Need help - body in central force-field

[chem_master]

[SOLVED] need help - body in central force-field

Hi,

I need help for the following issue:

A body B is falling freely on a straight line towards a body A, which emits a central force-field. The formula for acceleration is:

a(r)=Gm/r^2

I want to derive a formula both for:

v(t)=[?]

and

s(t)=[?]

where at t_0=0 the velocity of body B is v(t_0)=0 and distance s(t_0)=0 (the initial moment of body B falling on body A).

Thanks for any help in advance,

chemical master.

 

[Shooting Star]

Write a = dv/dt = (dv/dr)(dr/dt) = vdv/dr = Gm/r^2. Now you can integrate. (Are you sure there wasn't a minus sign on the RHS?)

 

[chem_master]

Write a = dv/dt = (dv/dr)(dr/dt) = vdv/dr = Gm/r^2. Now you can integrate. (Are you sure there wasn't a minus sign on the RHS?)

Really thanks, shooting star.

Yeah, I think it was a(r)=-Gm/r^2

I'm really ashamed of myself, but I now have a problem with doing that.
If I integrate: Integrate[-Gm/r^2, {r, -Infinity, Infinity}] i get a(r)=Gm/r and vdr/dt=Gm/r

First of all I doubt that it's correct, for second I don't know what to do next. How can I integrate -Gm/r^2 with respect to t if there's no t. Could you help?

 

[Shooting Star]

Why the plus and minus infinities? You have been given the initial conditions. If you integrate, you'll have the constants of integration, whcih you can find by puting in the given conditions.

I just noticed that you have written S=0 at t=0 . Look at the problem again. There was probably a given distance S at t=0 and v=0. The body is just falling from rest from a certain distance.

 

[chem_master]

First of all I would like to underline that this is not a homework nor a problem. I'm just very curious about the evolution of velocity from r=a (for instance) in time and the relation between the distance from a to b (on r) and time intervals. I knew from the first time it has something to do with derivatives, but still can't figure out how to integrate just do obtain the v(t) and s(t) of the falling body.

Any ideas?

 

[Shooting Star]

Formulate the problem a bit more precisely. Suppose S is the initial dist of the falling body at t=0, where you have just dropped it from rest, which means v = 0. You want to know how r and v changes with time. When the falling body reaches the pulling body, the speed becomes infinite. So, we will avoid putting r=0.

We got previously v.dv = -(GM/r^2)dr. Integrating, we get,

v^2/2 = -GM/r + A, where A is a constant.

Evaluate A by putting in the initial conditions. Then integrate again .

 

[chem_master]

Ok, I get

v^2/2 = -GM/r + v_0^2

v(r)=Sqrt(-2Gm/r + 2v_0^2)

Is that correct?

Suppose the initial r is 5 (r=5, t_0=0, v_0=0)

I want to get v(t) and r(t), not v(r). I want the formula to have a time variable.

Any thoughts?

Thanks.

 

[Shooting Star]

v = dr/dt, so you can integrate again. Your v(0) = 0 will make things simpler.

 

[chem_master]

How can I integrate: Integrate[-Gm/r, dt] if in "-Gm/r" there's no "t"?

 

[Shooting Star]

Bring all the 'r's to one side and take t to the other side.

( I'd posted this line more than one hour back, but somehow it didn't take it.)

 

[chem_master]

Can you give me a direct result, I'm getting more and more confused! Pls.

 

[chem_master]

Anybody?

 

[IITian]

v(r)=Sqrt(-2Gm/r + 2v_0^2)

now put v=dr/dt

dr/dt=sqrt(-2Gm/r + 2v_0^2)

or, dt=dr/sqrt(-2Gm/r + 2v_0^2) now solve...

 

[chem_master]

How? You cannot solve such an eqation! If so, show me how...

 

[rl.bhat]

Since the acceleration is the function of r only, you cannot exprexss v in terms of t only, but you can express v and s in terms of r only.
The first integration gives v^2/2 = -GM/r + A, where A is a constant. If B is falling from a distance R from A, v = o when r = R. So Constant A = GM/R and v^2 = 2(GM/R - GM/r) Similarly v = ds/dt = ds/dr*dr/dt = ds/dr*v of ds = dr. When you integrate you get s = r + C. When r = R, s = 0. So C = -R and s = r - R

 

[IITian]

thanks for this rl.bhat

 

[chem_master]

Since the acceleration is the function of r only, you cannot exprexss v in terms of t only, but you can express v and s in terms of r only.
The first integration gives v^2/2 = -GM/r + A, where A is a constant. If B is falling from a distance R from A, v = o when r = R. So Constant A = GM/R and v^2 = 2(GM/R - GM/r) Similarly v = ds/dt = ds/dr*dr/dt = ds/dr*v of ds = dr. When you integrate you get s = r + C. When r = R, s = 0. So C = -R and s = r - R

Ok, thanks, but suppose I wish to measure the time from when s_0=0 (v=0) to s_1. Just how could I calculate how long would it take?

