Question

A particle of mass m, having an energy E and angular momentum L, is in a parabolic trajectory around a planet of mass M. If the distance of the closest approach to the planet is r_m, which of the following statement(s) is(are) true?
(G is the Gravitational constant)

• A. E>0
• B. E=0
• C. \(𝐿 = \sqrt{2𝐺𝑀𝑚^2𝑟_m}\)
• D. \(𝐿 = \sqrt{2𝐺𝑀^2mr_m}\)

Answer 1

The Correct Option is B, C

Parabolic Trajectory and Total Mechanical Energy 

For a parabolic trajectory, the total mechanical energy of the particle is zero:

E = K + U = 0

where:

• K = (1/2) mv² is the kinetic energy.
• U = -GMm / r is the gravitational potential energy.

Thus, statement (B) is true.

Angular Momentum

The angular momentum L is given by:

L = m v_t r

where v_t is the tangential velocity at the closest approach r = r_m.

Velocity at Closest Approach

For a parabolic trajectory, the velocity at the closest approach r_m satisfies:

(1/2) m v² - (GMm / r_m) = 0

Solving for v:

v = √(2GM / r_m)

Since the motion is tangential at the closest approach:

v_t = √(2GM / r_m)

Substituting into Angular Momentum Expression

Substituting v_t into the angular momentum expression:

L = m v_t r_m = m √(2GM / r_m) r_m = √(2GM m² r_m)

Thus, statement (C) is true.

Evaluating Statements

• Statement (A) is false because the total energy E = 0 for a parabolic trajectory.
• Statement (D) is false because the correct angular momentum expression is L = √(2GM m² r_m), not L = √(2GM² m r_m).

Conclusion

The correct statements are (B) and (C).