
F1=(2R)2Gmm=4R2Gm2
F2=(2R)2Gmm=2R2Gm2
F3=(2R)2Gmm=2R2Gm2
⇒Fnet=F1+F2cos45∘+F3cos45∘
=4R2Gm2+2R2Gm221+2R2Gm221
=R2Gm2(41+21)=4R2Gm2(1+22)
Fnet=4R2Gm2(1+22)=Rmv2
⇒v=2G(1+22)
JEE Main 2021 — Physics Mechanics
Four identical particles of equal masses 1kg made to move along the circumference of a circle of radius 1m under the action of their own mutual gravitational attraction. The speed of each particle will be:
Held on 24 Feb 2021 · Verified 6 Jul 2026.
G(1+22)
2G(1+22)
2G(22−1)
2(1+22)G
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