Grasshopper
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?
rpenner
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
Grasshopper
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.

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?
Confused2
QUOTE (Grasshopper+)
Equation 1:

GM/r^2 = r(α')^2 - r''

I'm struggling too ..

I'm not sure we can take the acceleration for a circular orbit and just tag on a modification for the fact that it's not circular - I think this might need more thought

Time passes..

-C2.

Edit .. do we have a text convention for unit vectors? r hat ?

Edit2 .. I'm thinking GMm/r^2 = - mr'' is as far as one can go .. we have to get α' by some other means.
prometheus
QUOTE (Grasshopper+Mar 17 2009, 10:16 PM)
Energy equation:

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

If you want to figure out the equations of motion it is a lot easier to use the Euler Lagrange method. Now you have the energy you can easily work out the Lagrangian and use the Euler Lagrange equation to find the equations of motion.
Confused2
QUOTE (Prometheus+)
If you want to figure out the equations of motion it is a lot easier to use the Euler Lagrange method.

Humph. I can't even follow the notation of the wiki entry ( http://en.wikipedia.org/wiki/Euler-Lagrange_equation ).

Following the link suggested by rpenner leads to a nicely worked 'conventional' analysis :-

http://farside.ph.utexas.edu/teaching/301/...es/node155.html

-C2.
Grasshopper
QUOTE (prometheus+Mar 19 2009, 08:46 AM)
If you want to figure out the equations of motion it is a lot easier to use the Euler Lagrange method. Now you have the energy you can easily work out the Lagrangian and use the Euler Lagrange equation to find the equations of motion.

I actually have already used the Euler-Lagrange equation, if that is what you are referring to. I started out with a displacement function and a potential energy function, and used the Lagrangian to derive two equations, one which seems to imply that angular momentum is conserved.

According to the link below, I did get the correct two equations.

r'' - r(α')^2 = -GM/r^2

and

d/dt (m(r^2)α') = 0

In fact, to my great joy, the link appears to goes so far as to show how to get the equation of an ellipse.

http://farside.ph.utexas.edu/teaching/301/...es/node155.html

It looks like you've got to do a variable substitution and then solve a differential equation. In hindsight, that seems obvious, since in order to have the equation of an ellipse with an eccentricity, you need a cosine function to appear in a denominator of a term. You get cosines from solving differential equations, as I recall.

Anyway, thanks for all the help. This should serve me well next semester when I study classical mechanics (assuming I don't fail this semester )

I think I will figure this part out from the link, but if I have a question about how E < 0 relates to the eccentricity (and I think that is where I will have problems), I know who to call.
furyluzt
QUOTE (Grasshopper+Mar 16 2009, 05:44 AM)
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?

What exactly is s? You said that obviously dr/dt = ds/dt, but I can't find out what s is here. Thanks!