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brey | 9 years ago
a rocket takes off. let's say it's going upwards at 1 m/s at some short time after takeoff. why can't it then reduce (not to zero) acceleration such that it stays going upwards at 1 m/s all the way out of the gravity well? don't aim for escape velocity, keep the velocity at exactly 1 m/s until you're out.
yes, I know - Tsiolkovsky, reaction mass, delta-v etc etc clearly that's wildly impractical
but MUST the rocket achieve its 'escape velocity' in order to escape? why can't we have a casual stroll out of the gravity well?
escape velocity applies for an unpowered vehicle, right? throw a rock, it's gotta be going at least at the escape velocity or it won't escape. but throw a rock that's propelling upwards itself all the way out ...
if I want to climb stairs, I propel myself gradually on every step, I don't require that I launch myself off step 1 with the speed to get me to the top.
what am I missing here?
fessick|9 years ago
Escape velocity is essentially the velocity an object must start out with to move towards an asymptotically infinite distance from the gravity well without additional thrust. That last bit is what makes it escape velocity. The gravity well will slow the object to near zero velocity but the object's distance will eventually overcome the gravity before it reaches zero or negative velocity.
Your example of moving at a constant speed of 1 m/s is only achievable with constant thrust to cancel gravity, thereby stopping the object from slowing during its ascent. If the object has achieved its escape velocity, no additional thrust is needed. One is not better than the other, it's just a matter of what is more practical in terms of fuel or travel time.
brey|9 years ago
and if it's not, taking that to its logical conclusion - why do we NEED to go faster than the speed of light to escape from inside the schwarzschild radius of a black hole? can't we leave under constant thrust at 1 m/s?
unknown|9 years ago
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