HELP. Explain whether the rocket can still go faster from the point of view of an observer on the asteroid.
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HELP. Explain whether the rocket can still go faster from the point of view of an observer on the asteroid.

[From: ] [author: ] [Date: 11-09-14] [Hit: ]
-you have the right idea. at first the exhaust goes backward relative to the asteroid.but at some time the ship, and its exhaust, go FORWARD relative the asteroid. but the ship and most recent fired exhaust always go in opposite directions from the point of view of the shipignoring relativistic speed of light effects imagine it is done in three steps fire 1/3 fuel each time speed of exhaust is the same relative to the engine each firing mass ship = 1 ,......
Imagine that a rocket is launched from an asteroid in deep space and fires its engines until the speed of the rocket is relative to the asteroid is equal to the speed of the rocket’s exhaust. The exhaust ejected by the engine is now at rest with respect to the asteroid. If the engine continue to fire, will the rocket’s speed with respect to the asteroid still increase? (If it does, note that the exhaust will now move in the same direction as the rocket relative to the asteroid) Explain whether the rocket can still go faster from the point of view of an observer on the asteroid.

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you have the right idea. at first the exhaust goes backward relative to the asteroid. but at some time the ship, and its exhaust, go FORWARD relative the asteroid. but the ship and most recent fired exhaust always go in opposite directions from the point of view of the ship

ignoring relativistic speed of light effects

imagine it is done in three steps fire 1/3 fuel each time
speed of exhaust is the same relative to the engine each firing

mass ship = 1 , mass fuel 6 , exist Vel with respect to ship = 1000 speed units

everything observed from the asteroid.

motion with respect to the rest of the universe makes no difference
next to Asteroid pre fire momentum 0, ship and Fuel vel 0

first fire mass ( 1 + 4) V ship forward and 2 x 1000 backward exhaust
coast ship and fuel 4 move forward slowly, exhaust mass 2 moves back negative
second mass ( 1 + 2) V 2 gain over V1 forward and mass 2 at (V1000 - V1) backward

second coast mass ship and fuel 2 remaining forward faster and exhaust backward but less than 1000 v units

third fire ship mass 1 + V3 + V2 forward while remaining exhaust may even be moving forward, certainly much slower than the first exhaust

work the momentum numbers carefully, assume 5 firings if necessary ( i am too lazy to to do the math at almost mid night)
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