Halley’s comet moves about the Sun in an elliptical orbit, with its closest approach to the Sun being 0.586 A.U. and its greatest distance from the Sun being 35.5 A.U. (1 A.U. = the Earth-Sun distance). The comet’s speed at closest approach is 49 km/s. What is its speed when it is farthest from the Sun, assuming that its angular momentum about the Sun is conserved?

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JemdetNasr JemdetNasr · Answer 1 · 2019-08-09T08:18:21+02:00

Answer:

0.81 km/s

Explanation:

r₀ = distance at closest approach from sun = 0.586 AU

r = distance at greatest distance from sun = 35.5 AU

v₀ = speed of the comet at closest approach = 49 km/s

v = speed of the comet at greatest distance = ?

Using conservation of angular momentum

m v₀ r₀ = m v r

Inserting the values

(49) (0.586) = v (35.5)

v = 0.81 km/s

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shubhamchouhanvrVT shubhamchouhanvrVT · Answer 2 · 2022-03-19T13:10:27+01:00

The speed of the comet when it is farthest from the sun will be

[tex]V=0.81 \dfrac{km}{sec}[/tex]

It is given that

[tex]r_0[/tex] = distance at closest approach from sun = 0.586 AU

[tex]r[/tex] = distance at greatest distance from sun = 35.5 AU

[tex]v_0[/tex] = speed of the comet at closest approach = 49 km/s

[tex]V[/tex]= speed of the comet at greatest distance =?

Now from the statement that the angular momentum about the sun is conserved

[tex]mv_{0}r_{0}=mVr[/tex]

By putting the values

[tex](49)(0.586)=V(35.5)[/tex]

[tex]V=0.81\ \dfrac{km}{sec}[/tex]

Thus the speed of the comet when it is farthest from the sun will be

[tex]V=0.81 \dfrac{km}{sec}[/tex]

To know more about the Conservation of momentum follow

https://brainly.com/question/7538238