I have to derive the equations of motion describing a planet or comet orbiting the sun, then determine what quantity is conserved based upon those equations.


Bold denotes a vector, bold i and j are unit vectors.

Given:

Mass of comet = m

Mass of sun = M

Gravitational potential energy U = -GmM/r, G is the gravitational constant.

Position vector of comet r = rcos(α)i + rsin(α) j



Now, my plan was to find the kinetic energy, then use that to find the Lagrangian and then use the Euler-Lagrange equation to find the equations of motion.

Kinetic energy: Obviously I need to have the velocity dr/dt = ds/dt. My calculus skills are not quite up to par on that, but I was fortunate enough to have access to a formula that I believe works:

ds^2 = dr^2 + r^2dα^2

Which implies:

(ds/dt)^2 = (dr/dt)^2 + r^2 (dα/dt)^2

But I have two questions:

1) I'm not sure how to derive this from the position vector and the standard x = rcos(α) and y = r sin (α). If anyone can show me, that would be appreciated.

2.) When working with differentials, we can just take ds^2 = dr^2 + r^2 dα^2 and divide it all by dt^2? This is what appears to have happened.





Anyway, based upon that initial assumption:

T = (1/2)mv^2 = (1/2) m ((dr/dt)^2 + r^2 (dα/dt)^2)


Now, then therefore the Lagrangian is:

L = T - U

L = (1/2) m ((dr/dt)^2 + r^2 (dα/dt)^2) - (-GmM/r)


Now, I use the Euler-Lagrange equation on this, once for r and once for α: r' and α' are the time derivatives in this case, since I can't use dot notation here.

For r:

∂L/∂r' = mr' + 0 + 0


∂L/∂r = 0 + mr(α')^2 - GmM/r


d/dt (∂L/∂r') = mr''

So,

md^2r/dt^2 - mr(α')^2 + GmM/r = 0

==>

GM/r^2 = r(α')^2 - r''



For α:

∂L/∂α' = 0 + m(r^2)α' + 0


∂L/∂α = 0 + 0+ 0


d/dt(∂L/∂α') = d/dt (m(r^2)α')


So,

d/dt (m(r^2)α') - 0 = 0

==>

d/dt (m(r^2)α') = 0


Looking at this one, what's inside the outer parenthesis must be constant (because the derivative of it with respect to time is zero), so I am assuming that, unless m, r, α' is zero (which is useless), at least one of m, r or α' must be constant. Clearly r can change with time if the orbit isn't a perfect circle, so I am assuming that mα' is constant with respect to time. Is this not saying that angular momentum is conserved? I mean, if the "change in angular momentum with respect to time" is zero, is this not saying that it is constant with respect to time- i.e. it is conserved? Is this correct? (I would expect as much)


Anyway, are these correct? And can anyone help with the first part so this is a complete derivation?

One important conserved quantity is angular momentum which leads to Kepler's second law = equal areas in equal times is a type of conserved quantity = angular momentum about central point / mass
http://en.wikipedia.org/wiki/Orbit#Analysi..._orbital_motion

Angular momentum is a vector, which identifies the plane of the orbit.

There is another conserved vector quantity related to the eccentricity. This vector, the Runge-Lenz vector is also conserved.
http://en.wikipedia.org/wiki/Laplace-Runge-Lenz_vector

Okay, it seems I got the angular momentum one (if anyone can verify, it is the second bold equation I derived), but I can't see a way to get the second one you mentioned.

In fact, I can't even see how I can prove that the orbit is an ellipse from what I have posted. Intuitively, it seems that if E < 0 we would have an ellipse, because that would mean that the gravitational potential energy is negative and greater in absolute value than the kinetic energy- i.e. the force "pulling it" toward the sun is greater than the force "pulling it" away from the sun.

Does that sound about right?



So, how can I prove that the orbit is an ellipse based upon what I have derived about the energy and the two equations of motion I have found (if they are correct)? And where can I get the eccentricity?

Here's what I got:

Equation 1:

GM/r^2 = r(α')^2 - r''


Equation 2 (conservation of angular momentum):

d/dt (m(r^2)α') = 0


Energy equation:


E = (1/2) m ((dr/dt)^2 + r^2 (dα/dt)^2) + (-GmM/r)




Any ideas?