Question 376: Gravitation

Question

A hole is drilled half way to the centre of the earth. A body weighs 300 N on the surface of the earth. How much will, it weigh at the bottom of the hole?

Accepted Answer

Given, distance of bottom of hole from the surface of earth $(d)=$ half the radius of earth $=\frac{R_e}{2}$
If $g$ be the value of gravitational acceleration on the surface of earth, then weight of body
$$m g=300 \mathrm{~N}$$
If $g^{\prime}$ be the gravitational acceleration at the bottom of hole, then
$$g^{\prime}=g\left(1-\frac{d}{R_	heta}ight)=g\left(1-\frac{\frac{R_	heta}{2}}{R_e}ight) \Rightarrow g^{\prime}=\frac{g}{2}$$
$	herefore$ Weight of the body on the bottom of hole,
$$m g^{\prime}=\frac{m g}{2}=\frac{300}{2}=150 \mathrm{~N}$$

Answer

200 N

Answer

250 N

Answer

120 N

Answer

150 N

Gravitation

A hole is drilled half way to the centre of the earth. A body weighs 300 N on the surface of the earth. How much will, it weigh at the bottom of the hole?

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