Question

The value of 'g' at a particular point is $9.8m/{s^2}$. Suppose the earth suddenly shrinks uniformly to half its present size without any mass. The value of 'g' at the same point (assuming that the distance of the point from the center of the earth does not shrink) will become:
A) $9.8m/{\sec ^2}$
B) $4.9m/{\sec ^2}$
C) $19.6m/{\sec ^2}$
D) $3.1m/{\sec ^2}$

Answer 1

The gravity of Earth is defined as the net acceleration that Is applied due to the combined effect of the centrifugal force and gravitation of the objects. It is denoted by the letter g. Zero gravity does not exist.

Complete step by step solution:
Given data:
The value of g at a particular point$= 9.8m/{s^2}$
The acceleration due to gravity is given by the formula,
$g = \dfrac{{GM}}{{{r^2}}}$
Where G is the gravitational constant. M is the mass of the earth and r is the separation.
Hence if the volume is reduced without any change in the value of M and R, it is clear that the acceleration due to gravity remains constant.

Thus the value of g will be $9.8m/{s^2}$.

Note: 1. The acceleration due to gravity depends on the radius of the earth and the mass of the earth. The acceleration due to gravity is maximum at the poles and minimum at the equator.
2. The acceleration from gravity will remain always constant and is downward, even though the magnitude of the velocity and the direction change.
3. The acceleration is constant in the freefall, which is a special case of motion. For describing a freefall, the size, shape, and weight of the object are not considered as the factors. We cannot escape from the pull of gravity.
4. The amount of matter in a body is called mass. It does not change at any time. The force exerted on the object by gravity is called weight. The gravitational pull gives a downward acceleration for the object.