Question

If at depth ‘\(d\)’ the gravitational force acting on a particle is 300 N, then what is the force on a particle at depth ‘\(\frac d2\)’ ?

Answer 1

The correct answer is 150 N.

We know that the gravitational force acting on a particle is directly proportional to the mass of the particle and the acceleration due to gravity.
\(F = m \times g\), where F = Force, m = mass, g = acceleration due to gravity

Now, if we consider the situation where we have two particles, one at depth ‘d’ and another at depth ‘d/2’ we can use the formula above.
\(F_d=m\times g\)

Given: \(F_d= 300N\)
Similarly for the particle at \(\frac{d}{2}\):
\(F_{\frac{d}{2}}=m \times g _ {new}\)

The acceleration due to gravity changes with depth because the gravitational field strength decreases as we move away from the center of the Earth.
Therefore,
\(g'=G \times \frac{M}{r^2}\)
where, g’ = gravitational field strength, G = gravitational Constant, M = Mass of the Earth and r = distance from the center of the earth.

As the gravitational field strength at depth 'd/2' is half of the gravitational field strength at depth 'd', the force on the particle at depth '\(\frac{d}{2}\)' is half of the force at depth 'd'.

\(F_{\frac{d}{2}}= \frac{F_d}{2}\)
\(F_{\frac{d}{2}}= \frac{300}{2}\)
\(F_{\frac{d}{2}}= 150 N\)

Therefore, the force on the particle at depth \(‘\frac{d}{2}’ = 150 N\)

Answer 2

Given:
The force on a particle at depth d is \(F_d = 300 \, \text{N}\).
We need to find the force on a particle at depth \(\frac{d}{2}\).

Gravitational Field Strength and Depth
The gravitational field strength, g, varies with depth. As we go deeper into the Earth, the effective value of g decreases. The relationship can be approximated for small depths using the formula:

\(g' = G \times \frac{M}{r^2}\)

Force at Different Depths
For a particle at depth d:
\(F_d = m \times g\)

For a particle at depth \(\frac{d}{2}\):
\(F_{\frac{d}{2}} = m \times g_{\text{new}}\)

Given that the gravitational field strength at depth \(\frac{d}{2}\) is half of that at depth d, we can write:
\(g_{\text{new}} = \frac{g}{2}\)

Thus, the force at depth \(\frac{d}{2}\) is:
\(F_{\frac{d}{2}} = m \times \frac{g}{2}\)

Since \(F_d = m \times g\), we can substitute \(F_d\) into the equation:
\(F_{\frac{d}{2}} = \frac{F_d}{2}\)
\(F_{\frac{d}{2}} = \frac{300 \, \text{N}}{2} = 150 \, \text{N}\)

So, the answer is \(150 \, \text{N}\).