# My ballistic trajectory on the Moon looks wrong

1. Mar 20, 2017

### stationheister

Hi all,

I'm writing a science fiction story and I have managed to confuse myself. I have a rail line that accelerates a train in Tycho crater and tosses a heavy payload of ore to Descartes, about 1200 km, where it's supposed to land on another ramp... or something. I've used Google Earth to estimate my distance and ejection angle, then I fudged my figures into shape using this CalcTool:

http://www.calctool.org/CALC/phys/newtonian/projectile

Ejection angle: 9 degrees
Release velocity: 2500 m/s !?
g: 1.62 m/s^2

Which gives me

Max height: 49488.5 m
Distance traveled: 1192.2 km
Time taken: 482.2 s

The problem I see here is that 2500 m/s, mostly horizontal, is considerably more than the approximately 1000 m/s minimum orbital velocity of the Moon, and more than the 2380 m/s escape velocity of the Moon. It just doesn't seem right.

I'm aware that this calculation fails to account for the curvature of the Moon, its rotation, mass concentrations and the tenuous atmosphere that all these rockets would eventually create. But I had *hoped* that such considerations would not be significant enough to worry about.

(We also don't have to worry about such things because there is another railroad car parked across the tracks at the end of the ramp, and what we are really interested in is the size of the explosion it's going to create. Right now it looks... meteoric!)

I'd really appreciate some help in restructuring my calculations and helping me to understand what's going on here. Thank you in advance for your help.

2. Mar 20, 2017

### Staff: Mentor

Your distance is a significant fraction of the circumference of the Moon, the two points are much more than 9 degrees apart. It will be relevant.

As approximation for a flat trajectory, you can calculate the centrifugal acceleration seen by the projectile, and subtract that from the gravity of Moon to get an effective gravitational attraction. Iterate until you have a solution that fits.

Alternatively, use elliptic orbits to find a solution.