Suborbital Lunar Hops

[Hop_David]

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.

[CitizenSpace]

I'm guessing the reason why this got no replies is not many people know what your saying  What are you trying to say?

[AlanSE]

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.

[cordwainer]

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.

[ChrisWilson68]

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

[sdsds]

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!)

[AlanSE]

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.

[AlanSE]

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.

[Proponent]

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=180^o).

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.

[cordwainer]

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.

[Hop_David]

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 45º. 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.

[Hop_David]

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.

[Hop_David]

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.

[ChrisWilson68]

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.

[Hop_David]

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=180^o).

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.

[Proponent]

That's a brilliant and very helpful piece of geometric reasoning.

[AlanSE]

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.

[mheney]

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.

[KelvinZero]

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?

[Proponent]

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.

[Hop_David]

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.

This is one of the things I was interested in looking at.

Spirit and Opportunity landed on Mars using airbags. But after aerobraking and firing of retrorockets, the impact was only 24 meters/sec. That's about 86 km/hour or 54 miles per hour.

A suborbital hop from the lunar pole to equator would impact at about 1.53 km/s which is about 3400 mph.

Impact wouldn't fall to 54 miles per hour until the distance between departure and destination was shrunk to ~tenth of degree, about 400 yards.

But inert supplies would be much less fragile than a rover. For buik commodities a higher impact speed might be acceptable. If 250 mph crash is okay, the departure and destination could be 7.5 km apart. This might be a way to get stuff from the rim of Whipple Crater to the crater floor.

[Proponent]

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.

I didn't believe this at first, but I was wrong.  The details are worked out in the attachment (corrections welcome).  It also turns out that the tangent of the optimal launch angle is equal to the eccentricity of the optimal trajectory.

EDIT: Replaced attachment with corrected version.

[sdsds]

Wow, the geometric insights, confirmed by the calculus, are quite stunning.

Can I ask for confirmation of one underlying bit of the reasoning? Proponent writes: "The trajectory requiring the least initial speed is the trajectory
of least energy." That's because for all possible trajectories, the potential energy at the launch point is the same, so the only thing left to minimize is the kinetic energy, and minimizing kinetic energy is equivalent to minimizing speed. Yes?

Also, did I miss it? Is there a simple equation giving the initial velocity v0 as a function of theta, normalized with the circular orbital velocity at r=R (a constant, vC)? Obviously when theta=0 then v0=0, and as mentioned when theta=pi then v0=vC. And the function must be monotonically increasing on the interval. Plotting a few points it sure looks simple!

[Hop_David]

Wow, the geometric insights, confirmed by the calculus, are quite stunning.

Can I ask for confirmation of one underlying bit of the reasoning? Proponent writes: "The trajectory requiring the least initial speed is the trajectory
of least energy." That's because for all possible trajectories, the potential energy at the launch point is the same, so the only thing left to minimize is the kinetic energy, and minimizing kinetic energy is equivalent to minimizing speed. Yes?

Yes.

Specific orbit energy is v^2/2 - mu/r. First term is kinetic, 2nd potential. As you say, mu/r is fixed.
Mu/r = G * moon mass / 1738 km

Another way to see it is looking at the vis viva equation: v^2 = mu (2/r - 1/a). When a gets smaller, the term subtracted from 2/r gets bigger, which makes the whole thing smaller.

[Hop_David]

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.

I didn't believe this at first, but I was wrong.  The details are worked out in the attachment (corrections welcome).  It also turns out that the tangent of the optimal launch angle is equal to the eccentricity of the optimal trajectory.

I've found a geometrical way to demonstrate Alan SE's pretty result:



Using a well known reflective property of the ellipse it can be seen that
2 alpha + (pi - theta)/2 = pi.
A little straight forward algebra gives
alpha = (pi + theta)/4.

Angle between position vector and velocity vector is alpha + (pi - pheta)/2 which comes out to (3 pi - theta)/4

A horizontal velocity vector is pi/2 from the position vector. When pi/2 is subtracted from (3 pi - theta)/4 you get Alan's pleasing result.

I am still looking at my drawings searching for a visual demonstration of Proponent's nice find. Reading the first part of Proponent's pdf I can see Kepler style drawings but it becomes opaque to me when he starts using differential calculus.

[KelvinZero]

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.

