Animated Solution for Physics - Gravitation: Two bodies, each of mass $M$, are kept fixed with a separation $2L$. A particle of mass $m$ is projected from the mid-point of the line joining their centres, perpendicular to the line. The gravitational constant is $G$. The correct statement(s) is (are)

Imagine standing in deep space, watching a fascinating cosmic dance. We have two massive bodies, each of mass $M$, anchored firmly at a distance of $2L$ from one another.

Exactly at the midpoint of the line connecting these two giants, we place a tiny test particle of mass $m$. This midpoint, which we will call point $C$, lies at a distance of exactly $L$ from both of the larger masses.

Now, we give this tiny particle a sudden kick, projecting it with an initial velocity $v$ in a direction perpendicular to the line joining the two fixed masses. Our quest is to determine the minimum initial velocity required for this particle to break free from the gravitational embrace of these two massive bodies and escape to infinity.

Conservation of Energy

Before diving into the mathematics, let us step back and appreciate the physics at play. The only force acting on our test particle is gravity.

Because the gravitational force is a conservative force, the total mechanical energy of the particle—which is the sum of its kinetic energy ($K$) and gravitational potential energy ($U$)—must remain absolutely constant at every single point along its trajectory.

This fundamental truth immediately validates statement (d): the energy of the mass $m$ remains constant throughout its motion.

What does it truly mean to "escape" a gravitational field? To escape means to travel infinitely far away from the source masses, where their gravitational pull drops to zero.

By definition, we set the gravitational potential energy at infinity to be zero:

For the particle to escape with the absolute minimum initial velocity, it should just barely reach infinity. This means that upon arriving at infinity, its kinetic energy will also have dwindled to zero:

Thus, the total mechanical energy of the particle at infinity is exactly zero:

Since energy is conserved, the total mechanical energy at our starting point $C$ must also be exactly zero!

Let us find the gravitational potential $V_c$ at the midpoint $C$. Since potential is a scalar quantity, we can simply sum the potentials contributed by each of the two fixed masses:

Now, we can easily find the initial gravitational potential energy $U_c$ of our test mass $m$ at this midpoint by multiplying its mass by the potential:

The negative sign here is beautiful—it signifies that the particle is trapped in a gravitational potential well, bound to the two massive bodies.

Now, we bring everything together using our energy conservation equation:

Substituting our expressions for kinetic and potential energy, we get:

Let's rearrange this equation to solve for the velocity $v$:

Notice how the mass of the test particle, $m$, appears on both sides of the equation. It cancels out beautifully! This tells us a profound physical truth: the escape velocity is completely independent of the mass of the escaping object itself.

Taking the square root on both sides, we find the minimum escape velocity:

This perfectly matches statement (b).

Through the elegant application of energy conservation, we have shown that: 1. The minimum initial velocity required for the mass $m$ to escape is indeed $2\sqrt{\frac{GM}{L}}$, making statement (b) correct. 2. The total mechanical energy of the mass $m$ remains constant throughout its motion because gravity is a conservative force, making statement (d) correct.

Therefore, the correct options are (b) and (d).