(no subject)

[hutchingsmusic]

Here's one for the math geeks out there.

I'm working on a science-fiction premise. Yes, I know I don't write science fiction.

Anyway. If I have a sphere of radius r with a mass m such that the gravitational acceleration on its surface is 9.8 metres per second squared (m = 9.8 x r² / G, in other words, where G is the gravitational constant) then what is the impact velocity of a body falling from rest from a height of x metres above the sphere?

The acceleration at height x is G * M / (r+x)². What I need to find is the eventual velocity when the body gets to height 0 and hits the sphere's surface.

Problem is, the acceleration is dependent on the distance. The velocity depends on the acceleration. The distance depends on the velocity. And I haven't studied maths for over four years.

Any ideas, people?

Comments

[alwaschoen.livejournal.com]

In a vacuum?

[saraaaaaa.livejournal.com]

gah, ask me in a few months.

[derobrash.livejournal.com]

I posted an answer, but you've already got that one 8-)

[muphf.livejournal.com]

x = dv/dt = da/dt

You have an equation for acceleration as position X. Now ... don't think of position 0 as your ending position, but as your starting position ... take the derivative of your equation for acceleration, and solve for velocity at position x. That being said, I'm not sure you have the right equation there, but I live in the world of electrons and bits, so my Newtonian physics is a bit stale.

[skx.livejournal.com]

Bear in mind you have to account for real physical effects when you scale up from electrons.

So you'd have to factor in wind resistance/air density - something that normally gives you terminal velocity.

[chillies.livejournal.com]

Hah! Electrons don't exist! Einstein said so.

[sylk.livejournal.com]

Ideally, the sphere starts at some set height x and a velocity of 0. Shouldn't you be able to take an integral of the acceleration to find the velocity? You have the starting value for acceleration, since it occurs at a known x, and I think that could be used to find the constant since you know that velocity is equal to 0 when the object begins the fall. That's neglecting air resistance and everything, not to mention that in most cases r is so much greater than x that the change in x has little change on the overall acceleration.

You might have to do that a second time though to find out how long it takes the object to fall, since the way I would do it introduces time as a unit and you can't find the final velocity since it depends on t unless you solved the distance equation to find out when t is equal to 0 (impact). Again though, I think you have the values needed to solve that.

And if this makes sense, you understand more physics than me. It's quite possible you understand more physics than me even if it doesn't make sense, because I'm good at chemistry, not physics. Forget math.

[muphf.livejournal.com]

Yeah ... I'm backwards there ... if you have a linearly increasing velocity, for instance, you have a parabollically increasing position (position changes faster and faster as velocity incrases), making it a=dv/dt=d^2x/dt^2.

... and take that r term out of there ... it's a complication that you don't need unless you really care about the difference in gravitational pull caused by the difference between the surface of your body and the center of gravity.

Is this an invitation to be nerdy?

[willow-red.livejournal.com]

Based on how you phrased this problem, I will assume that x is large compared to r (though it would be much easier if it wasn't), and m is small compared to M.

This brings it down to a question of orbital mechanics, even though this "orbit" is really linear motion....

Okay, that deceptively simple problem had me going for the better part of an hour and involved cracking my diff eq book (something that hasn't happened since long before I graduated), since all my orbital mechanics books are on the shelf in my office right now.

The generic equation for orbital velocity is v = sqrt( 2 G M / r - G M / a ), where a is the semi-major axis.
Since your mass is initially at rest at radius (r+x), this means that a = ( r + x ) / 2. (That looks a little odd to me, but I'm going to go with it for now.)
Plugging that back into the first equation and setting r = radius of the sphere, that gives you an impact velocity of
v = sqrt( 2 G M x / ( r + x) ), assuming I did my math right.

I'm going to stop pulling my hair out now and go do something else now. I hope this helps you.

Re: Is this an invitation to be nerdy?

[randomchris.livejournal.com]

It's a useful point for when x is large compared to r, but most of the situations I'm considering are when x is of the same order as r.

Sorry to cause you to pull hair out! Thank you!

(Some dimensional analysis to check which of you and vyrin is correct: G is in m3 kg-1 s-2, M is in kg, x and (r+x) are both in metres: I think you might have missed out an r term somewhere when combining the fractions, or maybe you left it out as a typo rather than a math error. I'll assume it was the latter!)

Re: Is this an invitation to be nerdy?