Isn't that comparing the case without refueling? I know fuel goes up exponentially with delta-V, so couldn't a 70% delta-v easily work out to be about half the effort, and two 70% hops of equivalent effort?

When I saw that V^2 component my first guess was that distance travelled was proportional to kinetic energy no matter how you split it.. but such a simplistic approach would probably only work for travel by trebuchet

[AlanSE]

Also, did I miss it? Is there a simple equation giving the initial velocity v0 as a function of theta, normalized with the circular orbital velocity at r=R (a constant, vC)? Obviously when theta=0 then v0=0, and as mentioned when theta=pi then v0=vC. And the function must be monotonically increasing on the interval. Plotting a few points it sure looks simple!

I don't think anyone has posted this explicitly.  However, the information within this thread is sufficient to obtain it.  This is what I have:

1/2 v^2  = (mu/r) sin(theta/2)/(sin(theta/2)+1)
1/2 v^2  = (mu/r) x/(x+r)

I'm giving this in terms of the specific kinetic energy.  Here, x is the distance of the launch point from the y-axis in some of the illustrations used in this thread.  Suffice it to say that x=r sin(theta/2).  Also, mu=GM.  The radius of the moon is just r.  The following is a plot of fraction of mu/r that the kinetic energy at launch is.  To be more articulate, this is the ratio of the kinetic energy to the (negative of) gravitational energy at launch.



Or put into Wolfram alpha: plot sin(theta/2)/(sin(theta/2)+1)  for theta=0..Pi
Or Google

This is, as far as I'm concerned, the same plot as what was shown in the text I quoted.  The final 0.5 value is what's expected for a fully circular orbit, in the sense that v=sqrt(GM/r).  The requisite pieces of information are the following:

the distance f1->p->f2 is 2a.
2a = x + r
a = -mu / (2 eng)
eng = v^2 / 2 - mu / r

[Proponent]

Is there a simple equation giving the initial velocity v0 as a function of theta, normalized with the circular orbital velocity at r=R (a constant, vC)? Obviously when theta=0 then v0=0, and as mentioned when theta=pi then v0=vC. And the function must be monotonically increasing on the interval. Plotting a few points it sure looks simple!

See equation (12) in the PDF I wrote (and just revised to correct errors).  Just drop the factor GM/R to get speeds in terms of circular speed, of course.  There's something funny with that plot, by the way -- take-off speeds should always be less than circular.

EDIT:  Added missing closing bracket to quote block.

[Proponent]

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.

Isn't that comparing the case without refueling? I know fuel goes up exponentially with delta-V, so couldn't a 70% delta-v easily work out to be about half the effort, and two 70% hops of equivalent effort?

When I saw that V^2 component my first guess was that distance travelled was proportional to kinetic energy no matter how you split it.. but such a simplistic approach would probably only work for travel by trebuchet

True, if we're making the journey by rocket, then it's more complicated.  The answer will depend on specific impulse.  I suspect that, as a rule of thumb, the minimum propellant consumption occurs about when the delta-V for a hop (take-off delta-V plus landing delta-V) is about equal to effective exhaust velocity.  If the required per-hop delta-V is higher, then the exponential factor in the rocket equation starts to bite.

With circular speed at the lunar surface being about 1600 m/s and typical exhaust velocities twice that or better, I think it's likely that single hops are best.

[Hop_David]

Also of interest is the time of flight for the suborbital hop.

To look at fraction of area swept out, Kepler would vertically scale an ellipse by a factor of 1/sqrt(1-e²) to make it a circle.



I do the same. Without loss of generality, we can choose our units so a = 1. Things are simpler when working with a unit circle.

It's well known semi-latus rectum has length a(1-e²). When (1-e²) is scaled by a factor of 1/sqrt(1-e²) you get sqrt(1-e²).

We have a blue circular section with central angle acos(e). It's area is acos(e)/2.

We have a pink triangle of base e and height sqrt(1-e²). It's area is e * sqrt(1-e²)/2.

The circle section and triangle are mirrored below the major axis, so we can lose the denominator of 2. Total colored area is acos(e) + e * sqrt(1-e²).

The suborbital's fraction of the total period is (acos(e) + e * sqrt(1-e²))/pi

Orbital period is 2 pi * (a³/mu)^1/2. So time of flight is
(acos(e) + e * sqrt(1-e²)) *  (a³/mu)^1/2.

