How do you calculate the characteristics of an orbit of an object launched from a mass driver?

Question

What formula can I use to calculate the apoapsis of the orbit of an object launched from a mass driver on the moon at a given speed and angle to the surface (angle from horizontal)? The device will launch due east.

I've plugged some things into a Desmos calculator page. Expanding that with the rest of the formulae needed would be ideal.

I'm deciding how to configure the track of a mass driver for our virtual portrayal of a highly developed lunar town. I want to combine its use with the use of skyhooks, both vertical and rotating, to make it work well for trasferring goods between the town and manufacturing and construction projects in a wide range of orbits and locations in the Earth Moon system. Because the town is at the bottom of a large crater, and skyhooks are involved, it has seemed like a good idea to angle up the mouth of the mass driver so vessels launched rendezvous with the vertical skyhook in one of a couple of ways. Not necessary info, but yeah, that's what I'm trying to do.

Answer 1

Since we're in an elliptic orbit as soon as we leave the mass driver, we can use the velocity we have to set up the vis-viva equation;

$$v^2 = \mu \cdot \left(\frac{2}{r_{moon}} - \frac{1}{a}\right)$$

And then solve for the semi-major axis

$$a = \frac{1}{\frac{2}{r_{moon}} - \frac{v^2}{\mu}}$$

Having the semi-major axis, we can take advantage of Kepler's third law, the product of horizontal speed and radius being constant. The horisontal speed will match the orbital speed at exactly two points, the periapsis and the apoapsis. (Here we only care about the horizontal component $v_h$ at the mass driver)

$$\mu \cdot \left(\frac{2}{r_{AP}} - \frac{1}{a}\right) = \left(\frac{v_{h} \cdot r_{moon}}{r_{AP}}\right)^2$$

Yielding two solutions, one for the periapsis distance and one for the apoapsis distance:

$$r_{AP} = \frac{a \mu \pm \sqrt{a^2 \mu^2 - ar_{moon}^2 v_h^2 \mu}}{\mu}$$

Worth mentioning that since orbits repeat, the resulting orbit intersects with the ground, a sub-orbital trajectory.

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To find the optimal angle of the mass driver, consider time reversing the orbit, an object falling from the tether foot and impacting the Moon at some angle (the angle we want).

Solve for the semi-major axis of the orbit, same as above:

$$a = \frac{1}{\frac{2}{r_{tether foot}} - \frac{v_{tether foot}^2}{\mu}}$$

Find the velocity at impact, vis-viva again

$$v_{impact} = \sqrt{\mu \cdot \left(\frac{2}{r_{moon}} - \frac{1}{a}\right)}$$

And the horizontal component, Kepler's third law again:

$$v_{horizontal} = \frac{v_{tether foot} \cdot r_{tether foot}}{r_{moon}}$$

And the angle crystallises out by trigonometry:

$$\theta = \cos^{-1}\left(\frac{v_{horizontal}}{v_{impact}}\right)$$