Question

Two particles of mass m and 2m are initially at rest at a separation of 2d. Find their velocity of approach due to gravitational attraction, when separation is d.​

Answer 1

Answer:

10 m/s but as per ke it can change to 4.47 m/s

Answer 2

Given:

The mass of the two particles is m and 2m.

The distance between the particles is 2d.

To Find:

The velocity of approach due to gravitational attraction, when separation is d.​

Solution:

Initial kinetic energy, KE(initial) = 0

Initial potential energy, PE(initial) = -G(m)(2m) / 2d

= -2Gm² / 2d = -Gm² / d

Final kinetic energy = 1/2(m)v₁² + 1/2(2m)v₂²

Final potential energy = -G(m)(2m) / d = -2Gm²/d.

Since, there is no external force,

Initial potential energy = Final potential energy

m(0) + 2m(0) = m(v₁) + 2m(-v₂)

mv₁ = 2mv₂

which is, v₁ = 2v₂

By law of energy conservation,

Initial kinetic energy + Initial potential energy = Final kinetic energy + Final potential energy.

0 + (-Gm²/d) = 1/2(m)v₁² + 1/2(2m)v₂² +(-2Gm²/d)

since v₁ = 2v₂,

(2Gm²/d) - (Gm²/d) = 1/2m(2v₂)² + 1/2(2m)v₂².

Gm/d = 2v₂² + v₂².

Gm/d = 3v₂².

v₂ =$\sqrt{\frac{Gm}{3d} }$

now, v₁ = 2v₂ = $2\sqrt{ \frac{Gm}{3d}$

Hence, the velocities of particles m and 2m are, $2\sqrt{\frac{Gm}{3d} }$ and $\sqrt{\frac{Gm}{3d} }$ .