A planet moves around the sun (mass $M$) in an elliptical orbit. Its maximum distance from the sun is three times its minimum distance $r_p$. The speed of the planet when it is closest to the sun is
Answer: (C) $\sqrt{\dfrac{3GM}{2r_p}}$
Concept. At the nearest and farthest points the velocity is perpendicular to the radius. Use conservation of angular momentum and of mechanical energy between these two points.
Step 1. Angular momentum. With $r_a = 3r_p$:
$$v_pr_p = v_a(3r_p) \;\Rightarrow\; v_a = \frac{v_p}{3}$$
Step 2. Energy.
$$\frac12v_p^2 - \frac{GM}{r_p} = \frac12\cdot\frac{v_p^2}{9} - \frac{GM}{3r_p}$$
Step 3. Solve.
$$\frac12v_p^2\cdot\frac89 = \frac{2GM}{3r_p} \;\Rightarrow\; v_p^2 = \frac{3GM}{2r_p}$$
Answer. $\sqrt{\dfrac{3GM}{2r_p}}$ (Option C).
Solution by Sreeraj P, M.Sc Physics