(no subject)
Nov. 25th, 2004 12:05 am![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png)
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?
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?
no subject
Date: 2004-11-25 12:22 am (UTC)no subject
Date: 2004-11-25 12:25 am (UTC)no subject
Date: 2004-11-25 01:38 am (UTC)no subject
Date: 2004-11-25 01:51 am (UTC)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.
no subject
Date: 2004-11-25 08:52 am (UTC)So you'd have to factor in wind resistance/air density - something that normally gives you terminal velocity.
no subject
Date: 2004-11-25 11:40 am (UTC)no subject
Date: 2004-11-25 02:07 am (UTC)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.
no subject
Date: 2004-11-25 03:12 am (UTC)... 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?
Date: 2004-11-25 03:13 am (UTC)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?
Date: 2004-11-25 09:51 am (UTC)Sorry to cause you to pull hair out! Thank you!
(Some dimensional analysis to check which of you and
Re: Is this an invitation to be nerdy?
Date: 2004-11-25 12:31 pm (UTC)Re: Is this an invitation to be nerdy?
Date: 2004-11-25 06:16 pm (UTC)no subject
Date: 2004-11-25 04:38 am (UTC)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.
no subject
Date: 2004-11-25 09:52 am (UTC)no subject
Date: 2004-11-25 12:56 pm (UTC)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!
no subject
Date: 2004-11-25 03:28 pm (UTC)no subject
Date: 2004-11-25 06:19 pm (UTC)no subject
Date: 2004-11-25 10:16 am (UTC)Although I could be wrong because I'm drunk still.
And I'm assuming your sphere has an atmosphere.
no subject
Date: 2004-11-25 10:20 am (UTC)no subject
Date: 2004-11-25 10:28 am (UTC)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.
no subject
Date: 2004-11-25 10:53 am (UTC)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... :-)
no subject
Date: 2004-11-25 04:03 pm (UTC):: runs away ::
:: cries ::
no subject
Date: 2004-11-25 11:33 am (UTC)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.