Author Topic: Suborbital Lunar Hops  (Read 18991 times)

Offline Hop_David

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Suborbital Lunar Hops
« on: 04/04/2014 09:12 PM »
This problem has been vexing me for a few years: minimum energy ellipse between points on the Lunar surface.

Then it occurred to me that departure and destination points on the Lunar surface could be corners of a Lambert space triangle. Of course, the third point would be the moon's center.

And on page 65 of Prussing and Conway's 1993 edition of Orbital Mechanics is a description of the minimum energy ellipse.

In this case both r1 and r2 would be 1738 km, the moon's radius. θ would be the angle between the two locations.

The second focus of the ellipse would lie in the middle of the chord connecting departure and destination points. Major axis of the ellipse would be r(1+sin(θ/2)). e would be cos(θ/2) / (1+sin(θ/2)). Attached is a pic that might make this clear.

Also attached is a spreadsheet giving velocity and angle of suborbital launch from surface. In this spreadsheet I put the angular separation between the pole and an equatorial point - 90 degrees. This suborbital launch would be 1.53 km/s at an angle of 22.5 degrees. Typing into the pink cells user can change angle between departure and destination. The user can also type in radius and mass for other bodies.
« Last Edit: 04/05/2014 03:30 AM by Hop_David »

Offline CitizenSpace

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Re: Suborbital Lunar Hops
« Reply #1 on: 04/28/2014 10:48 AM »
I'm guessing the reason why this got no replies is not many people know what your saying  :P What are you trying to say?

Offline AlanSE

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Re: Suborbital Lunar Hops
« Reply #2 on: 04/28/2014 04:08 PM »
Then it occurred to me that departure and destination points on the Lunar surface could be corners of a Lambert space triangle. Of course, the third point would be the moon's center.

I think there are some prerequisite details that most people reading this won't get.  At least I didn't right away.  If you go back to the basic definition of an ellipse and orbital mechanics, it is true that:

 - The sum of the distances to the ellipse's foci from any point on the ellipse is a constant value
 - One focus of an elliptical orbit is the center of a planet

With these (and possibly some others), it seems quite possible that one could go about making a geometric proof to establish an algebraic formula to get the minimum-energy trajectory.

Mathematically this is interesting as a generalization of two more obvious optimization problems.  For one, if we're talking about short distances, the optimal trajectory would be a 45-degree toss in the absence of air resistance.  On the other extreme, getting to the exact opposite side of the moon could (theoretically) be done with a perfectly circular orbit at 0 altitude.

For the toss over short distances, I believe the second focus of the ellipse would lie on the surface itself.  Then it's obvious how the ellipse definition then leads to a parabola in this limit-case.  So over the full range of the problem, the second focus shifts from the surface of the moon to the center.  However, if you toss 1 meter or 100 meters, it seems that the focus is functionally on the surface, just at different horizontal displacements relative to the launch location.  This leads me to believe that changing the location of the second focus would be at least a 2nd order effect with respect to the separation of the launch and land points.  Sounds hard.
« Last Edit: 04/28/2014 04:25 PM by AlanSE »

Offline cordwainer

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Re: Suborbital Lunar Hops
« Reply #3 on: 04/29/2014 05:45 AM »
So basically it's like trying a to throw a baseball fast enough and at the right trajectory to reach the other side of the Moon. To put it in layman's terms. I suppose a mass driver could be used to effect a propellantless journey from on part of the moon to another in this way.

Offline ChrisWilson68

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Re: Suborbital Lunar Hops
« Reply #4 on: 04/29/2014 05:53 AM »
Why is this in Avanced Concepts?  I don't want to offend anyone, but honestly any good student in a first year university physics class should be able to solve this problem.

Edit: two things 
1) It may be easily solvable but the concept of minimum fuel takes a bit more than solving the equatiion.  Fine here.
2) if you see a problem with a topic, report it, don't post.
++Lar
« Last Edit: 04/29/2014 06:06 PM by Lar »

Offline sdsds

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Re: Suborbital Lunar Hops
« Reply #5 on: 04/29/2014 06:45 AM »
In the image I found it useful to shade in the Moon. Regarding what this is saying, it might be a way to find the angle and velocity at which to launch from point A so that you land at point B, using minimum energy. But I think it minimizes the launch energy, with the expectation that doing so also minimizes the sum of launch and landing energies, and that both launch and landing are instantaneous impulses.

Also, I would like to better understand the physics that supports the conclusion that the second focus is the center of the chord connecting the launch and landing sites!

(P.S.: I'm glad CitizenSpace found this topic and brought it back to life!)
-- sdsds --

Offline AlanSE

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Re: Suborbital Lunar Hops
« Reply #6 on: 04/29/2014 02:32 PM »
So basically it's like trying a to throw a baseball fast enough and at the right trajectory to reach the other side of the Moon. To put it in layman's terms. I suppose a mass driver could be used to effect a propellantless journey from on part of the moon to another in this way.

