**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 d

**r**/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?