Four masses ‘m’ each are orbiting in a circle of radius’r’ in the same direction under gravitational force. Velocity of each particle is
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a
Gmr(1+22)2
b
Gmr
c
Gmr1+22
d
Gm2r1+222
answer is D.
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Detailed Solution
F'net on A due to C and B=2Fcosθ2=2Fcos902=2Fcos45=2F12=2F=2Gm22r2=2Gm22r2=Gm22r2 ⇒Fnet=FAD+F'net ⇒Fnet provides centripetal force ⇒Gm22r2+Gm22r2=mv2r ⇒Gm4r+Gm2r=v2 ⇒v=Gmr14+12 ⇒v=Gmr14+22222=Gmr14+224==Gm4r1+22