rocket science

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rocket science

Postby oblivimous » Tue May 01, 2007 9:08 pm UTC

Imagine a rocket (mass = m1) that is able to exert a constant force (F) while maintaining a constant mass. This rocket is fired away from a planet (mass = m2) which exerts a gravitational force in the expected manner.

Is it possible to give an equation for the rocket's height at time=t? To simplify, we should consider the planet to be a point mass, and note that there are no other gravitational forces or planetary motion at play.
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Postby bigglesworth » Tue May 01, 2007 9:11 pm UTC

its force would be an acceleration towards the planet. This can then be put into a normal equation of motion (or as I like to say, SUVAT) that has been modified so that the acceleration changes over time. :) That's all i've got so far.
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Re: rocket science

Postby You, sir, name? » Tue May 01, 2007 9:13 pm UTC

oblivimous wrote:Imagine a rocket (mass = m1) that is able to exert a constant force (F) while maintaining a constant mass. This rocket is fired away from a planet (mass = m2) which exerts a gravitational force in the expected manner.

Is it possible to give an equation for the rocket's height at time=t? To simplify, we should consider the planet to be a point mass, and note that there are no other gravitational forces or planetary motion at play.


Functions of time are messy, since newton's second law for gravity becomes an unsavory differential equation. It's okay so long as you don't travel that far from the surface, then you can just use constant gravitational force and get a decent result, but as soon as you gain some height, you end up with expressions on the form a(t) = k/(s(t))^2

I -think- you can find a solution for them, but often it's easier to approximate numerically.
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Postby gmalivuk » Tue May 01, 2007 9:19 pm UTC

Is this newtonian physics, or relativistic?

Helpful Hint wrote:Recall that the radius of a timelike geodesic behaves as though it has mass 1 and is in a potential of Image
(Where L is r^2 phi', of course.)


Yes, I'm an ass, but is this homework?
Last edited by gmalivuk on Tue May 01, 2007 9:56 pm UTC, edited 2 times in total.
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Postby oblivimous » Tue May 01, 2007 9:21 pm UTC

I get to

p''=F/(M1) - G(M2)/(p^2)

p = position

And I'm stuck.
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Postby cmacis » Tue May 01, 2007 9:50 pm UTC

Our particle dynamics notes are online, we did stuff getting off the planet's surface.
http://www.maths.leeds.ac.uk/%7Elivermo ... _notes.pdf
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Postby Hawknc » Wed May 02, 2007 4:22 am UTC

I'm having a pretty summarily awful day so far, but I'll take a look when I get back. My instinct is to say that it would be difficult to give a simple equation for it, but what would I know, I'm just a rocket scientist. :P
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Postby SpitValve » Wed May 02, 2007 4:27 am UTC

Hawknc wrote:I'm having a pretty summarily awful day so far, but I'll take a look when I get back. My instinct is to say that it would be difficult to give a simple equation for it, but what would I know, I'm just a rocket scientist. :P


You've been waiting for a chance to use that line, haven't you? :D
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Postby Hawknc » Wed May 02, 2007 4:49 am UTC

...Maybe.

Edit: alright, time to put my money where my mouth is. I'm going to approach this from a Newtonian point of view, since I'm kinda weak on relativity and I don't think it's really relevant.

If the rocket is putting out a constant thrust F, that will be counteracted by the gravitational force Image. So the net force at any point is equal to:

Image

Solving this for acceleration gives:

Image

Obviously because it's not a constant acceleration we can't use Newton's equations of motion, so we integrate twice between 0 and some time t, which gives us:

Image

Now I'm almost guaranteed to have screwed something up here, so someone please check that what I'm doing isn't completely stupid.
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Postby skeptical scientist » Wed May 02, 2007 6:47 am UTC

Hawk, you're treating r as a constant, but really r depends on height, so you should really get a diffeq, not a simple integral.
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Postby Hawknc » Wed May 02, 2007 7:19 am UTC

...That'd be right. It did seem altogether too simple. ;)
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Postby SpitValve » Wed May 02, 2007 8:36 am UTC

It's alright, he's only a rocket scientist after all.
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Postby gmalivuk » Wed May 02, 2007 3:59 pm UTC

Hawknc wrote:since I'm kinda weak on relativity and I don't think it's really relevant.


Oh of course it's not. I was just being an ass because I don't like doing work that looks like someone else's assignment. (At least when they don't specify whether or not it's homework.)
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Postby You, sir, name? » Wed May 02, 2007 4:03 pm UTC

Hawknc wrote:...Maybe.

Edit: alright, time to put my money where my mouth is. I'm going to approach this from a Newtonian point of view, since I'm kinda weak on relativity and I don't think it's really relevant.

If the rocket is putting out a constant thrust F, that will be counteracted by the gravitational force Image. So the net force at any point is equal to:

Image

Solving this for acceleration gives:

Image

Obviously because it's not a constant acceleration we can't use Newton's equations of motion, so we integrate twice between 0 and some time t, which gives us:

Image

Now I'm almost guaranteed to have screwed something up here, so someone please check that what I'm doing isn't completely stupid.


Though, won't the rocket loose mass according to

dm/dt= F_thrust/v_matter_ejection

At least, that's what I think the formula is for thrust generated by matter streams. I think you need to factor that into the equation as well.
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Postby Vaniver » Wed May 02, 2007 6:36 pm UTC

Though, won't the rocket loose mass according to
It's a magic rocket.

Imagine a rocket (mass = m1) that is able to exert a constant force (F) while maintaining a constant mass.
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Postby oblivimous » Wed May 02, 2007 9:24 pm UTC

I started with a constant-mass rocket to create a more simple problem. Find P(t) when force of thrust is constant and force of gravity varies with the inverse of the square of p. Left me with a differential that I'm very stuck on.

Add in a mass that depends on time and I expect the problem gets more difficult.

Add in air resistance that depends on both position and the square of velocity and I'm not sure how the folks at NASA do it.
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Postby cmacis » Wed May 02, 2007 9:30 pm UTC

Solving point by point rather than by equation?
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Postby skeptical scientist » Wed May 02, 2007 9:36 pm UTC

Yeah, it's very easy to solve these numerically. But then, if you want to solve precisely, there are lots of books where you can look up whatever diffeq and find the solution.
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Postby Gelsamel » Thu May 03, 2007 5:27 am UTC

At 6am in the morning on the train to Uni I could implicitly define s(t) in the form of t(s), but not explicitly. Probably because I don't know enough math :S.
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Postby Spaz Funbag » Sat May 05, 2007 4:03 pm UTC

I don't know why, but for some reason my brain tells me that there is an all too obvious solution to this, but I just don't have it here right now.


But then I remember that F is not constant, and there it is again.

edit:working on it.....not actually looking for a solution, rather for the fun of it

I take for F=ma that m=m(rocket).

So I get x=k/mx² + [F2*t²/2m] with F2 being the thrust and k being G*m1*m2.

Which will get sort-ah messy when solving for x.


Can't we just say that G or one mass is zero? that would make things far easier ;-)


Crap.
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Postby Spaz Funbag » Sat May 05, 2007 4:58 pm UTC

Plotting would also be [awful] as it would (as i see it) require a n-dimensional graph, with n ranging somewhere between 5 and 8.



Other Idea: Can I say a(total) = a(due to Thrust) + a(due to Gravitation)

...It would be tempting to say that a(gravitation) is a free fall, but apparently that is not true.
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