Question 111: Gravitation

Question

Assuming the earth to be a sphere of uniform mass density, the weight of a body at a depth $$d=\frac{R}{2}$$ from the surface of earth, if its weight on the surface of earth is 200 N, will be:
(Given R = radius of earth)

Accepted Answer

The gravitational field strength, or equivalently, the weight of an object, decreases linearly from its surface value to zero at the center of a sphere of uniform mass density. This is because only the mass inside the radius at which the object is located contributes to the gravitational force at that location.
If ( $d = \frac{R}{2}$ ) is the depth below the surface of the Earth, then the radius of the sphere contributing to the gravitational force at that depth is ( $R - d = R - \frac{R}{2} = \frac{R}{2}$ ).
Since the gravitational force (or weight) decreases linearly with the radius in a sphere of uniform density, the weight of the object at depth ( $d = \frac{R}{2} $) is half its weight at the surface of the Earth.
So, if the weight of the body on the surface of the Earth is 200 N, its weight at depth ( $d = \frac{R}{2} $) is half of that, or 100 N.
Therefore, the correct answer is  100 N.

Answer

100 N

Answer

400 N

Answer

300 N

Answer

500 N

Gravitation

Assuming the earth to be a sphere of uniform mass density, the weight of a body at a depth $d=\frac{R}{2}$ from the surface of earth, if its weight on the surface of earth is 200 N, will be: (Given R = radius of earth)

Choose the correct answer: