A planet revolves about the sun in elliptical orbit of semi major axis 2 × 10¹² m. The areal velocity of the planet when it is nearest to the sun is 4.4 × 10¹⁵ m²/s. The least distance between planet and the sun is 1.8 × 10¹² m. Then find the minimum speed of the planet in km/s.
Physics · Mixed Chapter
JEE Mains

Question

A planet revolves about the sun in elliptical orbit of semi major axis 2 × 10¹² m. The areal velocity of the planet when it is nearest to the sun is 4.4 × 10¹⁵ m²/s. The least distance between planet and the sun is 1.8 × 10¹² m. Then find the minimum speed of the planet in km/s.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Kepler’s second law gives constant specific angular momentum h=2(dA/dt)=8.8×10¹⁵ m²/s. Step 2: The planet is slowest at aphelion. Its aphelion distance is rₐ=2a−rₚ=4×10¹²−1.8×10¹²=2.2×10¹² m. Step 3: At aphelion v is transverse and v_min=h/rₐ=(8.8×10¹⁵)/(2.2×10¹²)=4×10³ m/s. Step 4: Minimum speed=4 km/s.