Correct Answer - Option 4 : decreases, decreases
CONCEPT:
-
Gravitational acceleration: An acceleration that an object receives due to the force of gravity acting on it.
It is calculated by using the formula:
\(g = {GM\over R^2}\)
where g is the gravitational acceleration, G is the gravitational constant, M is the mass of the earth and R is the radius of the earth.
- When an object is at height 'h', gravitational acceleration: It is calculated by:
\(g_{_h} = {g \over (1+{h\over R})^2}\)
where gh is the gravitational acceleration at height h, g is the gravitational acceleration at the surface, and R is the radius of the earth.
- When an object is at depth 'd', gravitational acceleration: It is calculated by:
\(g_{_d} = g (1-{d\over R})\)
where gd is the gravitational acceleration at depth d, g is the gravitational acceleration at the surface, and R is the radius of the earth.
- As we go above the earth's surface, the gravitational acceleration decreases.
- As we go below the earth's surface, the gravitational acceleration decreases.
EXPLANATION:
Gravitational acceleration at height h is given by
\(g_{h} = {g \over (1+{h\over R})^2}\)
As we go above h increases, so \((1+{h\over R})^2\) increases.
As \((1+{h\over R})^2\) increases, \(g_{h} = {g \over (1+{h\over R})^2}\) decreases.
- So, as we go above the earth's surface, the gravitational acceleration decreases.
Gravitational acceleration at the depth d is given by:
\(g_{_d} = g (1-{d\over R})\)
As we go below the earth's surface, the depth 'd' will increase.
As d increases, \( (1-{d\over R})\) will decrease.
As \( (1-{d\over R})\) decreases, \(g (1-{d\over R})\) will also decrease.
So gd will decrease with depth.
- So as we go below the earth's surface, the gravitational acceleration decreases.
- Hence the correct answer is option 4.
