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Solving the Two-Body System

  1. Nov 26, 2011 #1
    Hey guys! So I'm looking at this textbook that attempts to describe to me how to get kepler's first law describing a two-body system in polar coordinates, and I have no clue how to get from one step to another because I have seen NO OTHER PLACE solve this problem this way.
    It starts off by describing angular momentum as L=m*r^2*w=m*r^2*do/dt and the total energy of the system as E=1/2mv^2-GMm/r. So far so good!
    But then out of nowhere, it pulls out that our velocity is v^2=(dr/dt)^2 + (r*do/dt)^2. I'm guessing the first component is the radial velocity and the second is the veolcity relating to the circular motion of the body?? But why is everything squared? Shouldn't it just be v=dr/dt + rdo/dt????
    Then I'm supposed to go through and replace do/dt using our trusty angular momentum equation and v using our energy equation to get rid of the time component and just have the diff eq dr/do = r sqrt((2mE/L^2)r^2 + (2GMm^2/L^2)r -1), which I worked through and got correctly.
    And somehow, by sheer magic, I am supposed to solve this differential equation to get r=p/1+ecos(o + c). How so I solve this differential equation?? Eek!
    So basically my questions are: How the heck does v^2=(dr/dt)^2+(r*do/dt)^2, and how the heck do I solve the resulting differential equation??
    My brain would much welcome some help :)
  2. jcsd
  3. Nov 28, 2011 #2
    Here's how to derive that result. Start with a polar-to-rectangular transformation:

    x1 = r*cos(ang)
    x2 = r*sin(ang)
    v1 = d(x1)/dt
    v2 = d(x2)/dt
    Then do some "trig" to find v2 = v12 + v22

    You can also get acceleration as
    a1 = d(v1)/dt
    a2 = d(v2)/dt
    You then set {a1,a2} equal to - GM/r2 * (radius unit vector)

    You will get second-order differential equations for r and ang, equations involving d2r/dt2 and d2(ang)/dt2. From them, find equations for the first-order ones: dr/dt and d(ang)/dt. You should be able to work out how they are connected to conservation of energy and angular momentum.

    At this point, you ought to try to find t and ang as functions of r; that's much easier than trying to find r and ang directly as functions of t. You'll see why when you get the solution. :D
  4. Nov 29, 2011 #3
    Hi. I've been trying to solve the first part but I guess I am just terrible at calculus :(
    I have
    x1 = r*cos(ang)
    x2 = r*sin(ang)
    v1 = d(x1)/dt
    v2 = d(x2)/dt
    as you said.
    This means v1= -(dr/dt)*sin(ang)*d(ang)/dt?? And likewise v2=(dr/dt)*cos(ang)*d(ang)/dt?
    and V=v1+v2=-(dr/dt)*sin(ang)*d(ang)/dt+(dr/dt)*cos(ang)*d(ang)/dt since its the vector sum of radial and circumferential components, right?
    But when I do my calculations, I get v1^2+v2^2=(dr/dt)^2 * (dAng/dt)^2 and when I calculate v^2 I get (dr/dt)^2*(dAng/dt)^2*(1-2sin(ang)*cos(ang). They are not the same :( Nor do they give me v^2=(dr/dt)^2 + (r*dAng/dt)^2 like the book says.
    I feel like this should be easy but its not. Sorry!
  5. Nov 29, 2011 #4
    Remember the sum and product rules of differentiation:
    d(f+g)/dt = (df/dt) + (dg/dt)
    d(f*g)/dt = (df/dt)*g + f*(dg/dt)

    v = dx/dt, a = dv/dt
    x1 = r*cos(ang)
    x2 = r*sin(ang)
    v1 = (dr/dt)*cos(ang) - r*(d(ang)/dt)*sin(ang)
    v2 = (dr/dt)*sin(ang) + r*(d(ang)/dt)*cos(ang)
    a1 = (d2r/dt2)*cos(ang) - 2*(dr/dt)*(d(ang)/dt)*sin(ang) - r*(d2(ang)/dt2)*sin(ang) - r*(d(ang)/dt)2*cos(ang)
    a2 = (d2r/dt2)*sin(ang) + 2*(dr/dt)*(d(ang)/dt)*cos(ang) + r*(d2(ang)/dt2)*cos(ang) - r*(d(ang)/dt)2*sin(ang)

