From a sphere of mass M and radius R, a smaller sphere of radius $\frac{R}{2}$ is carved out such that the cavity made in the original sphere is between its centre and the periphery (See figure). For the configuration in the figure where the distance between the centre of the original sphere and the removed sphere is 3R, the gravitational force between the two sphere is:

Question

Smart Achievers · Accepted Answer

(1)

Volume of removed sphere

$V_{remo} = \frac{4}{3} π {(\frac{R}{2})}^{3} = \frac{4}{3} π R^{3} (\frac{1}{8})$

Volume of the sphere (remaining)

$V_{remain} = \frac{4}{3} π R^{3} - \frac{4}{3} π R^{3} (\frac{1}{8}) = \frac{4}{3} π R^{3} (\frac{7}{8})$

Therefore mass of sphere carved and remaining sphere

Are at respectively $\frac{1}{8} M$ and $\frac{7}{8} M$

Therefore, gravitational force between these two sphere $≃ \frac{41}{3600} \frac{G M^{2}}{R^{2}}$

Solution

1 $\frac{41 G M^{2}}{3600 R^{2}}$

2 $\frac{41 G M^{2}}{450 R^{2}}$

3 $\frac{59 G M^{2}}{450 R^{2}}$

4 $\frac{G M^{2}}{225 R^{2}}$

Want Us to Email you About Special Offers & Updates?

Solution

1 41GM23600R2

2 41GM2450R2

3 59GM2450R2

4 GM2225R2

Ans. (1) Volume of removed sphere Vremo=43π(R2)3=43πR3(18) Volume of the sphere (remaining) Vremain=43πR3-43πR3(18)=43πR3(78) Therefore mass of sphere carved and remaining sphere Are at respectively 18M and 78M Therefore, gravitational force between these two sphere ≃413600GM2R2

Want Us to Email you About Special Offers & Updates?

1 $\frac{41 G M^{2}}{3600 R^{2}}$

2 $\frac{41 G M^{2}}{450 R^{2}}$

3 $\frac{59 G M^{2}}{450 R^{2}}$

4 $\frac{G M^{2}}{225 R^{2}}$