Question

Two balls, each of radius R and of equal mass and density, are placed in contact. Then, the force of gravitation between them is proportional to

Answer 1

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$\underline{\huge{\textbf{Solution}}}$

Given,
Two balls, each of radius R and of equal mass and density, are placed in contact.

$\underline{\large{\textbf{Step-1}}}$:
Find the Distance between the centre of two balls

Distance between the centre of two balls = Sum of their Radii.

Distance between the centre of two balls = R+R = 2R.

$\underline{\large{\textbf{Step-2}}}$:
Express the Mass of a ball as product of Density and Volume.
(This would be same for other ball ,given that two balls have equal mass)

Since, the shape of the ball is Sphere.

Volume of the ball $V = \frac{4}{3}\, \pi R^{3}$

So,
Mass of the ball $m = \rho \times \frac{4}{3}\, \pi R^{3}$.

$\underline{\large{\textbf{Step-3}}}$:
Find the force of gravitation between the two balls.

According to Newton's Law of Universal Gravitation

$F = G\, \dfrac{Mm}{r^{2}}$

Where,
F = Gravitational Force between two objects.
G = Gravitational constant
M = Mass of the first object
m = Mass of the second object
r = Distance between objects

Here,
$M = m = \rho \times \frac{4}{3}\, \pi R^{3}$.
$r = 2R$

Substituting Values

$\implies F =G\, \dfrac{m^{2}}{(2R)^{2}}$

$\implies F =G\, \dfrac{(\rho \times \frac{4}{3}\, \pi R^{3})^{2}}{4R^{2}}$

$\implies F =G\,\times (\rho)^{2} \times (\frac{4}{3})^{2} \times (\frac{1}{4}) \times (\dfrac{R^{6}}{R^{2}})$

$\implies F =G\,\times (\rho)^{2} \times (\frac{4}{3})^{2} \times (\frac{1}{4}) \times R^{4}$

$\implies F \propto R^{4}$

Therefore,
The force of gravitation between the two balls is proportional to $R^{4}$
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