Question

The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are K_A , K_B and K_C , respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then

• A. K_A<K_B<K_C
• B. K_B<K_A<K_C
• C. K_A>K_B>K_C
• D. K_B>K_A>K_C

Answer 1

The Correct Option is C

The problem revolves around the kinetic energy of a planet that is in an elliptical orbit around the Sun. To solve this and determine the order of kinetic energies at points A, B, and C, we must understand Kepler's Laws of Planetary Motion, particularly how velocity and distance from the Sun affect kinetic energy.

1. In an elliptical orbit, the velocity of a planet is highest at the point closest to the Sun (perihelion) and lowest at the point farthest from the Sun (aphelion). This variation is due to the conservation of angular momentum and Kepler's Second Law.
2. The kinetic energy (K) is given by K = \frac{1}{2}mv^2, where m is the mass and v is the velocity of the planet. Thus, higher velocity leads to greater kinetic energy.
3. Based on the arrangement, C is at the perihelion (closest to the Sun), so kinetic energy is the greatest at C.
4. A is at the aphelion (farthest from the Sun), so kinetic energy is the least at A.
5. B is positioned such that it lies at a level of intermediate kinetic energy between A and C, given it is neither at the perihelion nor at the aphelion.

Hence, the order of the kinetic energies is K_C > K_B > K_A, meaning the correct answer is K_A>K_B>K_C.