Submitted by Independent-Choice-4 t3_zy9lu2 in space
IsraelZulu t1_j24kb55 wrote
Reply to comment by EasyReader in A *dumb* question, for a mixup by Independent-Choice-4
That's about 5,324 MPH for Americans.
I'm curious about what this "escape velocity" actually means though. Is that just the raw velocity needed to escape lunar orbit as if the moon was the only thing in the universe, or is it just what you need to get out past L1 so that Earth's gravity becomes dominant enough to pull you away?
absoluteally t1_j24mxq3 wrote
Escape velocity is contextual so depends where you are and how fast you are already going. I this case i believe they are starting from static in the surface(usually what is meant).
What it actually means is for the moon the escape velocity is the velocity change you would need from where you are to reach infinity with no excess velocity. Assuming there is nothing else in the universe.
But given they way gravity tails off escape velocity is often not meaningfully different from the velocity to leave the bodies dominant influence.
zoicyte t1_j24x0mn wrote
*contextual relative to the object you're trying to escape.
you need more speed to escape from the gravity of the earth than the moon.
the escape velocity to exit the solar system is really, really high, because you're trying to escape the pull of the sun itself. to date i think we have only launched 4 probes that have achieved the solar system's escape velocity (pioneer 10/11, voyager 1/2). it's possible the new horizons craft has reached the escape velocity for the solar system, it's certainly truckin'.
rondonjon t1_j24m5ul wrote
Even we know that anything moving at multiple kilometers a second is fast. But thanks.
ExNihiloish t1_j24n5i7 wrote
Yep. I measure everything as either fast or not fast. Absolute numbers are irrelevant.
rondonjon t1_j24o6pd wrote
Well, there is also slow, which is not quite the same as not fast.
This_Username_42 t1_j25bsli wrote
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.
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.
HRDBMW t1_j24nu7q wrote
The escape velocity is dependent on the mass of the two objects close to each other. For one object being much more massive, like a moon and a baseball, the mass of the smaller object can be ignored. There is more roundoff error and truncation error in the math.
IsraelZulu t1_j24o1zu wrote
So, you're saying the baseball is too small for Earth's gravity to be relevant here?
lellololes t1_j24r4q6 wrote
No, they are saying that the baseball is too small for the baseball's gravity to be relevant here.
IsraelZulu t1_j24rn6d wrote
I fail to see how that had anything to do with my question then.
lellololes t1_j24tggk wrote
I was clarifying what they were saying. That's all.
Given the distance earth is from the moon, the gravity it imparts on a small mass like a baseball is also going to be a rounding error on a rounding error. It isn't nothing but it may as well be.
What they are saying is that if you have an object that is big enough to impart some gravitational pull it would then start affecting escape velocity for that object. E.g. The escape velocity of Phobos is 25 miles per hour or something like that ( it's big enough to have some gravitational effect). The escape velocity of Phobos from the surface of the moon would likely be a bit different than that of a baseball.
It all fits together, I don't think they misunderstood you.
HRDBMW t1_j27rlne wrote
Not exactly. What I am saying is that the mass of the ball is mostly irrelevant, and the only relevant mass for calculations of escape velocity are massive bodies very close to the ball. In this case, just the moon. The escape velocity is determined by the moon mass, not the position of other masses.
undercoveryankee t1_j24urhq wrote
The two-body escape velocity (if the moon and the projectile were the only things in the universe, would the projectile’s trajectory be unbounded?) is easy to define and calculate, so when you see a single number quoted as “the” escape velocity from a location it’s usually using the two-body definition.
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