2\times {10}^6)}^2}}={\displaystyle \frac{4.23\times {10}^{16}}{6.4\times {10}^{13}}}=661.5N$$

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.