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

To solve this problem, we need to understand the dynamics of a planet's movement in an elliptical orbit according to Kepler's laws of planetary motion. The essential concept here is the conservation of angular momentum and energy.

The kinetic energy of a planet in an elliptical orbit varies with its distance from the Sun due to the conservation of angular momentum and energy. The key points in an elliptical orbit are:

• A (Perihelion): The point of closest approach to the Sun, where the kinetic energy is highest.
• C (Aphelion): The point of farthest distance from the Sun, where the kinetic energy is lowest.
• B: A point at which the line from the Sun is perpendicular to the major axis AC.

According to the conservation of angular momentum:

\(mvr = \text{constant}\)

Here, \(m\) is the mass of the planet, \(v\) is the velocity, and \(r\) is the distance from the Sun.

At perihelion (A), the planet moves fastest, and at aphelion (C), it moves slowest. Therefore, the order of kinetic energies is:

• K_A: Maximum at A (perihelion).
• K_B: Intermediate at B.
• K_C: Minimum at C (aphelion).

Thus, the correct order of kinetic energies is:

K_A > K_B > K_C

Therefore, the correct answer is K_A > K_B > K_C.