**JEE Advanced 2019 Paper 1, Question 1**

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

**Solution**

Consider a small parcel of gas in the cloud with mass at a distance from the center of the cloud (see figure). The forces acting on the particle are

- The centrifugal force pulling outward:

(1)

where we have inverted the expression for kinetic energy to write in terms of known constants and .

- The force due to gravity of the spherical cloud of radius , pulling inward:

(2)

Note that the shell outside the region of radius exerts no force on . This is familiar to us from electrostatics, where we know that there is no electric field inside a uniformly charged shell. Both the Coulomb force and Newton’s law of gravitation are inverse square laws and share this property (see discussion below).

We are told that the parcel moves in circular orbits about the center. That means the net force in the radial direction is zero,

(3)

Substituting (1) and (2) and cancelling we get

(4)

Thus, we have arrived at an expression that relates all the variables given in the problem. However, the density is stuck inside an integral. To break it out we need to differentiate.

Since the density only depends on the radial co-ordinate we can write

(5)

To simplify the LHS we recall the definition of the derivative of a function ,

(6)

which means

(7)

The term in parantheses on the RHS is just the mass of a shell of radius and small thickness with uniform density . That is, . Therefore\footnote{hello},

(8)

using the identity . Subsituting this back into (5) and using we find

(9)

which is option A.

**Discussion**

*Gauss’s Law for gravity: *Recall that in electrostatics we determine the field inside a uniformly charged hollow shell using Gauss’s law,

In a similar way, we have that the surface integral of gravitational force is proportional to the mass enclosed within the surface,

We can therefore see that gravitational force inside the shell is zero by the same argument we use to show that electric field inside a charged shell.

There is another way to see this by a simple argument due to Newton. Consider a thin spherical shell as shown in the figure below. Consider the two small mass elements marked out by a conical surface with its apex at . The amount of mass in each element is proportional to its surface area. The area increases as (distance). However, the strength of the force varies as 1/(distance), where the distance is measured from the apex to the shell. Thus the forces of the two mass elements are equal and opposite, and cancel. We can pair up all the elements of the shell this way so the total force on is zero. A thick shell can be regarded as a succession of thin spherical shells, so the force inside is still zero.