Trollin' through Hop David poasts yesterday, I impacted the pogo. It's one more way around Tsiolkovsky, where the planetoid has low surface gravity and no atmosphere. If the maths added up . . . which we're here to check.
The maths aren't even about the pogo (I am preparing different maths for that); they're about getting off and getting back, by rocket if you have to. Hop has a spreadsheet, for the moon-intersecting ellipse. Travelling "just" 300 km means launching at 670 m/s and hitting the ground again at 670 m/s. Says he.
At stake here is if the ballistic jump will beat out an electric golf cart for that distance - especially if no roads. And it will all be even more feasible for lower-gravity worldlets like Ceres and Callisto.
Checking the spreadsheet, say you want to go one mile per leap, 1.60934 km. Hop claims 114 mph per impact; then, of course, you slow down until aposelene (r = a x (1+e)) and speed back to 114 mph on return. As a velocity this is exerted at a 45° angle. Cosine is 0.7071, so the horizontal vector started and ended 80.8 mph. So on a flat surface, which even on our small Moon we can assume paved for one measly mile, I should be taking 0.742 minutes.
Then I look at that "ToF" at cell D27. This has the flight taking 0.185726903 minutes. Where one mile / minute is 60 mph (famously), we've traveled that mile at 5.384 times 60 mph. A golf cart leaving and meeting the ballistic would be travelling an average 323 mph.
Too good to be true, I daresay. Something bad happened on the way to cell D27.
D15 e is the eccentricity and D16 a is the semimajor axis (in km) of my ballistic trajectory. Hold on to e; not using it yet, except to explain why a looks like only half the Lunar radius on a short hop.
Hop assumes a "planet" whose surface is a perfect sphere of wire mesh and whose whole mass is locked in a tiny ball in the centre. (As he should.) Let's grant to that core its actual 380 km radius. Now: consider falling through a hole in the mesh at a slight angle against that core. You drop 1360 km, blast past the core; get wrenched at a g high enough to break your Lunar-weakened bones and to send your internally-bleeding and concussed body back to the surface. That - Hop notes - is a Keplerian ellipse, eccentricity near 1 and the focus at the core. Contrast the Ceres airless mohole journey. That isn't Keplerian; the focus shifts as you fall, and the acceleration shifts (to zero), so your true a is diameter - not radius. Because... the mass of that 380 km radius isn't 100% of the lunar mass, it's 0.75–1.75%. But anyway: we're staying over the surface so don't touch D16.
If we don't touch D16 we move on to D17 surface velocity, in km/s. Here Hop, unbound by the Moon's internals, does vis viva SQRT($B$4*D9*(2/D10-1/D16)). $B$4*D9 is Lunar μ here 4900 km3/s2; 4904.8695 is the measured value but hey. D10 r = 1738 km which is where the Lunar disc intersects the ellipse; Hop doesn't want to be more precise than that in case of mountains and craters. D22 has the suborbital period, which is Kepler: =2*PI()*SQRT(D16*D16*D16/4904.8695) in seconds. For the near-perfect circle (so a=1738 km): D23 has 108.35 minutes. Lunar Reconnaissance Orbiter at 50 km so semimajor 1788 km, goes around at about 113 minutes (pdf). Kepler wins.
Time to look at D15, eccentricity. The equation checks out if I put 5458.5 (half circumference, as far as this XL will let me) as my travel-distance; it's almost a circle so e nears zero. And the time-of-flight is 54.17 minutes about half the suborbital period. Flight-time is bad only for short distances... meaning the distances we care about.
I take D26 to be the mean-anomaly against the mean-motion-constant. Last April I did some calculations starting with eccentric-anomaly at distance r given (cycler) orbits sent to Mars (and ideally back again). Hop instead has D26 = atan(sqrt((1-e)/(1+e))/e)/π, to be multiplied by the suborbital period. When it approaches a circle, this is very near 0.5; fine. Not so fine for a short hop. I have no idea whence Hop's D26 equation.
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