Question

A partical of mass m is thrown in a parabolic path towards earth having mass M and radius R. It comes to a greatest approach distance R/2 from surface of earth. Magnitude of total energy change required to make the particle moving in an elliptical path having eccentricity 1/2, with the greatest approach point either aphelion or perihelion are :

• A. $\frac{1}{3} \frac{GMm}{R}$
• B. $\frac{1}{6} \frac{GMm}{R}$
• C. $\frac{1}{2} \frac{GMm}{R}$
• D. $\frac{GMm}{R}$

Answer 1

Answer: (B), (C)

B, C

Answer 2

The particle is initially moving in a parabolic path towards the Earth. For a parabolic trajectory under a central gravitational force, the total mechanical energy of the particle is zero.

$E_{initial} = 0$.

The particle's point of greatest approach to the Earth's surface is $R/2$. This means the closest distance from the center of the Earth is $r_{min} = R + R/2 = 3R/2$. This point is the periapsis of the parabolic path.

The particle's trajectory is changed to an elliptical path with eccentricity $e = 1/2$. The point $r = 3R/2$ is given to be either the aphelion (apoapsis) or the perihelion (periapsis) of the new elliptical orbit. Let $a$ be the semi-major axis of the elliptical orbit. The total energy of a particle in an elliptical orbit is given by $E_{final} = -\frac{GMm}{2a}$.

Case 1: The point $r = 3R/2$ is the perihelion of the elliptical orbit.

The distance of the perihelion from the center is $r_p = a(1-e)$.

Given $r_p = 3R/2$ and $e = 1/2$.

$3R/2 = a(1 - 1/2) = a(1/2)$

$a = 3R$.

The total energy of this elliptical orbit is $E_{final, 1} = -\frac{GMm}{2a} = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R}$.

The change in energy required is $\Delta E_1 = E_{final, 1} - E_{initial} = -\frac{GMm}{6R} - 0 = -\frac{GMm}{6R}$.

The magnitude of the energy change is $|\Delta E_1| = \frac{GMm}{6R}$.

Case 2: The point $r = 3R/2$ is the aphelion of the elliptical orbit.

The distance of the aphelion from the center is $r_a = a(1+e)$.

Given $r_a = 3R/2$ and $e = 1/2$.

$3R/2 = a(1 + 1/2) = a(3/2)$

$a = R$.

The total energy of this elliptical orbit is $E_{final, 2} = -\frac{GMm}{2a} = -\frac{GMm}{2(R)} = -\frac{GMm}{2R}$.

The change in energy required is $\Delta E_2 = E_{final, 2} - E_{initial} = -\frac{GMm}{2R} - 0 = -\frac{GMm}{2R}$.

The magnitude of the energy change is $|\Delta E_2| = \frac{GMm}{2R}$.

The question asks for the magnitudes of the total energy change required, given that the point $r=3R/2$ can be either the aphelion or the perihelion. We found two possible magnitudes: $\frac{GMm}{6R}$ and $\frac{GMm}{2R}$.

Therefore, options (B) and (C) are correct.