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Central force to 2body problem: which values? 
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#1
Dec812, 06:21 AM

P: 5

Hi, I'm new to this forum and I've got a question.
In all articles I've found about the two body problem, they first start off by writing the distances between the masses as one function r(t) and one function R(t) (which is just simple the position of the reference frame). *note: i use initial position and speed to calculate the orbit equation* Then they go to the central force problem and solve the equation of the orbit of a reduced mass (μ). This is a pretty complicated calculation, but eventually they all end up with the same polar equation: [itex]r(θ) = \frac{p}{1+ε*cos(θθ0)}[/itex] where: [itex]p = L^2 / k[/itex] and [itex]L[/itex] is the angular momentum which is the cross product of speed and position wich is equal to [itex]L= r * v * sin(α)[/itex] where r and v are absolute values (the length's of the vectors). k is some constant [itex]G*M[/itex] in this case. ε and θ0 can be expressed with the same values r, v and α, shown on this site http://upload.wikimedia.org/wikipedi..._Mechanica.pdf page 242 (it's in Dutch, sorry). I evuentually ended up making a 'geogebra applet' which calculates the path with the orbit equation of the reduced mass. Now I want to go back to the two body problem, but I don't know which values r, v and α to use. Which values should I use?? Any help would be appreciated! (if anybody is interested, I will post the geogebra file too) Thanks in advance, physics2.0 


#2
Dec812, 07:19 PM

PF Gold
P: 1,165

The values of r,v, α determine the condition of the moving body at some point of space; you can choose the system of polar coordinates in such a way that this point has θ =0.
You can choose any values you want. The equation then gives the trajectory of the body as a function r(θ); pair [r(θ),θ] are polar coordinates of the points on the trajectory. 


#3
Dec912, 05:35 AM

P: 5

I still don't know which values to use. To be clear: I solved the kepler problem; the values I used were quite easy to determine, because there is only one radius and one velocity.
So my question is: what is the position and the velocity of the 'imaginary' reduced mass of the two body problem. How do I determine the trajectory of both masses? 


#4
Dec912, 10:26 AM

PF Gold
P: 1,165

Central force to 2body problem: which values?
Perhaps you need the definition of r in terms of radius vectors [itex]\mathbf r_1,\mathbf r_2[/itex] and masses [itex]m_1, m_2[/itex].
You can find this at https://en.wikipedia.org/wiki/Twobody_problem 


#5
Dec912, 10:27 AM

PF Gold
P: 1,165

Perhaps you need the definition of r in terms of radius vectors [itex]\mathbf r_1,\mathbf r_2[/itex] and masses [itex]m_1, m_2[/itex].
You can find this at https://en.wikipedia.org/wiki/Twobody_problem 


#6
Dec912, 10:28 AM

PF Gold
P: 1,165

Perhaps you need the definition of r in terms of radius vectors [itex]\mathbf r_1,\mathbf r_2[/itex] and masses [itex]m_1, m_2[/itex].
You can find this at https://en.wikipedia.org/wiki/Twobody_problem 


#7
Dec912, 10:29 AM

PF Gold
P: 1,165

Sorry for the repetition, my browser has some problems.



#8
Dec1012, 03:08 AM

P: 5

First of all thanks for the comments
Sorry I haven't said this, but I know that. The trajectory of the first mass is: [itex]x1(t) = R(t) + \frac{m2}{m1+m2}*r(t)[/itex] And the same holds for the second mass (but m2 will change in m1) Now I've learnt that the time dependent radius [itex]r(t)[/itex] cannot analytically be solved because the equation [itex]μr''=f(r)+ L^2/(μr^3 )[/itex] cannot be solved analytically. However when converting r as a function of [itex]\theta[/itex] you can get [itex]r(\theta)[/itex]. This is the formula I posted in my first post. From this point on I'm stuck. I don't know which values p,e and theta0 to use. Actually this boils down to two things I don't know: What is the speed of the reduced mass? Or what is the angular momentum of the reduced mass? If I know its speed I know its angular momentum because r0 is the distance between the to masse's at t=0; But maybe the angular momentum of the reduced mass is the sum of the angular momenta of the other two masses? Please help me out. 


#9
Dec1412, 07:29 PM

PF Gold
P: 1,165

What exactly do you want to find out?
It seems you want to find the mutual distance of two bodies as a function of the angle in polar coordinates. Let's say the observer is on the body 2 and observes the body 1, as it circles around 2. Let R be vector of 1 with respect to 2: [tex] \mathbf R =\mathbf r_1  \mathbf r_2 [/tex] This vector changes in time according to [tex] \ddot{\mathbf R} =  \frac{G(m_1+m_2)}{R^3}\mathbf R.~~~(*) [/tex] This should be easy to see from the original equations. This equation is the same as for any body of any mass orbiting around fixed centre of mass [itex]m_1+m_2[/itex]. So the trajectory and function R(θ) will be the same. The vector R circumscribes ellipse, and its magnitude will is given by [tex] R(\theta) = \frac{p}{1+\epsilon \cos (\theta\theta_0)}.~~~(\text{T}) [/tex] You can see that [itex]\theta_0[/itex] is the angle for which the radius vector has the smallest length  the body 1 is at perigeum. Because there are no other requirements on the orientation of the polar system of coordinates, we can orient it in any way. Particularly convenient choice is such that the perigeum is at [itex]\theta = 0[/itex], i.e. [itex]\theta_0 = 0[/itex]. The quantities [itex]p,\epsilon[/itex] in the above equation are connected to angular momentum and total energy of the body moving around the fixed centre. In order to find them, imagine you have a small planet that starts to move around fixed center according to *, with position R and velocity [itex]\dot {\mathbf R}[/itex]  which are given by the initial conditions. Find its energy and angular momentum and use the above mentioned formulae to find its p and [itex]\epsilon[/itex]. These are obtained in the course of derivation of the equation of trajectory (T), or you can find them on Wikipedia or in any book on classical mechanics. 


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