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CGPET 2011 Solved Paper
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© examsnet.com
Question : 16
Total: 150
A particle of mass
m
is placed at each of the three corners of an equilateral triangle of side a. The particles revolve in circular orbits under the influence of mutual gravitational force such that they always lie at the vertices of the triangle. The velocity of each particle is
√
Gm
a
√
2
Gm
a
√
G
m
2
a
√
3
2
Gm
a
Validate
Solution:
As the gravitational force between any two particles is
F
=
Gmm
a
2
, the resultant force on each.
Particle due to the other two
F
R
=
√
F
2
+
F
2
+
2
F
2
cos
60
∘
F
R
=
√
3
F
=
√
3
G
2
a
2
∵
AD
=
a
2
∴
BD
=
√
AB
2
−
AD
2
BD
=
√
a
2
−
(
a
2
)
2
=
√
3
a
2
and radius
BO
=
AO
=
OC
=
√
3
a
2
×
2
3
=
a
√
3
m
v
2
r
=
√
3
Gm
2
a
2
m
v
2
a
√
3
=
√
3
G
2
a
2
⇒
v
=
√
Gm
a
© examsnet.com
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