Time to fall from 200 miles

[Algr]

TL;DR

Freefall calculator gives strange answer. Gravitational acceleration.

I tried out this free fall calculator, and it says that it would take 4.3 minutes to fall to Earth from 200 miles up. (The altitude of the space shuttle, but no horizontal velocity.) This intuitively seems very fast to me, even assuming no atmosphere. Am I using it wrong?

It also gave me a very wrong answer for the moon falling to Earth. I suspect that this is because it does not adjust the gravitational acceleration to account for distance. Is that right? I think the correct answer should be about a week. (1/4 the time it takes for the moon to orbit the Earth.)

Finally, since the Moon is big enough to attract the Earth, would it fall faster than a 1 pound object at the same distance?
Thank you!
https://www.omnicalculator.com/physics/free-fall

 

[bob012345]

For the first question, assuming $g$ doesn’t vary much from its usual value on the surface, 4.3 minutes is about right. This assumes the object dropped is not in orbit, it just drops from rest 200 miles high. The Moon is in orbit so that is an entirely different situation. Think about what being in orbit means.

 

[Nugatory]

tried out this free fall calculator, and it says that it would take 4.3 minutes to fall to Earth from 200 miles up.

You can check this for yourself. From the SUVAT equations we have $s=\frac{1}{2}at^2$, it is not unreasonable to treat $a$ as constant over this distance (for extra credit, estimate the error resulting from using this approximation), solve for $t$.

 

[D H]

You can check this for yourself. From the SUVAT equations we have $s=\frac{1}{2}at^2$, it is not unreasonable to treat $a$ as constant over this distance (for extra credit, estimate the error resulting from using this approximation), solve for $t$.

On the other hand, assuming constant acceleration all the way out to the Moon is unreasonable. This assumption leads to 2.4592 hours. Perhaps the calculator is using a non-standard value of $g$ to arrive at 2.4595 hours. It shouldn't be arriving at that value at all; the assumptions leading to this value are invalid.

 

[Nugatory]

On the other hand, assuming constant acceleration all the way out to the Moon is unreasonable.

Yes, of course. I was responding to the first question, the one about the 200 mile fall - and for that one we can reasonably advise OP to calculate instead of relying on "intuitively feels very fast".

This assumption leads to 2.4592 hours. Perhaps the calculator is using a non-standard value of $g$ to arrive at 2.4595 hours.

When I see a discrepancy in the fifth significant digit I generally suspect careless rounding, limited precision arithmetic, or the like. But without access to the source code, it's hard to say.... I've just seen enough naive floating point to be reflexively skeptical

It shouldn't be arriving at that value at all; the assumptions leading to this value are invalid.

Yep.

 

[D H]

It also gave me a very wrong answer for the moon falling to Earth. I suspect that this is because it does not adjust the gravitational acceleration to account for distance. Is that right?

That is correct. Sometimes you do get what you pay for when you use a free site, which is nothing.

I think the correct answer should be about a week. (1/4 the time it takes for the moon to orbit the Earth.)

It's a bit less than a week; I get 4.83 days.

Finally, since the Moon is big enough to attract the Earth, would it fall faster than a 1 pound object at the same distance?

The correct version of Kepler's third law is $a^3\omega^2 = G(M+m)$, as opposed to $GM$ which is what one obtains ignoring the mass of the smaller body. In many cases, the smaller body's mass can be ignored. That is not the case with the Moon, whose mass is 0.0123 Earth masses.

 

[Algr]

Thank you!

Yes, in both cases I was assuming no horizontal motion. I was putting something together based on the observation that most people think that gravity stops in space. They don't realize that the horizontal speed is why things don't fall back to Earth from orbit.

 

[PeroK]

Finally, since the Moon is big enough to attract the Earth, would it fall faster than a 1 pound object at the same distance?

If you did an experiment where something the size and mass of the moon were "dropped" onto something the size and mass of the Earth, then the Earth would move significantly during the collision. The collision would take less time to happen than for a small object, where the Earth would effectively not move.

 

[Ibix]

If you did an experiment where something the size and mass of the moon were "dropped" onto something the size and mass of the Earth, then the Earth would move significantly during the collision. The collision would take less time to happen than for a small object, where the Earth would effectively not move.

To expand a little on this for the OP, there are no external forces here so the center of mass of the system does not move. For the Earth and a rock, the difference between the center of mass of the Earth and the center of mass of the system is immeasurably small and for all practical purposes the Earth doesn't move. At the other extreme, two identical planets have a center of mass half way between them and both will cover half the distance, initially each with the same acceleration the rock had, but as they both move the rate of change of acceleration will be higher.

Earth and the Moon is somewhere between the two extremes. Yes, the Earth will move in this case.