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This_Username_42 t1_j25bsli wrote

Reply to comment by IsraelZulu in A *dumb* question, for a mixup by Independent-Choice-4

There is a finite amount of potential energy that is stored by you being elevated in a gravitational field. That’s why going up a mountain is harder than going down a mountain, you’re putting energy into climbing, and getting it back when you descend

If you imagine being out in space, and you fell to earth (negating air resistance) — you’d be moving very very fast. When you started, there is a lot of energy stored — you’re at the very top of the “mountain”

That means that you have a certain amount of kinetic energy corresponding to your speed and mass.

When you “hit” earth, and imagine a soft landing where you aren’t obliterated — that kinetic energy is exactly equal to you “climbing” the mountain — I.e. being out in space.

When you want to climb the mountain to get out of earth’s gravitational field, then you need to use that much energy to get up to speed to jump out.

Escape velocity is simply how fast you’d have to start out at to get away from earth, negating air resistance.

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IsraelZulu t1_j25da60 wrote

Ok but we're talking about escape velocity from the moon, for an object flung towards Earth. At a certain distance from the moon, along such a trajectory, Earth's gravity has more influence on an object than lunar gravity, so it starts being pulled away from the moon without needing to expend more energy than is needed to reach that point.

My question is whether the escape velocity from the moon alone greater than the velocity required to reach the point where Earth's gravity can just take over.

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