From a uniform solid spherical planet of radius R,density a, a solid spherical material of radius 2R/3 is scooped out to leave an eccentric hole as shown.Acceleration due to gravity on the surface of the planet before scooping is g and mass of planet left after scooping is M. Aparticle of same mass M is kept at the end of the eccentric hole as shown in the figure and the system id released from rest. Assuming head-on elastic collision between particle and the planet and particle and Newton’s law of gravitation to be valid, after what time will the particle come back to the initial position for the first time?

To solve this problem, we need to analyze the gravitational interactions between the planet and the particle after the eccentric hole has been created. The key here is to understand how gravity behaves in this modified system and how the particle will move under the influence of gravitational forces.

Understanding the System

Initially, we have a uniform solid spherical planet with radius R and density ρ. The mass of the planet can be calculated using the formula:

• Mass of the planet (M): M = ρ × (4/3)πR³

When a solid spherical material of radius 2R/3 is scooped out, we need to consider the mass of the removed material as well. The mass of the scooped-out sphere (let's call it M_removed) is given by:

• Mass of the removed sphere (M_removed): M_removed = ρ × (4/3)π(2R/3)³ = ρ × (4/3)π(8R³/27) = (32/81)πρR³

After scooping out the material, the mass of the remaining planet becomes:

• Remaining mass (M_remaining): M_remaining = M - M_removed = (1 - 32/81)M = (49/81)M

Gravitational Force on the Particle

Now, consider the particle of mass M placed at the end of the eccentric hole. The gravitational force acting on the particle due to the remaining mass of the planet can be calculated using Newton's law of gravitation:

• Gravitational force (F): F = G(M_remaining * m) / r²

Here, G is the gravitational constant, m is the mass of the particle, and r is the distance from the center of the planet to the particle. Since the particle is at the end of the hole, r will be less than R, specifically:

• Distance (r): r = R - (2R/3) = R/3

Motion of the Particle

The particle will experience a restoring force due to gravity, which will cause it to oscillate back and forth. The motion can be modeled as simple harmonic motion (SHM). The effective gravitational acceleration (g_eff) acting on the particle can be derived from the gravitational force:

• Effective gravitational acceleration (g_eff): g_eff = G(M_remaining) / (R/3)² = 9G(M_remaining) / R²

Substituting M_remaining into this equation gives:

• g_eff: g_eff = 9G(49/81)M / R² = (441/81)(GM/R²)

Since g = GM/R², we can express g_eff in terms of g:

• g_eff: g_eff = (441/81)g

Calculating the Period of Oscillation

For a particle undergoing simple harmonic motion, the period (T) is given by:

• Period (T): T = 2π√(L/g_eff)

In our case, L is the distance from the equilibrium position to the maximum displacement, which is R/3. Thus, we have:

• Period (T): T = 2π√((R/3) / (441/81)g) = 2π√((81R) / (1323g)) = 2π√(R/(16.33g))

Time to Return to Initial Position

The particle will return to its initial position for the first time after completing one full oscillation, which is equal to the period T. Therefore, the time taken for the particle to return to its initial position after being released is:

• Time (t): t = T = 2π√(R/(16.33g))

In summary, the time it takes for the particle to return to its initial position after being released from the end of the eccentric hole is given by the formula above, which incorporates the effects of the gravitational changes due to the scooped-out material.