Gravitation Question 224
Question: A uniform ring of mass $ [m] $ and radius r is placed directly above a uniform sphere of mass $ [M] $ and of equal radius. The centre of the ring is directly above the centre of the sphere at a distance $ [r\sqrt{3}] $ $ [r] $ as . The gravitational force exerted by the sphere on the ring will be
Options:
A) $ [\frac{GMm}{8r^{2}}] $
B) $ [\frac{GMm}{4r^{2}}] $
C) $ [\sqrt{3}\frac{GMm}{8r^{2}}] $
D) $ [\frac{GMm}{8r^{2}\sqrt{3}}] $
Show Answer
Answer:
Correct Answer: C
Solution:
-
$ [dF=G\frac{Mdm}{4r^{2}}] $ $ [F=\Sigma dF\cos \theta ] $
$ [=\Sigma \frac{GMdm}{4r^{2}}\cos \theta ] $
$ [=\frac{GM}{4r^{2}}\times \frac{\sqrt{3r}}{2r}\Sigma dm] $
$ [=\frac{\sqrt{3}GMm}{8r^{2}}] $
Alternative solution:
The gravitational field due to the ring at a distance $ [E=\frac{Gm(\sqrt{3r})}{{{[r^{2}+{{(\sqrt{3}r)}^{2}}]}^{3/2}}}] $
or E=$ [\frac{\sqrt{3}Gm}{8r^{2}}] $ The required force is EM, i.e., $ [(\sqrt{3}Gm)M/8r^{2}] $ .
BETA
