First slide

Gravitational field intensity

Question

For the arrangement shown in the figure find the net gravitational field at point ‘O’ is

Difficult

A

$\frac{G M}{R^{2}} [\frac{12 - π}{2 π}]$ towards left
B

$\frac{G M}{R^{2}} [\frac{2 - π}{2 π}]$ towards right
C

$\frac{G M}{π R^{2}} [\frac{4 - π}{12}]$ towards left
D

Zero

Solution

Gravitational field due to arc $E_{1} = \frac{2 G (2 M) \sin (π / 6)}{\frac{π}{3} R^{2}}$
$E_{1} = \frac{6 G M}{π R^{2}} t o w a r d s l e f t$
Gravitational field due to a rod at the axial point
$E_{2} = \frac{G M}{R} [\frac{1}{R} - \frac{1}{2 R}]$
$E_{2} = \frac{G M}{2 R^{2}} t o w a r d s r i g h t$
Net gravitational field
$E = E_{1} - E_{2}$ towards left $[∵ E_{1} > E_{2}]$
$= \frac{6 G M}{π R^{2}} - \frac{G M}{2 R^{2}}$
$= \frac{G M}{R^{2}} [\frac{6}{π} - \frac{1}{2}]$
$E = \frac{G M}{R^{2}} [\frac{12 - π}{2 π}] t o w a r d s l e f t$

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