Question

A comet travels around the sun in elliptical orbit. Its mass is 10⁸ kg when 2.5 × 10¹¹ m away its speed is 2 × 10⁴ ms^–1. Find the change in KE when it has reached 5 × 10¹⁰ m away from the sun –

• A. 38 × 10⁸ J
• B. 48 × 10⁸ J
• C. 58 × 10⁸ J
• D. 56 × 10⁸ J

Answer 1

Answer: (B)

48 × 10⁸ J

Answer 2

v₁r₁= v₂r₂ Ž v₂ = $\frac { 2 \times 10 ^ { 4 } \times 2.5 \times 10 ^ { 11 } } { 5 \times 10 ^ { 10 } }$ = 10⁵

DKE = $\frac { 1 } { 2 }$ × 10⁸ [(10⁵)² – 4 × 10⁸] = 48 × 10⁸ J