    See if you can duplicate my results.
  6. Nov 29, 2011 #5
    of course! haha so silly of me. yes, i got everything you got.
    one last question. I think im finally getting this.
    I want to find a^2=a1^2+a2^2, correct? then I set a=-GM/r^2?
  7. Nov 29, 2011 #6
    You won't need that for the solution. Instead, what you need is

    acceleration (2nd time deriv. of position) = acceleration of gravity
    (vector equality)

    Acceleration of gravity = (inverse-square law for magnitude) * (direction vector of source relative to to object acted on)
    or GM/r2[/sub] * (- (x1,x2}/r)

    a1(gravity) = - GM/r3 * x1
    a2(gravity) = - GM/r3 * x2

    Try verifying this one. You set the accelerations equal by setting their components in each direction equal:

    a1(2nd t deriv) = a1(gravity)
    a2(2nd t deriv) = a2(gravity)


    (d2r/dt2)*cos(ang) - 2*(dr/dt)*(d(ang)/dt)*sin(ang) - r*(d2(ang)/dt2)*sin(ang) - r*(d(ang)/dt)2*cos(ang) = -GM/r2 * cos(ang)
    (d2r/dt2)*sin(ang) + 2*(dr/dt)*(d(ang)/dt)*cos(ang) + r*(d2(ang)/dt2)*cos(ang) - r*(d(ang)/dt)2*sin(ang) = -GM/r2 * sin(ang)

    You can now factor out the "trig" to get some simpler-looking equations.
  8. Nov 30, 2011 #7
    eek! now I am confused again.

    this is kind of a question from before, when I was v, but to get rid of the cosines and sines in v1 and v2, we set the angle=0? why can we just do that without loss of generality?
    should I do the same for the acceleration?

    then with those acceleration equations, do i take the derivative of dr/do = r sqrt((2mE/L^2)r^2 + (2GMm^2/L^2)r -1) with respect to time and plug in using the accel, momentum, and energy equations to get r=p/1+ecos(ang)?
  9. Nov 30, 2011 #8
    You don't need to set (ang) to anything. I'll demonstrate with v2 = v12 + v22

    v1 = (dr/dt)*cos(ang) - r*(d(ang)/dt)*sin(ang)
    v2 = (dr/dt)*cos(ang) + r*(d(ang)/dt)*cos(ang)

    v12 = (dr/dt)2 * cos(ang)2 - 2 * r * (dr/dt) * (d(ang)/dt) * cos(ang)*sin(ang) + r2 * (d(ang)/dt)2 * sin(ang)2
    v22 = (dr/dt)2 * sin(ang)2 + 2 * r * (dr/dt) * (d(ang)/dt) * cos(ang)*sin(ang) + r2 * (d(ang)/dt)2 * cos(ang)2

    v2 = ((dr/dt)2 + r2*(d(ang)/dt)2) * (cos(ang)2 + sin(ang)2)


    From the two acceleration-component equations, one can get two simpler ones:

    (d2r/dt2) - r*(d(ang)/dt)2 = - GM/r2
    r*(d2(ang)/dt2) + 2*(dr/dt)*(d(ang)/dt) = 0

    That should be easy to verify.

    Now to integrate them, and that's rather tricky.

    The second one is the easiest. Divide it by r*(d(ang)/dt):
    (d2(ang)/dt2)/(d(ang)/dt) + 2 * (dr/dt)/r = 0

    That is easy to integrate:
    log(d(ang)/dt) + 2*log(r) = constant


    d(ang)/dt = h/r2
    where h is a constant of integration, the specific angular momentum.
    The angular momentum L = m*h

    Now take on the radial equation, substituting in the angular-momentum equation:
    (d2r/dt2) - h2/r3 = - GM/r2

    Multiply it by (dr/dt) and integrate:
    (1/2)*(dr/dt)2 + (1/2)(h2/r2) = W + GM/r
    where W is another constant of integration, the specific energy.
    The energy E = m*W

    It is easy to show that
    (1/2)*v2 = W + GM/r
  10. Nov 30, 2011 #9
    So r = GM/((1/2)v^2-W)?
    Why did we need to do all that work to find this? All the acceleration integration stuff makes sense.....but I don't get where I am supposed to use the result :(
  11. Nov 30, 2011 #10
    Yes indeed.

