How do you determine (what is the formula?) the delta-V for a sub-orbital 'hop' from one location on the Moon to another? Is it different from a typical orbital calculation?
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1$\begingroup$ The hop is complicated by powered descent and by lunar gravity field. It is necessary to take into account landing accuracy and obstacle avoidance requirements. $\endgroup$– Deer HunterCommented May 2, 2014 at 1:45
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3 Answers
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There was an earlier question about suborbital hops. I will reuse some of the diagrams and explanation from that answer.
A minimum energy ellipse between departure and destination corners of a Lambert space triangle is described on page 65 of the 1993 edition of Prussing and Conway's Orbital Mechanics textbook.
In this particular Lambert Space triangle, both $r_1$ and $r_2$ would be the radius of the moon, 1738 km. The 3 points of the triangle would be moon's center and the departure and destination points on the lunar surface. $\theta$ would be the angle between the two points.
The second focus of this minimum energy ellipse would lie on the center of the chord connecting the points on the lunar surface.
Distance between foci, $2e\cdot a$, is $r \cos(\frac\theta2)$. The major axis of this ellipse ($2a$) is $r(1 + \sin(\frac\theta2))$.
Knowing $r$ (1738 km) and $a = r(1 + \sin(\frac\theta2))$, the vis viva equation can be used to get $\Delta v$ for take off as well as soft landing at other end of suborbital hop.
The vis viva equation is $$v = \sqrt{GM\left(\frac2a - \frac1r\right)}$$
Another useful piece of info is what angle you should depart from the moon's surface. If the destination is near, the angle will be close to $45^\circ$. As the angle between departure and destination approaches $180^\circ$, the flight path angle will approach $0^\circ$, that is, horizontal.
I made a spreadsheet for this. User can input data into the colored cells. I set it for Luna, but a user could also input mass and radius of other bodies, Ceres and Mercury for example.
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$\begingroup$ @ Deer Hunter, Hobbes, HopDavid amd FraserOfSmeg. Thanks everyone. I appreciate all of the input! $\endgroup$ Commented May 2, 2014 at 13:50
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There is one difference with typical orbital calculations: the Moon's gravity field is uneven, due to mass concentrations in various places. If your trajectory crosses one of these mass concentrations, your trajectory will be changed a bit. The gravity value differs by about 0.3% across the Moon's surface.
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1$\begingroup$ An afterthought from this is that it isn't really any different from accounting for the zonal harmonics of the Earths mass concentrations (J2 is typically taken into account, higher harmonics less frequently). $\endgroup$ Commented May 2, 2014 at 21:41
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In its simplest form it isn't different from any other type of orbital calculation. The only difference you may notice is that the perigee of the orbit may be within the surface of the moon. For a quick thought experiment, imagine drawing an ellipse onto a 2D picture of the moon. If you make this ellipse pretty small (and pretty circular) you can connect two close by points by drawing about half of the ellipse, starting at your take off and stopping at your landing point. As you go for a larger hop you draw more and more of the ellipse.
Things get a little more complicated when you take into account how you want to take off and land, but in essence it's just like any other type of orbit.
