HRDBMW t1_j24nu7q wrote on December 29, 2022 at 5:10 PM

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

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 on December 29, 2022 at 5:12 PM

So, you're saying the baseball is too small for Earth's gravity to be relevant here?

lellololes t1_j24r4q6 wrote on December 29, 2022 at 5:32 PM

No, they are saying that the baseball is too small for the baseball's gravity to be relevant here.

IsraelZulu t1_j24rn6d wrote on December 29, 2022 at 5:35 PM

I fail to see how that had anything to do with my question then.

lellololes t1_j24tggk wrote on December 29, 2022 at 5:47 PM

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 on December 30, 2022 at 6:35 AM

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.