Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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 spheres is :

Option: 1
$\frac{41GM^{2}}{3600R^{2}}$

Option: 2
$\frac{41GM^{2}}{450R^{2}}$

Option: 3
$\frac{59GM^{2}}{450R^{2}}$

Option: 4
$\frac{GM^{2}}{225R^{2}}$

Answers (1)

best_answer

As we discussed in

Newton's Law of Gravitation -

$F\; \alpha\; \frac{m_{1}m_{2}}{r^{2}}$

$F\; = \frac{G\, m_{1}\, m_{2}}{r^{2}}$

$F\rightarrow$ Force

$G\rightarrow$ Gravitalional constant

$m_1,m_2\rightarrow$ Masses

$r\rightarrow$ Distance between masses

- wherein

Force is along the line joining the two masses

Volume of removed sphere

V_removed = $\frac{4}{3}\pi (\frac{R}{2})^{3}=\frac{4}{3}\pi R^{3}(\frac{1}{8})$

Volume of the sphere (remaining)

V_remain = $\frac{4}{3}\pi R^{3}-\frac{4}{3}\pi R^{3}(\frac{1}{8})=\frac{4}{3}\pi R^{3}(\frac{7}{8})$

Therefore the mass of removed sphere and remaining sphere are at respectively $\frac{1}{8}M$ and $\frac{7}{8}M$

$F_{net}=\frac{GM\frac{M}{8}}{{9R}^{2}}\:-\:\frac{G\frac{M}{8}\times\frac{1}{8}M}{(\frac{25}{4}R)^{2}} =\frac{41}{3600}\:\frac{GM^{2}}{R^{2}}$

Posted by

avinash.dongre

View full answer

Similar Questions

0.1 M acetic acid solution having pH =3 is titrated against 0.05 M NaOH solution. Calculate pH at 1/4th stage of neutralization of acid.Option: 1

0.25×60+0.10x=0.15×(60+x)

Trending Articles/News

anam-khan

BITSAT 2025 session 1 scorecard out on bitsadmission.com for BE, MSc, and BPharm admission

anam-khan

BITSAT session 2 slot booking 2025 begins; exams from June 22 to 26

anam-khan

BITSAT 2025 session 2 application correction window opens at bitsadmission.com; slot booking from June 16

Latest Question