Question

A funnel is dug along a diameter of the earth.The force on a particle of mass m placed in a tunnel at a distance x from the centre is

Answer 1

Hi there!
Here's the answer:

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¶¶¶ POINTS TO REMEMBER:

¶¶ According to Newton's law of Gravitation,
$F = G\frac{m_{1}m_{2}}{R^{2}}$

Where,
F = Gravitational Force between two objects (kg-m/s)

G = Gravitational constant (N-m²/kg²)

$m_{1}$ = Mass of the first object (kg)

$m_{2}$= Mass of the second object (kg)

r = Distance between objects (m)

¶¶ $Density = \frac{Mass}{Volume}$

¶¶ Volume of the sphere = $\frac{4}{3} \pi R^{3}$

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¶¶¶ SOLUTION:

Let M be Mass of the Earth
and R be Radius of the Earth

Given,
The tunnel is dug and is at x distance from the centre of Earth

Assume an Imaginary sphere with this distance 'x' which will touch the tunnel edge.

• $Mass = Density × Volume$

Mass of the Earth , M = $\frac{4}{3} \pi R^{3}\rho$

Mass of the Imaginary sphere having radius x , M' = $\frac{4}{3} \pi x^{3}\rho$

• Expressing M' in terms of M,
$\frac{M'}{M} = \frac{ \frac{4}{3} \pi x^{3}\rho}{ \frac{4}{3} \pi R^{3}\rho}$

$\implies \frac{M'}{M}=\frac{x^{3}}{R^{3}}$

$\implies M' = \frac{x^{3}}{R^{3}}×M$

Given, m is the mass of particle placed x distance from centre

Since,
$m_{1}$ = M
$m_{2}$ = m
$R$ = x

• Gravitational force
$F = G\frac{MM'}{x^{2}}$

$F = \frac{GM}{x^{2}} × M'$

$F = \frac{GM}{x^{2}} × \frac{x^{3}}{R^{2}}×m$

$\implies F = \frac{GMmx}{R^{2}}$

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