Free fall to earth from lunar distance, how long does it take?

[Kerbin Dallas Multipass]

Wondered how long it would take the Moon to fall down.

How would I look up such a thing on this interwebz thingy? Or do I have to do the maths myself (don't know how).

This silly and hypothetical question popped up in my head. I would have guessed it takes a couple of days or maybe a week, but the moon is so far out that earth's gravity is only something like 44mm/s². So this could take a while longer.

[Dust]

Sounds like a question for http://what-if.xkcd.com/

[Brotoro]

4.83 days, if you stopped its sideways motion instantly and let it fall.

[goldenpeach]

Brotoro, how can we calculate the time it would take(taking in account the change of gravity)?

It's not that I don't trust you but I want to calculate something else that would need that equation.

PS: When I was writing the second line of the post, I almost wrote "thrust" instead of "trust"

[Brotoro]

I used Keplers 3rd Law.

P²=d³

Where p is the orbital period and d is the semimajor axis (the average distance of the Moon from the center of the Earth)

If the Moon is stopped in its orbit and falls straight down to Earth, it's essentially now in a very skinny elliptical orbit that has a major axis equal to the Moon's distance from the Earth.

The orbital period of the Moon is p=SQRT(d³)

For the falling Moon, the orbital period would be SQRT((d/2)³)

because it's SEMImajor axis is now d/2.

So the time to fall will be half of that.

So take the orbital period of any object, and the time to fall straight down would be that orbital period divided by 5.66

Sidereal period of the Moon is 27.322 days, so it would fall in 4.8 days.

Note that this is the time it would take to reach the center of the Earth, not its surface, but it's going to be moving pretty fast by the time it hits the atmosphere and the shock wave annihilates you, so you won't have time to complain about the small error.

Edited April 11, 2014 by Brotoro
Typo, and clarification

[goldenpeach]

Thank you!

Do you know how to calculate the speed of an object when it's acceleration isn't constant but grow up(english isn't my maternal language so I didn't found anything better) at a constant rate(in relation to distance travelled/time etc...)

Also, how can I Know what are those equations? Do I need to integrate something?

EDIT: if you want to know why I want to know that, here's the reason:

When I was a young student, I accidentally dropped my collation on someone under me(on a stair). The teacher punished me(probably because he(or she, I can't remember) tough I did it on purpose). While I was waiting on a chair while the others were outside, a teacher who heard I told I didn't did it on purpose, explained me that it could be dangerous even if I didn't did it on purpose(of course, for something else than my collation). As an example, he told me that my collation could weight a ton if I dropped it from hight enough.

Now, I want to know at which height it would need to be dropped in order that relativity raise it's mass to an ton.

EDIT2: please, don't give me the answer, I want to calculate the height by myself.

Edited April 11, 2014 by goldenpeach

[Brotoro]

If you drop something onto the Earth, it will never move fast enough for relativistic effects to become important no matter how high you drop it from, so you won't be able to calculate such a height.

[peadar1987]

If you drop something onto the Earth, it will never move fast enough for relativistic effects to become important no matter how high you drop it from, so you won't be able to calculate such a height.

Although if it was high enough in the sun's gravity well and had been falling for a LONG time, and the earth just happened to passing through the wrong part of space at the rong time...

[Brotoro]

Although if it was high enough in the sun's gravity well and had been falling for a LONG time, and the earth just happened to passing through the wrong part of space at the rong time...

No. If you fall from a standing start, even from a very great distance, you will only be going at escape velocity when you reach the Earth (or Sun if you want to fall onto that). The escape velocity of the Earth, and even of the Sun, are far from relativistic speeds.

That's what escape velocity is...the speed you have to go away from an object if you want to barely keep moving away at greater and greater distances. So falling in from far away simply reverses the process.

Edited April 11, 2014 by Brotoro

[Doozler]

Do you know how to calculate the speed of an object when it's acceleration isn't constant but grow up at a constant rate...

... Do I need to integrate something?

For the above question, Yes. Start with constantly increasing acceleration a=z*t. Then integrate to get velocity, then integrate again to get position. Use the position equation to find the time of impact, and put that back into the velocity equation to get the velocity. I can supply more details (or the answer) if you want.

However, I'm not sure that's what you want:

he told me that my collation could weight a ton if I dropped it from hight enough.

I expect the teacher was not referring to relativistic effects! He was probably trying to say something like "a rapidly falling collation could do as much damage as a 1 ton weight"

You could calculate this a few different ways, I'm not sure which is best. An easy way is "from what height would the collation have to fall to have a kinetic energy equal to the potential energy of 1 ton weight?" I doubt that's the right way to calculate ~"as much damage" but it's an easy place to start.

[Kerbin Dallas Multipass]

@Brotoro

Cool, thanks!

[peadar1987]

No. If you fall from a standing start, even from a very great distance, you will only be going at escape velocity when you reach the Earth (or Sun if you want to fall onto that). The escape velocity of the Earth, and even of the Sun, are far from relativistic speeds.

That's what escape velocity is...the speed you have to go away from an object if you want to barely keep moving away at greater and greater distances. So falling in from far away simply reverses the process.