I'm going to do a blog entry Transportation on Airless Worlds. It will link to a spreadsheet that will include Luna, Mercury, Ceres, and a few other round, airless worlds. I will toss in Mars because it's nearly airless.

Proponent and AlanSE would you mind if I mentioned you? Proponent, I'd like to upload your pdf and link to it.

[Hop_David]

I know fuel goes up exponentially with delta-V, so couldn't a 70% delta-v easily work out to be about half the effort, and two 70% hops of equivalent effort?

When you have propellents with exhaust velocities on the order of 4 km/s, there's not a whole lot of incentive to break delta V budgets of 2*1.6 km/s (or less) into smaller hops. This is even more true if using rail guns or trebuchets.

There may be other reasons to break a big suborbital hop into smaller hops, though. If your destination is close to pi radians (180 degrees) away, flight path angle is close to zero. So crashing into mountains becomes a problem. Getting the flight path angle big enough to avoid obstacles might be a reason to break it smaller hops.

On Mars, atmosphere might become a hassle for longer suborbital hops that require greater velocity. And with lower flight path angle, the payload spends more time in Mars' lower (and denser) atmosphere.

[Proponent]

Please feel free to use the PDF anyway you like.

I suspect the most efficient way to clear obstacles on long hops would be to use a slightly off-optimal trajectory with a larger apoapsis.  If that doesn't do the trick and we're using rockets rather than mass drivers or something like that, then use a two-impulse trajectory.  For example, if a 10-km obstacle is very close to the origin (worst case), boost vertically at 180 m/s, coast up to 10 km, then boost horizontally.

I could see multi-hop trajectories being useful if some sort of ISRU propellant of low specific impulse were being used.

[AlanSE]

The question of energy efficiency is also intriguing.  As others have pointed out, the optimal number of hops is trivial for this problem.  Traveling the entire distance in one hop requires the least amount of energy.  But this is jumping the gun a little bit, because the optimal solution for the limit-case of small hops isn't known to begin with!  Here is my math:

Small Hops:
h(t) = v0 t - 1/2 g t^2   = { 0 = t ( v0 - g t / 2 ) }
t = 2 v0 / g   (the useful solution)
v0 = sqrt(2) v
distance = v0 t = 2 v0^2 / g = 4 v^2 / g
(mass-distance/energy) = distance / (1/2 v^2) = 8 / g

for 1,000 kg:
(distance/energy) = 8 / ((1.622 m/s^2)*(1,000 kg)) = 3.065e-6 mile/J
 x (1.3e8 J/gallon) = 398.4 mpg
with 20% conversion efficiency --> 80 mpg

In the first block of math, the 8/g result contains nothing about the distance of the hop.  That means that any number of hops is equally energy efficient.  I won't do the general case which takes into account the planet's curvature, but I'll compromise by entertaining the largest possible hop.

Full 180 degree movement:
1/2 v^2 = 1/2 GM/R
(mass-distance/energy) = Pi R / (1/2 GM/R) = 2 Pi R^2 / (GM)

for 1,000 kg:
(distance/energy) = Pi*(radius of the moon)^2*2/(G*(mass of the moon)*(1000 kg)) = 2.4e-6 mile/J
 x (1.3e8 J/gallon) = 312 mpg
with 20% conversion efficiency --> 62.4 mpg

We see that this is clearly more energy efficient, which suggests that the efficiency of the hop improves monotonically as the angle increases.  However, this isn't predicted by the flat-surface Newtonian mechanics.  The improvements are only due to the benefit of centripetal acceleration.

Of course, the miles-per-gallon figure is artificial.  Obviously nothing would be running on gas on the moon.  However, this still gives a reference to compare the method's efficiency to the drivetrain efficiency of common vehicles.  Actually, the above numbers are extremely generous.  Even the 2012 Toyota Camry weighs close to 1,500 kg.  That puts it into surprisingly close parity with the above numbers.

Actually, since the Camry performs that well on Earth, we would suspect a moon-car to be even more efficient.  The wheel and drivetrain friction are mostly constant energy losses, whereas the drag increases with speed.  We discussed this a great deal on Physics Stack Exchange.  For normal automobiles, their efficiency suffers at low speeds due to reduced engine efficiency, which is why optimal speed is somewhere around 55 mph.  That means that the Earth mpg is a very poor predictor of efficiency of a moon vehicle, because air resistance is still relatively high at 55 mph.

