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JEE Advanced · Physics · 9. Gravitation

Consider a spherical gaseous cloud of mass density ρr in free space where r is the radial distance from its center. The gaseous cloud is made of particles of equal mass m moving in circular orbits about the common center with the same kinetic energy K . The force acting on the particles is their mutual gravitational force. If ρr is constant in time, the particle number density nr=ρr/m is: [ G is universal gravitational constant]

  1. A Kπr2m2G
  2. B 3Kπr2m2G
  3. C K6πr2m2G
  4. D K2πr2m2G
Verified Solution

Answer & Solution

Correct Answer

(D) K2πr2m2G

Step-by-step Solution

Detailed explanation

Given that all the particles of gaseous cloud has mass m and are moving due to mutual attraction with same kinetic energy K .
Now in that cloud, let us consider a sphere of radius r containing a group of particles adding to a total mass M .
Now, another particle of mass m is moving with speed V due to gravitational attraction of M, in a circle of radius r .
Centripetal force is here given by M
GMmr2=mv2r ....(1)
But, kinetic energy, K=12mv2
mv2=2K
Put it in (1)
GMmr2=2Kr
M=2KrGm
Differentiating it we get,
dM=2KGmdr .....(2)
Now, if density of cloud is ρ,
Then elementary mass,
dM=4πr2drρ=2KGmdr
ρ=K2πGmr2
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