"the right trajectory" being the minimum energy trajectory, yes.

Mass drivers and tether tossers are possible.  I've read a lot about these, but in real life, it seems like the act of catching the projectile would be considerably more challenging than throwing it in the first place.

Offline AlanSE

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Re: Suborbital Lunar Hops
« Reply #7 on: 04/29/2014 04:09 PM »
The second focus of the ellipse would lie in the middle of the chord connecting departure and destination points. Major axis of the ellipse would be r(1+sin(θ/2)). e would be cos(θ/2) / (1+sin(θ/2)). Attached is a pic that might make this clear.

Also attached is a spreadsheet giving velocity and angle of suborbital launch from surface. In this spreadsheet I put the angular separation between the pole and an equatorial point - 90 degrees. This suborbital launch would be 1.53 km/s at an angle of 22.5 degrees. Typing into the pink cells user can change angle between departure and destination. The user can also type in radius and mass for other bodies.

I've had a look at this spreadsheet, and I grabbed all the calculations and reduced them.  In the spreadsheet you perform the following calculation:

Given angular separation between points of: theta = 90 degrees
Found optimal launch angle of: 22.5 degrees

Allow me to propose an alternative closed-form version of this.  This follows directly from the calculations in your spreadsheet.  In fact, it is just a reduced form of all the formulas there.

optimal launch angle = (Pi - theta) / 4
 = (180 - 90) / 4 = 22.5

This doesn't give you the other values, such as eccentricity and whatnot.  However, it would seem that the optimal launch angle would be established by a really really simple calculation, given that you know the angle between the release and land points.
« Last Edit: 04/29/2014 04:10 PM by AlanSE »

Offline Proponent

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Re: Suborbital Lunar Hops
« Reply #8 on: 04/29/2014 04:47 PM »
Also, I would like to better understand the physics that supports the conclusion that the second focus is the center of the chord connecting the launch and landing sites!

Recall that the energy of an elliptical orbit of semi-major axis a is -0.5GM/a.  Therefore, minimizing the energy of the transfer orbit is equivalent to minimizing a.  Let be T the angle between the major axis and a point on the ellipse as seen from the moon's center, where the other focus is at T=0 (in other words, T is theta/2 in sdsds's diagram, above).  Then the radius of a point on the ellipse is r = (1 - e2) a / (1 - e cos T), where e is the eccentricity (usually the denominator is 1 + e cos T, but the other focus is taken to be at T=180o).

Now set r to the moon's radius.  In fact, it's convenient to measure distances in units of the moon's radius, so r=1.  Solve the radius equation for a as a function of e and minimize.  The distance from one focus to another is 2ea, which, at minimum a, turns out to be cosT.  QED.
« Last Edit: 04/29/2014 06:51 PM by Proponent »

Offline cordwainer

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Re: Suborbital Lunar Hops
« Reply #9 on: 05/01/2014 10:22 PM »
Well if your only sending bulk materials then you could place them in an oblong shaped travel container like a rhombus or caltrop shape and place airbags or a woven TPS over the exterior to protect the container from crash damage. You don't have to worry about catching something if your just crashing it into the surface, and the amount of energy for one of these trajectories on the Moon wouldn't necessarily require a great deal of velocity or kinetic energy.

Offline Hop_David

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Re: Suborbital Lunar Hops
« Reply #10 on: 05/02/2014 07:09 AM »
Why is this in Avanced Concepts?

It's about transportation between points A and B on a roughly spherical body like the Moon. It could also be applied to other airless, spherical bodies like Ceres, Mercury, Enceladus, etc.

Taking off from one points A to B on the lunar surface assumes a railgun, sling, or perhaps ISRU propellant. Else, why not just go directly from earth to point B?

I'd like to see suborbital hops on the moon become routine. But right now it's Advanced Concepts, in my opinion.


I don't want to offend anyone, but honestly any good student in a first year university physics class should be able to solve this problem.

Some questions came up in a space forum: how much delta V would it take to go from one lunar pole to the other? How much ffrom a pole to the equator? How much from a north pole to a point 10 degrees south at an 80 degree latitude?

There seemed to arise a consenus: Near points could be reached with little delta V taking off at close to 45. Far points (like from north pole to south pole) would take off horizontally at a close to circular orbit at nearly 1.6 km/s. But I didn't see anyone offer a method for determining exactly what the departure speed and flight path angle were. There were a number of competent people participating. I don't recall if was NasaSpaceflight or another forum.

Recently I saw a similar question come up in the Space Stack Exchange. For a suborbital flight between Sidney and London, how far below the surface was perigee? Mark Adler gave the correct answer but he used numerical methods, tools not easily accessible to many space enthusiasts. But some of Adler's wording reminded me of a phrase "minimum energy ellipse" from Prussing & Conway's chapter of Lambert space triangles. I looked up the chapter and, voila, a nice, simple method unfolded.