Put together, it looks like a normal vehicle would use less energy than the orbital hops, although they would obviously take longer.  The delivery of energy is an issue, but could be done by an electrified road.  If you just built a (mostly) conventional railroad, this would demolish the efficiency of the alternatives.  Train tracks seem impractical.  A hardened surface would seem to be the most practical road.  Even if you had a space cannon for suborbital hops, you would still send the bulk materials along this road if I understand the physics well enough.

[Hop_David]

Put together, it looks like a normal vehicle would use less energy than the orbital hops, although they would obviously take longer.  The delivery of energy is an issue, but could be done by an electrified road.  If you just built a (mostly) conventional railroad, this would demolish the efficiency of the alternatives.  Train tracks seem impractical.  A hardened surface would seem to be the most practical road.  Even if you had a space cannon for suborbital hops, you would still send the bulk materials along this road if I understand the physics well enough.

Yes, I am hoping there would eventually be roads. Or, better yet, railroads.

But initially there will be no transportation infrastructure on other bodies. It's my belief that transportation between the first mines and manufacturing centers would need to be done by suborbital hops.

I like to imagine very advanced infrastructure where underground tunnel rails connect points in straight line chords. Here's an exercise: If there were a mohole linking Ceres' north and south poles, how much energy would it take to travel from one pole to the other?

[IslandPlaya]

Assuming the distances between the COM are different between the North and South poles and you have some way of converting linear motion into a storeable form then a round trip would take no energy at all. (Neglect losses in the system of course.)

[Proponent]

I was wondering if, with very good guidance you could use electromagnetic mass drivers for both launch and landing.  With electromagnetic braking on landing, it would in principle be possible to recover much of the energy.

[IslandPlaya]

Good point. All true.
Formidable engineering though!

[Hop_David]

... a round trip would take no energy at all. (Neglect losses in the system of course.)

That is my opinion. Here is a model. If memory serves I got it from Dr. John Stockton:



I believe the round trip time would be the same as orbital period at surface. In the case of Ceres, that'd be about 3.4 hours.

[Hop_David]

I did a blog post on this. My thanks to AlanSE and Proponent.

[Proponent]

Assuming the distances between the COM are different between the North and South poles and you have some way of converting linear motion into a storeable form then a round trip would take no energy at all. (Neglect losses in the system of course.)

And, in fact, if the body is spherically symmetric and of uniform density, then travel along a straight line connecting any two points on the surface takes no energy, if friction can be avoided.  And the trip time is always the same, regardless of the distance between the two points on the surface.

[IslandPlaya]

Very true! Good point.

[Hop_David]

Assuming the distances between the COM are different between the North and South poles and you have some way of converting linear motion into a storeable form then a round trip would take no energy at all. (Neglect losses in the system of course.)

And, in fact, if the body is spherically symmetric and of uniform density, then travel along a straight line connecting any two points on the surface takes no energy, if friction can be avoided.  And the trip time is always the same, regardless of the distance between the two points on the surface.

Wendy Kreiger also told me trip time is always the same. I'm willing to take your word as well as Wendy's. But I would be happier if I knew why.

[Hop_David]

And, in fact, if the body is spherically symmetric and of uniform density, then travel along a straight line connecting any two points on the surface takes no energy, if friction can be avoided.  And the trip time is always the same, regardless of the distance between the two points on the surface.

It's a free energy ride if we're willing to let gravity do all the work. But then we're stuck with trip time on order of half an hour to an hour for most rocks.

But we can shorten trip time by investing energy to accelerate at the beginning of the trip. But then the payload would need to be slowed before it reaches its destination. The deceleration at the end of the trip could put energy back into the system. So it seems to me faster trips could also be done for very little energy.

[Proponent]

Wendy Kreiger also told me trip time is always the same. I'm willing to take your word as well as Wendy's. But I would be happier if I knew why.

Without using calculus, we can easily show how the problem reduces to simple harmonic motion.

Consider a particle constrained to travel along chord parallel to the y-axis.  The acceleration of a particle along the chord is -(GM/R³)y.  (I'm sure you can figure this out yourself: just calculate the acceleration of gravity at a given x and y, and then determine its y-component.)  The form of the acceleration is the same as for a mass on a spring or for a small-amplitude pendulum (restoring force is proportional to displacement).  In both of those cases, the period of oscillation is independent of amplitude.  The acceleration is also independent of x, so it doesn't matter which chord (i.e., between which two points on the surface) we are travelling.

