Question

A planet of mass \(m\) moves in an elliptical orbit around the Sun of mass \(M\). Let \(r_{1}\) be the distance at aphelion and \(r_{2}\) be the distance at perihelion. If \(v_{1}\) and \(v_{2}\) are the velocities of the planet at these points, then ratio of kinetic energy at aphelion to kinetic energy at perihelion is

• A. \(\frac{r_{1}}{r_{2}}\)
• B. \(\frac{r_{1}^{2}}{r_{2}^{2}}\)
• C. \(\frac{r_{2}}{r_{1}}\)
• D. \(\frac{r_{2}^{2}}{r_{1}^{2}}\)

Hint:

Angular momentum is conserved, so the product of speed and distance is the same at both points. Then square the speed ratio.

Answer 1

The Correct Option is D

Step 1: Areal speed idea
Kepler's second law says the planet moves slower when farther away, so $v_{2}$ is greater than $v_{1}$ at aphelion.

Step 2: Eliminate options
Kinetic energy depends on the square of speed. Options (A) and (C) have a single power, so they cannot be right.

Step 3: Choose between (B) and (D)
With $v\propto\dfrac{1}{r}$, $K\propto\dfrac{1}{r^{2}}$, so $\dfrac{K_{1}}{K_{2}}=\dfrac{r_{2}^{2}}{r_{1}^{2}}$, which is below 1. Option (D) matches.

Final Answer:
The ratio equals r2 squared by r1 squared. \[ \boxed{\text{(D) }\dfrac{r_{2}^{2}}{r_{1}^{2}}} \]