Can the escape velocity be achieved at any angle?

AI Thread Summary
The discussion centers on the concept of escape velocity and whether it can be achieved at any angle. It is established that escape velocity, calculated as vesc = sqrt(2GM/r), does not include an angle term, suggesting that it can theoretically be achieved at any angle. However, it is emphasized that for an object to escape, its velocity must be directed radially outward; otherwise, conservation of angular momentum prevents successful escape. The conversation also touches on the implications of pointing velocity directly at the mass, noting that this scenario does not align with the physical definition of escape velocity. Ultimately, the consensus is that while escape velocity can be achieved at any angle, it is most effective when directed outward.
nhmllr
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So I looked at a neat derivation for what the minimun escape velocity is, and It was pretty clever. Because this is conserved:
KE + PE = 1/2 *mv^2 + -GMm/r
You can find what the velocity would have to be to get to infinity with zero velocity
1/2 *mvesc^2 + -GMm/r = 1/2 *m(0)^2 + -GMm/(infinity) = 0
1/2 *[STRIKE]m[/STRIKE]vesc^2 = GM[STRIKE]m[/STRIKE]/r
vesc = sqrt(2GM/r)
However, there is no "angle" term included in this. Does this mean that it can travel at this velocity at any angle and it will escape? What if the velocity vector is pointed right at the thing it's orbiting?

Thanks
 
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I believe that velocity has to be directed radially outwards. The escape velocity is defined as the velocity required for an object to "get to infinity" with zero remnant velocity. If the velocity is not radially, then that cannot happen from the conservation of angular momentum.
 
Yuqing said:
I believe that velocity has to be directed radially outwards. The escape velocity is defined as the velocity required for an object to "get to infinity" with zero remnant velocity. If the velocity is not radially, then that cannot happen from the conservation of angular momentum.

Nope - it works just fine regardless of angle. Keep in mind that from a conservation of angular momentum point of view, v can still go to zero (as r -> inf).
 
cjl said:
Nope - it works just fine regardless of angle. Keep in mind that from a conservation of angular momentum point of view, v can still go to zero (as r -> inf).

But mathematically, if you pointed the velocity vector straight at the object with mass M, the force would be infinite when r = 0, and the object would be stationary. Right? Or does the velocity also grow to be near infinite, kind of making them "cancel out?"
 
cjl said:
Nope - it works just fine regardless of angle. Keep in mind that from a conservation of angular momentum point of view, v can still go to zero (as r -> inf).

Ah yes, forgot about the fact that r -> inf.

@nhmllr It doesn't really make physical sense to point the velocity towards the object. Escape velocity is defined in terms of the object "escaping" the planet.
 
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