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

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.

• The first thing to do is get rid of your parabolic approximation. Instead work with acceleration vector referenced to the center of mass, and the resulting elliptical motion about it. May 3 at 23:59
• @RyanC Alright. Is the parabola resulting from the angle of launch not relevant at all? It creates an arc with a peak a significant fraction of the radius of the moon. Wouldn't that be relevant to locating the origin of the vector? I don't know if I'm saying that right. May 4 at 0:03
• I can see how it isn't important in the case linked to, but there is also this other case, and I'm trying to combine the two. desmos.com/calculator/e076jaggam May 4 at 0:06
• Launching an unpowered projectile to rendezvous with a skyhook seems like it could get messy if anything goes wrong. May 4 at 0:23
• That's the nature of skyhooks in general. Working with a vertical skyhook helps. I'm not concerned with safety systems at the moment though. The case I'm working on would leave the payload vessel in a highly eccentric orbit if it missed, btw. And the other case isn't that far above the ground, much easier to create a backup for. May 4 at 0:25

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.

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:

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

• Thank you very much, good guy. (I'll take that as a reasonable short form of your tag.) I made the noob mistake of not locating the focus properly in trying to make a visual sketch. The rest I can add to the spreadsheet and try to get used to it. I also notice you fixed the issue in your tether calculator page so it again displays the graphic. I still get a lot out of that tool and am quite happy I've been able to circle back to include it in our modelling again. May 4 at 17:14
• I worked through this as a Desmos page, but the result turned out a bit different than what I could get by playing with the tether tool. So I'm taking the tether tool as authoritative, now that I understand the impact angle is equivalent to the needed launch angle. Playing around shows that a mass driver track with an angle of 28.2 degrees can be used both to launch to the foot at 20 km height, at low acceleration, and to a station at 1512 km height, at higher acceleration. That's what I was looking to design around. Thanks again. May 5 at 1:03