Does that help?

[Proponent]

And, in fact, if the body is spherically symmetric and of uniform density, then travel along a straight line connecting any two points on the surface takes no energy, if friction can be avoided.  And the trip time is always the same, regardless of the distance between the two points on the surface.

It's a free energy ride if we're willing to let gravity do all the work. But then we're stuck with trip time on order of half an hour to an hour for most rocks.

But we can shorten trip time by investing energy to accelerate at the beginning of the trip. But then the payload would need to be slowed before it reaches its destination. The deceleration at the end of the trip could put energy back into the system. So it seems to me faster trips could also be done for very little energy.

I agree.

[AlanSE]

random question:

What is the highest altitude that one of these hops reaches?

[Proponent]

See equation (9) of the analysis attached to this post.

[AlanSE]

See equation (9) of the analysis attached to this post.

With that equation I find the following:

Angle at which maximum apoapsis happens: theta = Pi/2

Radius of this apoapsis: R/( 2 (sqrt(2)-1) )

So subtract R from that radius to get the altitude.  Numerically, I find this altitude to be about 0.2071 R.  On the moon, this would be about 360 km.

[Hop_David]

Wendy Kreiger also told me trip time is always the same. I'm willing to take your word as well as Wendy's. But I would be happier if I knew why.

Without using calculus, we can easily show how the problem reduces to simple harmonic motion.

Consider a particle constrained to travel along chord parallel to the y-axis.  The acceleration of a particle along the chord is -(GM/R³)y.  (I'm sure you can figure this out yourself: just calculate the acceleration of gravity at a given x and y, and then determine its y-component.)  The form of the acceleration is the same as for a mass on a spring or for a small-amplitude pendulum (restoring force is proportional to displacement).  In both of those cases, the period of oscillation is independent of amplitude.  The acceleration is also independent of x, so it doesn't matter which chord (i.e., between which two points on the surface) we are travelling.

Does that help?

Yes, that's helpful. Thanks.

[Proponent]

A quick geometrical demonstartion that second focus lies on center of chord connecting two points....

Usually these days one uses mathematics to derive or prove geometrical statements.  As you demonstrate, it is possible to go the other way.  In that spirit, let me mention I book I've recently come across that you might enjoy:  The Mathematical Mechanic by Mark Levi.  Levi, for example, uses the simple fact that a triangular aquarium filled with water does not spontaneously tend to rotate to prove the Pythagorean theorem.

[JasonAW3]

So, if I'm reading what's been written so far correctly...

A lunar rover, driving on the surface of the moon would take longer to get from point A to point B, but it would be more efficent, energy wise than a lunar ballistic hop.  Where as, a Lunar Ballisic Hop would get from point A to point B faster, but use more energy than a Rover would.

In this case, I guess that the use of a rover versus a ballistic hop would greatly depend upon how quickly one needs to get from point A to B.

But, to throw a monkey wrench into the equation, there maybe terrain between point A and B that may make use of a rover inpractical, unless equiped with short range ballistic jump capibility.  In this case, you may wind up using as much or more energy than simply using a Ballistic Hop in the first place.

In this latter case, perhaps a ballistic hopper with limited rover capibilities may be more appropriate, as most likely, any sort of situation that would require a dedicated sub-mission using either a rover or ballistic hopper, would likely have a certain time constraint to it.

  Therefore, which system to use, Hopper or Rover, would depend on proximity of points A and B to one another as well as the urgency of the time constraints to get to point B.

(not to meantion if there are points C, D, Etc. to likewise get to).

[TrevorMonty]

NASA Developing drones/hoppers for Mars, asteriods and Moon.

http://www.parabolicarc.com/2015/08/09/nasa-mars-drones/

For the Moon a small H2O2 hopper could operate off one the Xprize landers eg MoonExpress  MX-1 which uses H2O2.
One big plus of hopper using lander fuel is that all the lander's reserve fuel can be used for exploration.
A 10kg hopper + 2kg of H202 (12Kg total) with 150 ISP thrusters would have 10Km radius.
4kg fuel would give 30km radius, ideal for exploring around Shackleton crater (21km diameter).
 
http://space.stackexchange.com/questions/4413/lunar-sub-orbital-trajectories
NB 1degree is 30Km.