I cannot calculate it neither from v nor from a since it's changing in terms of r.

Any hints?

 

[Shooting Star]

Since the acceleration is the function of r only, you cannot exprexss v in terms of t only, but you can express v and s in terms of r only.
The first integration gives v^2/2 = -GM/r + A, where A is a constant. If B is falling from a distance R from A, v = o when r = R. So Constant A = GM/R and v^2 = 2(GM/R - GM/r) Similarly v = ds/dt = ds/dr*dr/dt = ds/dr*v of ds = dr. When you integrate you get s = r + C. When r = R, s = 0. So C = -R and s = r - R

What is 's' and 'r' that you've used?

 

[chem_master]

I would like to indicate that i need a solution to velocity and distance from r=R to the center IN TERMS OF TIME, not r. Any suggestions would be gratly appreciated.

Thanks,

chemical master

 

[Shooting Star]

All right, let's start from where we had left.

v^2/2 = GM/r + A. If v=0 at r=R, then A=-GM/R. This means that at a dist of R from the centre, you are simply allowing the body to fall with initial speed zero, ina st line towards the centre of force. I presume that's what you had wanted.

v = sqrt(2GM(1/r -1/R) => sqrt(2GM/R)dt = dr*sqrt(r/(R-r)), (where v= dr/dt)

Can you do this integral by substituting r=Rsin^2y? You’ll get ‘t’ in terms of r. Unfortunately, that eqn can’t be inverted analytically. :grumpy:

I might as well give you the answer, since I’ve spent quite some time doing it.

Sqrt(2GM/R)*t = arcsin(sqrt(r/R)) –sqrt[r(R^2-r^2)/R^(3/2)] - pi/2.

You may recheck my calculation to find any careless algebraic mistakes.

 

[chem_master]

Sqrt(2GM/R)*t = arcsin(sqrt(r/R)) –sqrt[r(R^2-r^2)/R^(3/2)] - pi/2.

Thanks for your effort shooting star.
Does this mean that:
t(r)=(arcsin(sqrt(r/R)) –sqrt[r(R^2-r^2)/R^(3/2)] - pi/2)/Sqrt(2GM/R)
and I've got to invert this function to obtain r(t)?

Does this not seem too complicated to you?

I feel threre should be a simpler solution.

 

[Shooting Star]

I'm sorry to hurt your "feelings", but no, there isn't any simpler one. You can't invert it analytically. Have you ever tried to solve for the simplest two body problem in GR? Then you would hug this equation to death.

Once you plug this into a computer or one of the better pocket calculators, then you can immediately get the r for t. You can also draw a graph to get an idea how the test particle falls. Note that though the speed becomes infinite at r=0, the time taken to fall to the centre is finite. That's why I said that a graph would be a good idea.

 

[chem_master]

My ""feelings"" are even worse. I think that there is a simpler solution, but you can't admit there is one. I don't mean to be rude, but it seemed like your post was a bit unfriendly so I decieded to mark this topic as solved although I don't think it is.

 

[Shooting Star]

I was just playing with words. I had no intention of actually hurting your feelings in any way. And I'm telling you the truth. If that solution is correct, you just cannot invert it to get r as a function of t. I am surprised that you considered me to be unfriendly, considering that I really did spend quite some time integrating the eqn after your request. (It took me time because I'm not so much used to these things nowadays.)

Hoping I have removed all misconceptions.:!)

 

[chem_master]

No problem, maybe I've understood you wrong (I'm a bit "word-sensitive").

Btw, I think I found what I was looking for:

http://www.dartmouth.edu/~phys44/lectures/Chap_4.pdf"

p. 4 (---> r(t)=r_0*Sqrt(... )

What do you think?

 

[Shooting Star]

Hi,

I need help for the following issue:

A body B is falling freely on a straight line towards a body A, which emits a central force-field. The formula for acceleration is:

a(r)=Gm/r^2

In your original post, you had asked for the formula for one dimensional motion of a body falling toward the centre of force. The link you have mentioned is for the general two dimensional motion of a body under a central force. It is very useful because it covers the most general case.