JEE Challenger
More from Gravitation
PhyMedAdv 2013 • Paper 2

Escape Velocity of Particle Projected from Midpoint of Two Fixed Masses

Two identical bodies, each having a mass MM, are held fixed at a separation distance of 2L2L. A small particle of mass mm is launched from the midpoint of the line segment joining their centres in a direction perpendicular to this line. Taking the universal gravitational constant to be GG, select the correct statement(s):

Options

A

The minimum launch speed required for the mass mm to completely escape the gravitational field of both bodies is 4GML4\sqrt{\frac{GM}{L}}.

B

The minimum launch speed required for the mass mm to completely escape the gravitational field of both bodies is 2GML2\sqrt{\frac{GM}{L}}.

Correct
C

The minimum launch speed required for the mass mm to completely escape the gravitational field of both bodies is 2GML\sqrt{\frac{2GM}{L}}.

D

The total mechanical energy of the mass mm is conserved throughout the motion.

Correct

Topics & Concepts

GravitationGravitation

Step-by-Step Solution

To determine the correct statements, we analyze the motion of the particle of mass mm moving in the gravitational field of two fixed bodies of mass MM.

1. Conservation of Mechanical Energy

The two bodies of mass MM are held fixed at coordinates (−L,0)(-L, 0) and (L,0)(L, 0) along the xx-axis. The gravitational force exerted by these fixed masses on the particle mm is a conservative force: F=−∇U\mathbf{F} = -\nabla U Because there are no non-conservative forces (such as friction or drag) acting on the particle, and the gravitational potential is static (time-independent), the total mechanical energy EE of the particle of mass mm is conserved throughout its motion: E=K+U=constantE = K + U = \text{constant} Thus, statement (D) is correct.

2. Gravitational Potential Energy

The particle of mass mm is projected from the origin (0,0)(0, 0) along the perpendicular bisector (the yy-axis). At the initial position (0,0)(0, 0), the distance from the particle to each of the masses is LL. The initial gravitational potential energy UiU_i of the system (mass mm in the field of the two fixed masses) is: Ui=−GMmL−GMmL=−2GMmLU_i = -\frac{GMm}{L} - \frac{GMm}{L} = -\frac{2GMm}{L}

If the particle is launched with an initial speed vv, its initial kinetic energy is: Ki=12mv2K_i = \frac{1}{2}mv^2

The total initial mechanical energy is: Ei=Ki+Ui=12mv2−2GMmLE_i = K_i + U_i = \frac{1}{2}mv^2 - \frac{2GMm}{L}

3. Escape Velocity

For the particle to completely escape the gravitational field of both bodies, it must reach infinity (y→∞y \to \infty) where the gravitational potential energy vanishes: Uf=lim⁡y→∞(−2GMmL2+y2)=0U_f = \lim_{y \to \infty} \left(-\frac{2GMm}{\sqrt{L^2 + y^2}}\right) = 0

To reach infinity, the kinetic energy at infinity must be non-negative: Kf≥0K_f \ge 0

Applying the principle of conservation of mechanical energy: Ei=EfE_i = E_f 12mv2−2GMmL=Kf+Uf=Kf≥0\frac{1}{2}mv^2 - \frac{2GMm}{L} = K_f + U_f = K_f \ge 0

12mv2≥2GMmL\frac{1}{2}mv^2 \ge \frac{2GMm}{L}

v2≥4GMLv^2 \ge \frac{4GM}{L}

v≥2GMLv \ge 2\sqrt{\frac{GM}{L}}

Thus, the minimum launch speed required for the mass mm to escape is: vmin=2GMLv_{\text{min}} = 2\sqrt{\frac{GM}{L}}

Therefore, statement (B) is correct, while statements (A) and (C) are incorrect.

Conclusion:

The correct options are B and D.

PreviousNext