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

Answer 1

Answer: (B)

48 × 10⁸ J

Answer 2

v₁r₁ = v₂r₂ or

v₂ = $\frac { 2 \times 10 ^ { 4 } \times 2.5 \times 10 ^ { 11 } } { 5 \times 10 ^ { 10 } } = 10 ^ { 5 }$

∆KE = $\frac { 1 } { 2 }$ × 10⁸[(10⁵)² – 4 × 10⁸]

= 48 × 10⁸J