We discuss problems, which highlight certain aspects of the study leading to gravitational field. The questions are categorized in terms of the characterizing features of the subject matter :

Problem 3 : A small hole is created on the surface of a spherical shell of mass, “M” and radius “R”. A particle of small mass “m” is released a bit inside at the mouth of the shell. Describe the motion of particle, considering that this set up is in a region free of any other gravitational force.

Figure 1: A particle of small mass “m” is released at the mouth of hole.

Gravitational force

Solution : The gravitational potential of a shell at any point inside the shell or on the surface of shell is constant and it is given by :

V = - G M R V = - G M R

The gravitational field,”E”, is :

E = - ⅆ V ⅆ r E = - ⅆ V ⅆ r

As all quantities in the expression of potential is constant, its differentiation with displacement is zero. Hence, gravitational field is zero inside the shell :

E = 0 E = 0

It means that there is no gravitational force on the particle. As such, it will stay where it was released.

Problem 4 : A ring of mass “M” and radius “R” is formed with non-uniform mass distribution. Find the minimum work by an external force to bring a particle of mass “m” from infinity to the center of ring.

Solution : The work done in carrying a particle slowly from infinity to a point in gravitational field is equal to potential energy of the “ring-particle” system. Now, Potential energy of the system is :

W F = U = m V W F = U = m V

The potential due to ring at its center is independent of mass-distribution. Recall that gravitational potential being a scalar quantity are added algebraically for individual elemental mass. It is given by :

V = - G M r V = - G M r

Hence, required work done,

⇒ W F = U = - G M m r ⇒ W F = U = - G M m r

The negative work means that external force and displacement are opposite to each other. Actually, such is the case as the particle is attracted into gravitational field, external force is applied so that particle does not acquire kinetic energy.

Problem 5 : Imagine that a hole is drilled straight through the center of Earth of mass “M” and radius “R”. Find the speed of particle of mass dropped in the hole, when it reaches the center of Earth.

Solution : Here, we apply conservation of mechanical energy to find the required speed. The initial kinetic energy of the particle is zero.

K i = 0 K i = 0

On the other hand, the potential energy of the particle at the surface is :

U i = m V i = − G M m R U i = m V i = − G M m R

Let “v” be the speed of the particle at the center of Earth. Its kinetic energy is :

K f = 1 2 m v 2 K f = 1 2 m v 2

The potential energy of the particle at center of Earth is :

U f = m V f = - 3 G M m 2 R U f = m V f = - 3 G M m 2 R

Applying conservation of mechanical energy,

K i + U i = K f + U f K i + U i = K f + U f

⇒ 0 − G M m R = 1 2 m v 2 - 3 G M m 2 R ⇒ 0 − G M m R = 1 2 m v 2 - 3 G M m 2 R

v = G M R v = G M R