CGPET 2015 Solved Paper

Consider the movement of three bodies along a circle centered at as shown in the diagram alongside.Gravitational force between A and B
FBA=

Gm2

(AB)2

Gravitational force between B and C
FBC=

Gm2

(BC)2

From geometry of the figure, we can write BO‌cos‌30°+CO‌cos‌30°=BC
⇒AB=BC=CA=2R‌cos‌30°=√3R
Net force on B,
FB=FBA‌cos‌30°+FBC‌cos‌30°
=

2Gm2

(AB)2

×

√3

2

=

2Gm2

(√3R)2

×

√3

2

=

2Gm2

3R2

×

√3

2

This force will provide centripetal force for circular motion of B
FB=

2Gm2

3R2

×

√3

2

=

mvB2

R

=

G

√3R

=vB2
⇒vB=√

G

√3R