You're right, I miscalculated the solar escape velocity as 600,000km/s instead of 600,000m/s. Rookie mistake!

[Technical Ben]

4.83 days, if you stopped its sideways motion instantly and let it fall.

about 60 seconds if your in the Star Trek Into Darkness Universe.

(PS, yes, yes I'm BITTER!!!)

[K^2]

Moon's orbit is pretty elliptic, by the way. It can take more than 5 days from a high point. But it's still in the ballpark of the figure Brotoro specified and it's computed exactly as he stated.

[goldenpeach]

For the above question, Yes. Start with constantly increasing acceleration a=z*t. Then integrate to get velocity, then integrate again to get position. Use the position equation to find the time of impact, and put that back into the velocity equation to get the velocity. I can supply more details (or the answer) if you want.

However, I'm not sure that's what you want:

I expect the teacher was not referring to relativistic effects! He was probably trying to say something like "a rapidly falling collation could do as much damage as a 1 ton weight"

You could calculate this a few different ways, I'm not sure which is best. An easy way is "from what height would the collation have to fall to have a kinetic energy equal to the potential energy of 1 ton weight?" I doubt that's the right way to calculate ~"as much damage" but it's an easy place to start.

I know, but I want to take him litterally.

[Doozler]

Fair enough! Have you worked out the percentage of c that's needed to get your mass up to one ton?

Edit: I think you're asking the question "what height should I drop something from to reach velocity v"? This isn't too hard - set potential energy equal to kinetic energy and solve for height.

You won't be able to get enough speed from the earth or the sun. I suspect you'll need a super massive black hole....

Edited April 12, 2014 by Doozler

[goldenpeach]

Good idea!

But I want a general way to do it: what is the equation to calculate speed in relation of time when the acceleration raise constantly?

Also, just for curiosity, what is the equation to calculate the distance travelled in relation with time when the acceleraion raise constantly?

[N_las]

In this one - dimensional case:

Gravitational Force: F = - G*m*M_earth/x^2

Newton: F = m*a

acceleration and position can be connected by: a = (dx)²/(dt)²

putting everything togheter:

-G*M_earth/x² = (dx)²/(dt)²

the start position: x(t=0s) = x_moon;

and start velocity: dx/dt = 0;

You have to learn how to solve differential equations to learn the general way. But I am just lazy and solve something like this numerically with my computer.

Edited April 12, 2014 by N_las

[Doozler]

The difficulty is that, when falling, acceleration is not changing "constantly" in the sense of a=z*t, but changes depending on position (as described above). This is why you have to solve differential equations to find the answer. There's a derivation of escape velocity at Wikipedia which I think is the calculation you want, but it's not very illuminating:

http://en.wikipedia.org/wiki/Escape_velocity#Derivation_using_G_and_M

But now I have a question: If the "escape velocity" of a black hole is the speed of light, does that mean an object falling from infinity would reach lightspeed at the event horizon? That doesn't feel right to me but seems to follow from the definition of escape velocity.

[FREEFALL1984]

Has anyone considered how much the earth moves towards the moon? the moon is a not too insignificant mass, as the moon approaches, the earth will be pulled towards it at ever increasing velocity, which would cause the moon to increase its velocity even faster.

There are of course an abundance of variables which would also effect the rate of fall such as relative position of celestial bodies and relative position during fall, but they're generally too insignificant to consider.

[Nikolai]

Has anyone considered how much the earth moves towards the moon?

The orbital calculation does that, since the period and semimajor axis of the orbit are dictated by the mutual attraction of the objects.

[TheDataMiner]

I can see the ads now:

Ultimate Spacediving!

4.83 days of free fall!

Just think of all formations you could do with all that time!

It's Fun for the whole family!

Sign up now for only $7,999,999,999.99

[FREEFALL1984]

The orbital calculation does that, since the period and semimajor axis of the orbit are dictated by the mutual attraction of the objects.

Firstly I'm no physicist, and I'm not particularly good at maths.

Indeed the semimajor axis is determined by the mutual attraction of the objects.

But the movement of the earth isn't accounted for in any of these equations. Normally the movement of the parent body isn't considered because the effect is so small, but the barycentre of the earth and moon is only about 1700km below the surface of the earth meaning the effect of the moons pull on the earth is not insignificant. In fact due to the mass of the moon and the exponential nature of the equations the effect would be so significant it would throw off brotoros 4.83 days by quite a large margin.

[ZetaX]

Do not call something "exponential" that isn't. Never.

Apart from that, Keplers laws are perfectly valid if taking the barycenter as focal point of the eclipse. And as you said, this barycenter is inside the earth , thus the moon reaching it or earth is actually less of a difference than the one already allowed above by ignoring the radiuses of both.

[FREEFALL1984]

Apologies for my foul language lol, I'm still having difficulties grasping how standard orbital mechanics can be applied to a situation where two bodies are initially orbiting each other and one of those bodies suddenly stops moving. Do both bodies simply move towards the barycentre as though its a fixed point in space or does the location of the barycentre become irrelevant due to the fact that the bodies are no longer in orbit. In fact I think I'll put this in a new thread since my curiosity is piqued and its beginning to stray off topic