Substituting the single quantity r for r1 and r2 in Conway's equations quickly led to the the second focus of the minimum energy ellipse lieing on the center of the chord connecting desparture and destination points. From there high school trigonomety and algebra give semi-major axis a.

And yes, freshmen aerospace gives the vis viva equation.

Simple, yes. Obvious? Well, then you are being insulting as it took a few years for this to occur to me.
« Last Edit: 05/02/2014 07:26 AM by Hop_David »

Offline Hop_David

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Re: Suborbital Lunar Hops
« Reply #11 on: 05/02/2014 07:23 AM »
In the image I found it useful to shade in the Moon. Regarding what this is saying, it might be a way to find the angle and velocity at which to launch from point A so that you land at point B, using minimum energy. But I think it minimizes the launch energy, with the expectation that doing so also minimizes the sum of launch and landing energies, and that both launch and landing are instantaneous impulses.

Quite so.

Also, I would like to better understand the physics that supports the conclusion that the second focus is the center of the chord connecting the launch and landing sites!

I first arrived at that conclusion by substituting the single quantity r for r1 and r2 in Prussing and Conway's equations for a minimum energy ellipse. But a nice simple geometrical explanation has occured to me. Hope I will have time and energy to do the drawings in the near future. Thanks for shading the moon for me, looking at it I can see both seem nearly circular which is confusing. When I do the drawings, I'll shrink the angle between departure and destination points which will make the suborbital hop more elliptical.

Offline Hop_David

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Re: Suborbital Lunar Hops
« Reply #12 on: 05/02/2014 07:36 AM »
Allow me to propose an alternative closed-form version of this.  This follows directly from the calculations in your spreadsheet.  In fact, it is just a reduced form of all the formulas there.

optimal launch angle = (Pi - theta) / 4
 = (180 - 90) / 4 = 22.5

That's pretty. Thanks!

It's late and I'm tired. At the moment don't have energy to reduce the equations, but plugging different quantities for theta into my spreadsheet does indeed consistently give (pi - theta) / 4.

Offline ChrisWilson68

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Re: Suborbital Lunar Hops
« Reply #13 on: 05/02/2014 07:54 AM »
Why is this in Avanced Concepts?

It's about transportation between points A and B on a roughly spherical body like the Moon. It could also be applied to other airless, spherical bodies like Ceres, Mercury, Enceladus, etc.

Taking off from one points A to B on the lunar surface assumes a railgun, sling, or perhaps ISRU propellant. Else, why not just go directly from earth to point B?

I'd like to see suborbital hops on the moon become routine. But right now it's Advanced Concepts, in my opinion.

Different ideas for how to get around on the Moon is potentially a legitimate subject for "Advanced Concepts".

But the original post in this thread had nothing like that.  It was simply addressing a question of how to calculate the answer to a specific question.  That question can be easily answered with basic classical mechanics.  It's definitely not "Advanced Concepts" by any stretch of the imagination.

And yes, freshmen aerospace gives the vis viva equation.

Simple, yes. Obvious? Well, then you are being insulting as it took a few years for this to occur to me.

How long you personally thought about it before a solution came to you is irrelevant.  It's still a question from a first-year physics class.  It's still not "Advanced Concepts".

Discussions of how to solve basic mechanics problems can be interesting to some people.  I'm just suggesting that a forum area entitled "Advanced Concepts" is not the place for them.

Offline Hop_David

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Re: Suborbital Lunar Hops
« Reply #14 on: 05/02/2014 09:19 AM »
A quick geometrical demonstartion that second focus lies on center of chord connecting two points.

First, I'll quote Proponent as he's already done some work for me:

Also, I would like to better understand the physics that supports the conclusion that the second focus is the center of the chord connecting the launch and landing sites!

Recall that the energy of an elliptical orbit of semi-major axis a is -0.5GM/a.  Therefore, minimizing the energy of the transfer orbit is equivalent to minimizing a.  Let be T the angle between the major axis and a point on the ellipse as seen from the moon's center, where the other focus is at T=0 (in other words, T is theta/2 in sdsds's diagram, above).  Then the radius of a point on the ellipse is r = (1 - e2) a / (1 - e cos T), where e is the eccentricity (usually the denominator is 1 + e cos T, but the other focus is taken to be at T=180o).

Now set r to the moon's radius.  In fact, it's convenient to measure distances in units of the moon's radius, so r=1.  Solve the radius equation for a as a function of e and minimize.  The distance from one focus to another is 2ea, which, at minimum a, turns out to be cosT.  QED.

Of special interest is energy of an elliptical orbit of semi-major axis a is -0.5GM/a. Minimum energy ellipse between two points is the one with the shortest a.

