Deriving Mechanics Equations for the Moon

[Dime]

Alright, so I have a few questions.
a) Consider a moon of mass m orbiting a planet of mass M in an elliptical orbit. The equation for the radius of an elliptical orbit is r=a(1-e²) / (1+ecos$$\vartheta$$). Derive equations for the orbital radius of the moon when it closest to and farthest from the planet.
My answer turns out to be periapse: r=a(1-e)
and for apoapse: r=a(1+e) a is the semi major axis of the ellipse and e is the eccentricity

b) Combing your result from a with angular momentum (L=mvr), derive an equations for the ratio of the moon's speed at periapse and apoapse in terms of eccentricty, that is v_p/v_a.

I ended up getting v_p/v_a=(1+e) / (1-e)​

c) Write down the equation which equates the total energy of the moon at apoapse with its total energy at periapse using the equation E_p=-GMm / r
I am getting confused with this one. So the total energy would be the gravitational potential energy plus the kinetic energy right? So for the energy at periapse would look like

E=(-GMm / r) + 1/2mv²​

the velocity being the one derived earlier.
then using the equations derived in the first couple parts I end up getting

E=(-2GMm + mv(1-e)) / (2a(1-e)​

And somehow I think that is wrong. The next part asks me to show that at periapse

v²=(GM / a)[(1+e)/(1-e)]​

but I may be over complicating things for part c and trying to solve d at the same time I am not sure. Anyone have any ideas if I made a mistake earlier or if they can offer some advice in solving it?

 

[Dime]

nevermind I figured it out

 

[kerimek]

I would approach this problem by first reviewing the equations and understanding the variables and their meanings. The equation for the radius of an elliptical orbit is r=a(1-e2) / (1+ecos\vartheta), where a is the semi-major axis and e is the eccentricity of the orbit. This equation can be derived from Kepler's laws of planetary motion.

To derive equations for the orbital radius of the moon at periapse and apoapse, we can use the fact that at these points, the velocity of the moon is perpendicular to the radius vector. This means that the angular momentum (L=mvr) is constant at these points. So, we can equate the angular momentum at periapse and apoapse to get:

mvp(1-e)=mva(1+e)

Solving for vp and va, we get:

vp=va(1+e) / (1-e)

This is the ratio of the moon's speed at periapse and apoapse, and it is dependent on the eccentricity of the orbit.

Moving on to the third part, we need to equate the total energy of the moon at periapse and apoapse. The total energy includes both kinetic and potential energy, so we can write:

Ep=Ek+Ep

At periapse, the kinetic energy is zero since the moon is at its closest point to the planet and has no velocity. So, we can write:

Ep(periapse)=-GMm / r(periapse)

Using the equation for the radius at periapse (r=a(1-e)), we get:

Ep(periapse)=-GMm / a(1-e)

Similarly, at apoapse, the potential energy is zero and the kinetic energy is at its maximum. So, we can write:

Ep(apoapse)=Ek(apoapse)=1/2mva2

Using the equation for the radius at apoapse (r=a(1+e)), we get:

Ep(apoapse)=-GMm / a(1+e)

Setting these two equations equal to each other, we get:

-GMm / a(1-e) = 1/2mva2

Solving for a, we get:

a=GM / 2va2

Substituting this value of a into the equations for the radius at periapse and apoapse, we get:

r(periapse)=GM / 2