Suppose astronomers watching the planet Jupiter observe an asteroid almost colli

Question

Suppose astronomers watching the planet Jupiter observe an asteroid almost collide with the massive planet. The asteroid is far away but initially had a trajectory that would give it a perpendicular distance from Jupiter of r_i = 125000 km. When the asteroid is viewed as it passes the planet it is at a height of 14000 km above the surface of Jupiter. The radius of Jupiter is 71490 km and its mass is 1.9 times 10^27 kg. The mass of the asteroid is later determined to be 5250kg. (a)What is the asteroid's velocity as it passes 14000 km above the surface of the planet? (b) Directly behind the first asteroid is a second asteroid that is traveling on the same line as the first, but is ten times more massive. How will the final velocity of the second asteroid be different than the first, smaller asteroid? Why? (c)Assume that you have the technology to control the asteroid's velocity. What would the velocity of the first asteroid need to be at 14000 km above the surface of Jupiter for it to be in a stable circular orbit? d) Assuming that same technology, what would the minimum velocity of the asteroid need to be at 14000 km above the surface of Jupiter for it to escape?

Explanation / Answer

3)

ri = initial distance = 1.25 x 108 m

rf = final distance = 14000 + 71490 = 8.55 x 107 m

m = mass of asteroid = 5250 kg

Vi = inital velocity

Vf = final velocity

using conservation of angular momentum

m Vi ri = m Vf rf

Vi (1.25 x 108) = Vf (8.55 x 107)

Vf = 1.46 Vi

Vi = Vf/1.46                       eq-1

using conservation of energy

energy at far away = energy at height 14000 km

(0.5) m Vi2 = (0.5) m Vf2 - GMm/rf

(0.5) Vi2 = (0.5) Vf2 - GM/rf

(0.5) (Vf/1.46)2 = (0.5) Vf2 - (6.67 x 10-11) (1.9 x 1027)/(8.55 x 107)

Vf = 7.5 x 104 m/s

b)

mass of the asteroid did not play any role in calculation of speed at height above the planet since mass "m" got cancelled out hence the final velocity for more massive asteroid will remain same as that of light asteroid.

c)

for stable orbit:

orbital velocity = sqrt(GM/(rf)) = sqrt((6.67 x 10-11) (1.9 x 1027)/(8.55 x 107)) = 3.85 x 104 m/s

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