Question

A stone weighs 100 N on the surface of the earth. The ratio of its weight at a height of half the radius of the earth to its weight at a depth of half the radius of the earth will be approximately
A. 3.6
B. 2.2
C. 1.8
D. None of these

Answer 1

Hint: One simple formula for acceleration due to gravity g is obtained when we equate the gravitational force with the weight of stone (mg). Though, the mass of the earth involved in the formula beneath the earth's surface is going to get reduced.

Formula used:
The g felt by any object at a distance r from the center of the earth is given as:
$g = \dfrac{GM}{r^2}$ .

Complete answer:
If the radius of the earth is R then at the surface of the earth, then
$g = \dfrac{GM}{R^2}$ .
Now finding out the variation in g for the given two conditions:

1. As we go at a height of h above the value of g has to be:
$g'_{above} = \dfrac{GM}{(R+h)^2}$
We are given that h = R/2, so we may write:
$g'_{above} = \dfrac{GM}{(R+(R/2))^2}$ ;
taking R as a common factor from the denominator, we get:
$g'_{above} = \dfrac{GM}{(1+(1/2))^2 R^2}$
$g'_{above} = \dfrac{4GM}{9R^2}$.
In this expression we can substitute the value for g at the surface. So we get:
$g'_{above} = \dfrac{4}{9} g$

2. As we go at a depth of d beneath the surface of Earth, the separation from the centre will surely reduce but so will the effective mass of Earth responsible for creating a force of gravity on the stone.
The effective mass is calculated as:
$M' = density \times \dfrac{4}{3} \pi (R-d)^3$
If we assume earth we have uniform density throughout, we may write:
$M' = \dfrac{M}{\dfrac{4}{3} \pi R^3} \times \dfrac{4}{3} \pi (R-d)^3$ ;
where M is the total mass of earth i.e., mass of sphere of radius R.
Upon substituting d = R/2, we get the effective mass to be:
$M' = \dfrac{M}{8}$.

Now, the value of acceleration due to gravity at a depth of d beneath the surface of the Earth is transformed as:
$g'_{below} = \dfrac{GM'}{(R-d)^2}$.
Keeping the value of M' and d in this we obtain:
$g'_{below} = \dfrac{GM}{2R^2}$
or we can write,
$g'_{below} = \dfrac{1}{2} g$ .
The ratio of the weights above and below is just going to be:
$\dfrac{mg'_{above}}{mg'_{below}} = \dfrac{4/9}{1/2} = \dfrac{8}{9} = 0.889$

Therefore, the correct answer is option (D). None of these.

Note:
There are direct formulas to find variation in g at an altitude of h above the Earth surface and at a depth d below the Earth's surface. By using the altitude formula here one gets g = 0. So, it is recommended to not use those formulas as those formulas are an approximation for the case when h and d are very small as compared to the radius of the Earth.