Question

Consider a classical particle subjected to an attractive inverse-square force field. The total energy of the particle is \( E \) and the eccentricity is \( \epsilon \). The particle will follow a parabolic orbit if

• A. \( E>0 \) and \( \epsilon = 1 \)
• B. \( E<0 \) and \( \epsilon<1 \)
• C. \( E = 0 \) and \( \epsilon = 1 \)
• D. \( E<0 \) and \( \epsilon = 1 \)

Hint:

For a parabolic orbit, the total energy \( E = 0 \) and the eccentricity \( \epsilon = 1 \). This represents the boundary between elliptical and hyperbolic orbits.

Answer 1

The Correct Option is C

Step 1: Understanding the conditions for parabolic orbits.
In orbital mechanics, the parabolic orbit corresponds to a specific case where the total energy \( E \) is zero and the eccentricity \( \epsilon \) equals 1. This condition applies to orbits that are at the threshold between bound elliptical orbits and unbound hyperbolic orbits.

Step 2: Analyzing the options.
(A) \( E>0 \) and \( \epsilon = 1 \): Incorrect. Positive energy with eccentricity 1 corresponds to a hyperbolic orbit, not a parabolic one.
(B) \( E<0 \) and \( \epsilon<1 \): Incorrect. Negative energy with eccentricity less than 1 corresponds to an elliptical orbit.
(C) \( E = 0 \) and \( \epsilon = 1 \): Correct. For a parabolic orbit, the total energy is zero and the eccentricity is 1.
(D) \( E<0 \) and \( \epsilon = 1 \): Incorrect. Negative energy with eccentricity 1 corresponds to a bound hyperbolic orbit.

Step 3: Conclusion.
The correct answer is (C) because the conditions for a parabolic orbit are \( E = 0 \) and \( \epsilon = 1 \).