Halley’s comet moves about the Sun in an el- liptical orbit, with its closest ap

Question

Halley’s comet moves about the Sun in an el- liptical orbit, with its closest approach to the Sun being 0.59 A.U. and its greatest distance from the Sun being 35.9 A.U. (1 A.U. = the Earth-Sun distance). The comet’s speed at closest approach is 47 km/s. What is its speed when it is farthest from the Sun, assuming that its angular momen- tum about the Sun is conserved? Answer in units of km/s. Halley’s comet moves about the Sun in an el- liptical orbit, with its closest approach to the Sun being 0.59 A.U. and its greatest distance from the Sun being 35.9 A.U. (1 A.U. = the Earth-Sun distance). The comet’s speed at closest approach is 47 km/s. What is its speed when it is farthest from the Sun, assuming that its angular momen- tum about the Sun is conserved? Answer in units of km/s.

Explanation / Answer

m1=0.59*1.67*10-27kg v1=471000m/s m2=35.9*1.67*10-27kg v2 Applying the law of conservation of momentum we have as, m1v1=m2v2 From this we get the velocity at farthest point. Hence we get by it. "Hope this helps!Best of luck for the rest of yourcoursework."

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.