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This may be a dumb question, however I was confused with escape velocity and in the end determined that escape velocity is based off of energy, whereby (i think assuming air resistance and all that is negligible and hence won't do work on the projectile), at that specific escape velocity, the projectile will's kinetic energy is sufficient enough to overcome Earth's gravitational attraction whereby its GPE will be zero. (can someone confirm please?)

However, my main part of the question is that if this whole escape velocity thing is energy dependent, what's stopping us from saying if we fire a projectile directly down at Earth, technically it should have sufficient kinetic energy to escape, but intuitively, obviously this doesn't happen. Why is this the case? (unless it actually does happen that I'm not aware of)

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  • $\begingroup$ "should have sufficient kinetic energy to escape, but intuitively, obviously this doesn't happen", sorry, your intuition is failing you here. You are failing to apply the "assuming air resistance and all that is negligible" bit of your ruleset. Firing down at escape velocity will cause the projectile to escape, assuming it doesn't encounter friction for, say, hitting your foot. Or the core of the Earth, etc. $\endgroup$
    – PcMan
    Commented Nov 20, 2021 at 16:04

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You are correct. Since its value is obtained from an energy argument, escape “velocity” does not have a directional component or dependence - it is really an escape speed, and not a velocity vector at all.

If you fire a ballistic projectile in any direction, if it does not impact the ground, and if it is not acted on by any forces apart from gravity (air drag, friction etc.) then it will “escape” (i.e. enter an unbounded parabolic or hyperbolic trajectory) if its speed is greater than or equal to the escape velocity at the point where you fire it.

If you drill a tunnel through the centre of the Earth, use a vacuum pump to remove the air from it, and then fire a projectile “down” the tunnel at escape velocity or faster then the projectile will exit the other end of the tunnel with a speed that is the same as the firing speed (but upwards) and will not return to Earth. But please don’t try this at home.

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  • $\begingroup$ I suspect if you could get this working at home you would be able to get a pretty good job with your country's space agency :) (or put on a list) $\endgroup$
    – Andrew
    Commented Nov 20, 2021 at 20:53
  • $\begingroup$ Thanks for the reply and everyone else it makes so much more sense now, but I kept thinking about this and had something to add on: I realised that in deriving the escape velocity, we said total energy is zero right, ie kinetic energy decreases as GPE increases to zero, so the KE is being converted to GPE. However, if we fire straight down, the KE isn't being converted to GPE anymore, if anything the GPE decreases, hence can we still say escape velocity would still work? Because doesn't this conflict with an underlying assumption whilst deriving escape velocity? $\endgroup$
    – physicsphil
    Commented Nov 21, 2021 at 1:00
  • $\begingroup$ @physicsphil If you fire the projectile straight down the conversion is initially working in the opposite direction. GPE is being converted into KPE, and the projectile increases speed until it passes the centre of the Earth. After that point GPE starts increasing again, KPE is converted back into GPE, and the projectile slows down. We see a similar two-way exchange between GPE and KPE on a rollercoaster. $\endgroup$
    – gandalf61
    Commented Nov 21, 2021 at 7:36
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The other answers are right. I just wanted to point out that for them to be applicable, you can’t measure projectile velocity in the way you normally might—relative to the lab. If you have a launcher on the equator that will just get a projectile to escape velocity, if you launch east, the projectile will escape, but if you launch west it won’t. This makes it seem like direction matters. But it is only because, in a non-rotating reference frame fixed relative to the center of the earth, the projectile moves faster when fired east: the rotational speed of earth’s surface gets added.

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escape velocity is based off of energy, whereby (i think assuming air resistance and all that is negligible and hence won't do work on the projectile)

This is correct. Escape velocity is based off energy. It is the velocity corresponding to 0 total energy, where gravitational potential energy is referenced to infinity. So a projectile acted on only by gravity will continue out to an infinite distance, never stopping and falling back.

if this whole escape velocity thing is energy dependent, what's stopping us from saying if we fire a projectile directly down at Earth, technically it should have sufficient kinetic energy to escape, but intuitively, obviously this doesn't happen. Why is this the case?

As you said above, we are “assuming air resistance and all that is negligible and hence won't do work on the projectile”. A collision with the ground typically does work on the projectile and is included in “air resistance and all that”. If your projectile, for instance, were made of neutrinos then you could indeed fire it down at the earth and it would pass through without interacting and continue on to escape through the other side of the earth. It really is direction independent, but it does depend on the accuracy of the assumption. A projectile fired downward at escape velocity will escape, provided it does not experience resistance. This is uncommon, but not impossible.

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Yes, the escape velocity works in any direction. The reason you wouldn't escape if you went in the direction of the ground is simply that you'd hit the ground first. If you could somehow move at the escape velocity towards the ground without hitting the ground, you'd still escape.

If you want to actually get anywhere, you have to be able to keep moving at escape speed. So it's usually phrased as "you can move in any direction that's away from the ground" for this reason

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