Question 115: Gravitation

Question

The weight of a body on the earth is $$400 \mathrm{~N}$$. Then weight of the body when taken to a depth half of the radius of the earth will be:

Accepted Answer

The gravitational field inside a uniform spherical body varies linearly with distance from the center. If we consider the Earth to be such a body, then the gravitational field strength (and hence weight) of an object would decrease linearly as we go deeper inside the Earth.
The weight of an object at a depth $d$ from the Earth's surface is given by:
$W_d = W_e (1 - \frac{d}{R})$,
where:
$W_d$ is the weight at depth $d$,
$W_e$ is the weight at the Earth's surface (i.e., the weight of the object), and
$R$ is the radius of the Earth.
In this case, we're given that $W_e = 400 \, 	ext{N}$, and we're asked to find the weight at a depth of $d = \frac{R}{2}$. Substituting these values into the formula, we get:
$W_d = 400 \, 	ext{N} (1 - \frac{1}{2}) = 200 \, 	ext{N}$.
Therefore, the weight of the body when taken to a depth half of the radius of the Earth is 200 N.

Answer

300 N

Answer

200 N

Answer

100 N

Answer

Zero

Gravitation

The weight of a body on the earth is $400 \mathrm{~N}$. Then weight of the body when taken to a depth half of the radius of the earth will be:

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