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Gravitation Part 1

Section: Physics

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Question : 31 of 100

Marks: +1, -0

A solid sphere of radius

R

gravitationally attracts a particle placed at

3 R

from its centre with a force

F_{1}

. Now, a spherical cavity of radius

\left(\frac{R}{2}\right)

is made in the sphere (as shown in figure) and the force becomes

F_{2}

F_{1}:F_{2}

[25 Feb 2021 Shift 1]

$25:36$
$36:25$
$50:41$
$41:50$

Solution:

Given, radius of sphere

=R

Distance of particle from centre of Earth

=3 R

Force between sphere and particle

=F_{1}

Radius of cavity

=\frac{R}{2}

Let mass of sphere

=M

Mass of sphere with cavity

=M^{\prime}

Mass of particle

=m

Now,

M^{\prime}=\rho \cdot \frac{4}{3}\pi \left(\frac{R}{2}\right)^{3}=\frac{M}{8}

By using concept of gravitational force,

F_{1}= \frac{G M m}{(3R)^{2}}=\frac{G M m}{9R^{2}}

. . . (i)

and

F_{\text{cavity}} = \frac{G M^{\prime} m}{(AB)^2} = \frac{2}{25 \times 8 R^{2}}

= \frac{4 G M m}{25}

. . . (ii)

\therefore F_{2}=F_{1}-F_{\text{cavity}}=\frac{G M m}{9 R^{2}}-\frac{4}{25 \times 8} \frac{G M m}{R^{2}}

=\frac{G M m}{R^{2}}\left(\frac{1}{9}-\frac{1}{50}\right)=\frac{41}{50 \times 9} \frac{G M m}{R^{2}}

\because \frac{F_{1}}{F_{2}}=\frac{50}{41}

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