    This acceleration integration is how you find the conservation laws -- the conserved quantities emerge as integration constants.

    Here's how one can derive the conservation laws without expanding into components. I'll derive them for the specific angular momentum and energy: h, W (L = m*h, E = m*W)

    Bold = vector
    r = |x| = sqrt(x.x)
    v = |v| = sqrt(v.v)
    dx/dt = v
    dv/dt = a = - GMx/r3

    Angular momentum: h = (x)x(v)
    ()x() = cross product

    Does it change? dh/dt = (v)x(v) + (x)x(a) = - GM/r3 * (x)x(x) = 0

    Energy: W = (1/2)v2 - GM/r

    Does it change? dW/dt = v.a + GM/r3 * (x.v) = 0

    This discussion works for any central force. There's an additional conserved quantity that the inverse-square case has, the Laplace-Runge-Lenz vector.

    R = (v)x(h) - GMx/r = xv2 - v(x.v) - GMx/r

    Does it change? dR/dt =2x(v.a) - a(x.v) - v(x.a) - GMv/r + GMx(x.v)/r3 = 0

    It's an additional conserved quantity, and when divided by GM, it yields the "eccentricity vector", (eccentricity)*(closest-distance direction)

    However, R is not completely independent of W and h, however. One can show that R2 = 2Wh2 + (GM)2 and that R.h = 0. So it's only one additional conserved quantity.
  12. Nov 30, 2011 #11
    Let's see where we are at:

    v2 = (dr/dt)2 + r2(d(ang)/dt)2
    (1/2)v2 - GM/r = W
    r2*(d(ang)/dt) = h

    The second is easy:

    d(ang)/dt = h/r2

    Plug it into the first one:
    (1/2)((dr/dt)2 + h2/r2) - GM/r = W

    Solve for dr/dt:
    (dr/dt)2 = 2(W-GM/r) - h2/r2

    You get an integral for t as a function of r:
    t = integrate dr/sqrt(2(W-GM/r) - h2/r2)

    For (ang), you need to integrate over r:
    d(ang)/dr = (d(ang)/dt) / (dr/dt)

    ang = integrate (h/r2) * dr/sqrt(2(W-GM/r) - h2/r2)

    This integral is easier if one integrates over 1/r instead of over r.
  13. Dec 1, 2011 #12
    I'll now show how to do the integrals. It's very awkward to do them when in the form of my previous post, but there are some simplifications that make life easier.

    The distance goes from a*(1-e) to a*(1+e), and the radial velocity must be zero at both distances.

    (dr/dt)2 = (1/r2)((2(W*r + GM)*r - h2)
    (the - in the previous post is a typo)

    One gets two equations, and they can be solved for W and h2:

    W = - GM/(2a)
    h2 = GM*a*(1-e2)

    These give
    (dr/dt)2 = GM/(a*r2)*((a*e)2 - (r-a)2)

    or t = integrate sqrt(a/GM) * r * dr / sqrt((a*e)2 - (r-a)2)

    It's easy to do this integral with the help of substitution of variables:
    r = a*(1 - e*cos(ea))
    where ea is the "eccentric anomaly".

    It produces t = t0 + sqrt(a3/GM) * ma
    where ma = ea - e*sin(ea)
    is the "mean anomaly"
    One also gets Kepler's Third Law.

    Now consider the angle integral

    ang = integrate a*sqrt(1-e2)/r dr / sqrt((a*e)2 - (r-a)2)

    That's more difficult, but let us try r = 1/z. After a little algebra, we get

    ang = integrate a*(1-e2) dz / (e2 - (z*a*(1-e2)-1)2)

    Substitute variables again:
    z = (1 + e*cos(ta))/(a*(1-e2))
    or r = (a*(1-e2))/(1 + e*cos(ta))
    where ta is the "true anomaly".

    One gets
    ang = ang0 + ta
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