[randomchris.livejournal.com]

Er, former. I mean former. D'oh.

Re: Is this an invitation to be nerdy?

[willow-red.livejournal.com]

Erk! Yes, I did leave out a 1/r term, but it was a typo, honest! When I checked my written notes this morning, it was there. I think I accidentally dropped it because I did my calculations with the stand-in variables I'm used to, then switched to your variables while typing. Thanks to vyrin for catching that!

[vyrin.livejournal.com]

I'll think about this some more and post a better answer, but my initial thought is that you can find an upper bound for the velocity by just looking at the escape velocity needed to launch something from the surface of the sphere out into space. This is the velocity needed to completely escape the gravity well, and thus by symmetry, it is the maximum velocity that an object released in space to drop onto the sphere will ever see.

The escape velocity can be found by:

Ve = sqrt(2GM/R)

For objects that are significantly far apart this upper bound will be a very good approximation of the final terminal velocity.

Oh wait a second! We've already got the approach to solve this. You can simply look at this problem as a conversion between potential energy and kinetic energy.

Looking at the problem backwards (escape velocity still), then we start out with potential energy of GMm/r and we want to lift the object to a potential energy of GMm/(r+x). To do that we need kinetic energy of 1/2mV^2.

1/2mV^2 = GMm/r - GMm/(r+x)

V^2 = 2*(GM/r - GM/(r+x))

V = sqrt( 2GM * (1/r - 1/(r+x)))

V = sqrt( 2GMx/(r^2+rx))

where r is the sphere radius and x is the distance above the surface.

Which is almost the same thing that willow_red got, but with an extra 1/r term in the sqrt. Huzzah.

[randomchris.livejournal.com]

A quick check with dimensional analysis proves you correct and declares you the winner. Well done! And many thanks.

[andrewwyld.livejournal.com]

Aha!  I saw your next post and thought, GPE.

Interesting fact:  The GPE inside a uniform hollow sphere is constant, so there is no gravitational force inside one.  Since we know what happens oustide a uniform solid sphere, if you're falling down a tunnel into one, you can split the sphere into hollow shells "above" you, and a solid sphere "below" you.  Since the shells exert no force on you, the only force you experience is from the sphere below; hence, at the centre of a planet, you experience no gravitational force whatsoever.

Tada!

[randomchris.livejournal.com]

Oooh, hollow spheres. That might work rather well...

[willow-red.livejournal.com]

Those typos get me every time! Anyway, thanks for using the energy method. I thought about that too, but had already been working on the other way far too long. But since two people used different methods to come up with the same answer (if I could freakin' type!), that means we must be right.

[ellielabelle.livejournal.com]

My initial feeling is that if falling from orbit, if the thing didn't burn up in the atmosphere, it would reach a terminal velocity limited by drag forces on it.

Although I could be wrong because I'm drunk still.

And I'm assuming your sphere has an atmosphere.

[randomchris.livejournal.com]

It's not falling from orbit - we're looking at something suddenly falling from a rest position from several metres above the sphere (which is of the order of a few hundred metres.) There is an atmosphere involved, but the velocities I'm ending up with are only a few metres per second - I want to limit the impact velocity to the equivalent of a short fall on Earth.

[ellielabelle.livejournal.com]

eeuuurgh, it sounds like Death Stars or mountaineering on Deimos to me.

My field of experience is more things with ridiculous orders of magnitude, so I'm not sure. I think the derivation in the thread above is right, though.

[axamendes.livejournal.com]

M = 9.8x^2/G

The force generated by M outside of itself on a mass m at radial distance r is given by:

F = -GMm/r^2

and acts in the radial direction only.

Substituting all this into F=ma, and rearranging, we get:

r^2 d/dt(dr/dt) = -9.8 x^2

Integrate wrt t to get velocity equation.
Integrate wrt t again to get position equation.

Using the position equation, constants of integration are found by setting r = a+x when t = 0. Set r = x to find time tx when the m reaches the surface of M.

Plug tx into the velocity equation to find the velocity at that time. Voila.

I won't do the integrations, because they are nasty... :-)

[saraaaaaa.livejournal.com]

oh god, integrations.
:: runs away ::
:: cries ::

[ashfae.livejournal.com]

the not-a-math-geek slips in quietly to the back

Yes, I know I don't write science fiction.

No, all you know is that you never have before. Not the same thing at all.