Here's a pic of one possilbe ellipse between two points:



The major axis is 2a which is the sum of the black leg and red leg.

Now here's 3 possible ellipses all passing through the same two points on the lunar surface:



So with three ellipses we have three major axi, each a sum of a black leg and colored leg. In all three, the black leg is the same.

So the one with the shortest a is the one with the shortest colored leg. Thinking of various possibilities, the shortest possible colored leg seems to be going to the focus on the center of the chord connecting the two points.

Offline Proponent

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Re: Suborbital Lunar Hops
« Reply #15 on: 05/02/2014 03:21 PM »
That's a brilliant and very helpful piece of geometric reasoning.

Offline AlanSE

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Re: Suborbital Lunar Hops
« Reply #16 on: 05/02/2014 05:23 PM »
Thinking of various possibilities, the shortest possible colored leg seems to be going to the focus on the center of the chord connecting the two points.

Really?  Do you "know" this, or is it simply an intuitive guess, as your wording might be indicating?  This would fairly completely describe the problem with just geometry.  Let me play with this:

Start with:  Assume the focus lies on the cord, as Hop suggests.

Now, for launch angle, we need to entertain the local angle of the ellipse at the release/catch point.  This would be difficult with calculus, but I wonder if we can get a shortcut.  An ellipse is a collection of points for which the sum of distances to the two foci is constant.  We might be able to determine the angle by that fact.

 - draw a circle around both foci, which passes through the launch point
 - the angle of the orbit at the launch point will be half-way between the directions of the circles there, in order to preserve the distance sum between the foci
 - the center focus' circle just traces the path of the planet, so that direction is along the surface
 - the other focus is on the cord, which is the same y-position as the launch point
 - that means the angle of that circle at the launch point is directly up
 - angle of the surface (relative to the horizontal) is theta/2 (via triangles)
 - angle between the two circles (at launch point) is thus pi/2-theta/2
 - that means that the direction of the orbit at launch will be (pi-theta)/4 (via the 2nd point)

I think this is fairly robust logic, and it shows that the optimal launch angle (which was previously proven by minimization of the semi-major axis by Hop, which is the formula I teased out of his spreadsheet) can be obtained by making the suggested assumption that the focus lies on the cord, as suggested.

So I guess you could formally prove the fact if you could simply reverse the above steps.
« Last Edit: 05/02/2014 05:26 PM by AlanSE »

Offline mheney

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Re: Suborbital Lunar Hops
« Reply #17 on: 05/02/2014 06:36 PM »
Actually, the "proof" relies on the properties of an ellipse and the fact that you're trying to minimize the semi-major axis (a).

Let's start by placing the center of the body we're considering at (0,0), and placing the major axis of any orbital ellipses we construct along the Y axis (x=0).

Start with these facts (assuming a spherical body)
1:  The ellipse has two foci - f1 and f2
2:  One of the two foci (call it f1) lies at the center of the body.
3:  For any point p on an ellipse with semi-major axis a, the distance f1->p->f2 is 2a.
3.  For any point p on the ellipse that also intersects the surface of the body, the distance f1->p is r (the radius of the body).

That means that in order to minimize the semi-major axis (a), you need to minimize the distance p->f2 (which has value 2a-r).

p is a fixed point on the surface, at (x1, y1).  The focus f1 is fixed, at (0,0).
What can vary is the focus f2 - which, since it's on the y axis, is at (0, y2).

We want to minimize the distance between p and f2, which is sqrt( (x1 - 0)2 + (y1 - y2)2). 
The only non-constant value in there is y2, so the only part we can minimize is (y1 - y2)2, which is 0 if y1 = y2.

So yes, you minimize a (and therefore the total energy) if f2 is on the chord between the launch and impact sites.

Offline KelvinZero

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Re: Suborbital Lunar Hops
« Reply #18 on: 05/03/2014 11:33 AM »
Now that you guys have done the math,
Suppose in some future suborbital lunar hops were used regularly to travel around the moon, much like jets are used on earth. Would it be more efficient to make all journeys in a single hop or would there be reason to break larger trips into two or more smaller hops with refueling?

Offline Proponent

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Re: Suborbital Lunar Hops
« Reply #19 on: 05/03/2014 02:03 PM »
It's definitely more efficient to do it in a single hop.

Consider, for example, the case where the distance to be travelled is much smaller than the moon's radius, so we can just pretend the moon is flat.  Then, by Tartaglia's formula, the range of a hop is v2/g, where v is the take-off speed and g is the local acceleration of gravity.  A take-off speed of 100 m/s will get you a certain distance.  To cover the same distance in two equal hops, each take-off would have to be at 70.7 m/s, giving a total delta-V of 141 m/s.  And that's ignoring the braking needed at the end of each hop.
« Last Edit: 05/03/2014 02:05 PM by Proponent »

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