carl0ymunc8ess
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I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's average distance from the sun was about 5km/s. What was the orbital speed of this material when it was 40,000 AU from the sun (before it fell inward with the collapse of the nebula)? Use the law of conservation of angular momentum. I have the angular momentum = m x v x r. Is this correct?

Answer :

The angular momentum of a particle in orbit is 

l = m v r 

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2" 

m_1 v_1 r_1 = m_2 v_2 r_2 

Assuming that the mass did not change, conservation of angular momentum demands that 

v_1 r_1 = v_2 r_2 

or 

v1 = v_2 (r_2/r_1) 

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have 

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore,  the orbital speed of this material when it was 40,000 AU from the sun is 4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
Parrain

When the material was 40,000 AU from the sun, its orbital speed was 0.004875 km/s.

What was the orbital speed?

Use the law of conservation of angular momentum which is:

Mass of Pluto x Velocity of Pluto x Distance of Pluto = Mass of material x Velocity of material x Distance of material

The mass will not change and so can be removed as a constant. The average distance of Pluto is 39 AU.

Solving gives:

5 x 39 = Velocity of material x 40,000

Velocity of material = 195 / 40,000

= 0.004875 km/s

In conclusion, the speed is 0.004875 km/s.

Find out more on the law of conservation of angular momentum at https://brainly.com/question/7538238.

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