JEE Main 2024 · 5 April · morning · Single correct · +4 / −1
Match List I with List II :
(Where a= radius of planet orbit, r= radius of planet, M= mass of Sun, m= mass of planet)
Choose the correct answer from the options given below :
| LIST I | LIST II | ||
|---|---|---|---|
| A. | Kinetic energy of planet | I. | -GMm / a |
| B. | Gravitation Potential energy of sun-planet system | II. | GMm / 2 a |
| C. | Total mechanical energy of planet | III. | Gm/r |
| D. | Escape energy at the surface of planet for unit mass object | IV. | -GMm / 2 a |
- (a) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
- (b) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
- (c) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
- (d) (A)-(I), (B)-(IV), (C)-(II), (D)-(III)
Answer and solution
Answer (a) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Solution
Physics sir · 4 steps
Get the kinetic energy from gravity supplying the centripetal force, and the other three follow.
- Kinetic energy. The Sun's pull supplies the centripetal force: GMm/a² = mv²/a, so mv² = GMm/a. That makes KE = 1/2mv² = GMm/2a, which is positive.
- Potential and total. The potential energy of the Sun-planet system is -GMm/a. Adding the two gives the total energy.
E = GMm/2a - GMm/a = -GMm/2a
- Escape energy. On the planet's surface, an object of unit mass has potential energy -Gm/r, where m is the planet's mass. To escape, it needs exactly enough energy to bring this to zero, which is Gm/r.
Quantity Expression List II A. Kinetic energy +GMm/2a II B. Potential energy -GMm/a I C. Total energy -GMm/2a IV D. Escape energy (unit mass) Gm/r III - So. The match is A-II, B-I, C-IV, D-III, which is option (a).
The idea to keep. For any circular orbit, KE : PE : total energy = 1 : -2 : -1. Find the KE, then double it with a minus sign for PE, and flip its sign for the total energy.
Answer from the NTA final answer key. Solution by PrepWiser.