The kinetic energies of a planet in an elliptical orbit about the Sun, at positions, A , B and C are K A , K…

The kinetic energies of a planet in an elliptical orbit about the Sun, at positions, A, B and C are KA, KB and KC, respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shows in the figure. Then
  1. $K_{B} < K_{A} < K_{C}$
  2. KA>KB>KC
  3. $K_{A} < K_{B} < K_{C}$
  4. KB>KA>KC

Solution

By angular momentum conservation about the sun, $L = I \omega = \text{constant}$ $KE = \frac{L^2}{2I}$ $\because S_A \angle S_B \angle S_C$ $\Rightarrow I_A < I_B < I_C$ $KE_A > KE_B > KE_C$

Asked in: NEET 2018

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