[brey]

this one always bugged me a little

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]

"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." <-- that will totally work, but isn't escape velocity.

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]

yes - my question is whether escape velocity is necessary to leave a gravity well.

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?

[PeterWhittaker]

There are a few things to tease apart in this. First, maintaining a constant speed of 1 m/s requires constant application of force to overcome the forces working against you. On a road surface, these are air resistance, rolling resistance, etc. It takes very little force to oppose these at this speed (wind resistance increases with the square of speed, so the faster you go, the more force you need to keep going that fast, which is why gas mileage drops as you move above your engine's optimum operating range).

Heading up from Earth, you are facing a deceleration of 9.8 m/s/s, so you need to apply enough force to counter that. The amount of force depends on payload, e.g., a little plastic drone don't need so much, a Saturn V carrying something to high orbit needs a whole lot.

Escape velocity is the velocity at which no further thrust is required. That's the key point: The object is travelling away from the well so fast that it is already going fast enough to overcome all of the rest of the gravitational drag it will experience until it hits micro/zero-G.

Rockets need constant thrust (constant force and acceleration) to overcome the constant force and acceleration of gravity).

If you accelerated a human being to escape velocity at the surface of the Earth, you would concuss them to death (and likely organ damage them to death).

A photon starts life at C, whatever C is in its current medium.

[brey]

Understood - I appreciate why light, or any unaccelerated object, cannot leave a black hole.

But does that same restriction prevent a rocket which wants to coast at 1m/s - under constant acceleration - from the surface to infinity?

I.e.: is the schwartzchild radius only a point of no return to light, not rockets?

And if not a hard limit to rockets - oh, I guess that's fun to know :-)

(Yes clearly utterly impractical - trying to understand if there's any fundamental reasons at play here)

[al452]

Yes, the fundamental reason is conservation of energy. If there is not enough energy stored in the rocket's fuel to escape the gravity well, it doesn't matter whether you expend it all at once (and fail to reach escape velocity) or expend it over time as in your example. "Escaping the gravity well" from any point takes a well-defined amount of energy (which we call "gravitational potential energy") so your fuel use strategy won't help you in the end.

[fessick]

Yep. Along similar lines of thought, there's the issue that the equations and laws governing classical mechanics fall apart in such settings.

Even so, assuming that we can apply classical thought to the problem: the simplest roadblock to rockets escaping from a black hole is that their change in mass per time unit multiplied by their velocity must at least partially exceed the gravitational pull.

Either you utilize ungodly amounts of fuel per second or you already have a crazy-high velocity or some combination of both. The math just makes it infeasible. There's limits on mass you can have and eject, and limits on attainable speed.