[A_M_Swallow]

NASA Developing drones/hoppers for Mars, asteriods and Moon.

http://www.parabolicarc.com/2015/08/09/nasa-mars-drones/

For the Moon a small H2O2 hopper could operate off one the Xprize landers eg MoonExpress  MX-1 which uses H2O2.
One big plus of hopper using lander fuel is that all the lander's reserve fuel can be used for exploration.
A 10kg hopper + 2kg of H202 (12Kg total) with 150 ISP thrusters would have 10Km radius.
4kg fuel would give 30km radius, ideal for exploring around Shackleton crater (21km diameter).
 
http://space.stackexchange.com/questions/4413/lunar-sub-orbital-trajectories
NB 1degree is 30Km.

The 10 lbf and 60 lbf hydrogen peroxide rocket engines developed by MSFC for the Mighty Eagle may still get to the Moon.
https://en.wikipedia.org/wiki/Mighty_Eagle

[Dalhousie]

This might be of interest:

"Developing an Aerial Transport Infrastructure for Lunar Exploration" by David L. Akin

Abstract http://www.lpi.usra.edu/meetings/leagilewg2008/pdf/4096.pdf

Presentation http://www.lpi.usra.edu/meetings/leagilewg2008/presentations/oct29pmSalonIII/Akin4096.pdf

[Warren Platts]

This might be of interest:

"Developing an Aerial Transport Infrastructure for Lunar Exploration" by David L. Akin

Abstract http://www.lpi.usra.edu/meetings/leagilewg2008/pdf/4096.pdf

Presentation http://www.lpi.usra.edu/meetings/leagilewg2008/presentations/oct29pmSalonIII/Akin4096.pdf

I see a possible discrepancy. Hop's blog has the ΔV = sqrt(μ(1/a - 1/r)), whereas the presentation Mr. Dalhousie linked to has it as ΔV =2 * sqrt(μ(2/r - 1/a)). What gives?

What I would like is a formula for the Moon that gives me the ΔV just in terms of the angular separation θ...

The main problem with suborbital hops is that they often are even more expensive than going to orbit and back. Consider a base that's also a major ISRU propellant manufacturer. The goal of the base is to support the scientific exploration of the rest of the Moon.

Do you see the problem?

Presumably there would be a depot-station at L1 or L2. ΔV to there is ~2.5 km/sec. Therefore, for the SS lander to fly from the L2 depot down to a remote location on the Moon, and then back to the depot-station, total ΔV = ~5 km/sec, as there would be no opportunity for refueling at the remote site.

So, you ask, why not simply fly from the Lunar base to the remote location and back? Suborbital hops have got to be cheaper/easier than flying to orbit and back, Right?

Not necessarily. Because you've got four major burns when going on a sortie from your main base at Whipple Plateau: (1) the burn to take off; (2) the burn to land at the remote location; (3) the burn to take off from the remote location; and (4) the burn to land at the home base at Whipple. IOW, whatever the ΔV formula is, you have to multiply that by four for the total ΔV for a suborbital round trip!

Take our scientific-expedition SS-lander with a nominal ΔV = ~5 km/sec. According to Hop's blog post, the ΔV for a 45° suborbital hop is 1.25 km/sec. Therefore, 45° is the largest separation our lander is capable of doing. The formula for the area of a (unit) spherical cap defined by an angle θ  is: 2π(1 - cos θ); for 45°, the area is 1.84 compared to 12.57 for the whole sphere, or about 1/7th of the whole sphere. IIRC, 1 degree on the Moon corresponds to 30 km, so a 45° angle is only a distance of 1350 km. Anything beyond this limit, you'd be better off flying from Whipple to L2, refuel, then go where you're going, then back to L2, refuel again, then return to base. Thus one is tempted to do away with the base altogether. Or rather, the scientific outpost itself should be based at L2, rather than at the ISRU plant. A disturbing implication IMO....

PS I also vote that we ask for this thread to be moved over to the Moon section, as all this suborbital stuff has been worked out long ago in excruciating, book-length detail by the US Air Force back in the 1950's so we could bomb the crap out of the Russians with suborbital hops.

(Extra credit: We want to destroy the Chinese base at their equatorial location from our base at Whipple, but they have their anti-missile defenses pointed at our base, so the plan is to launch a surprise attack up their backside with a 270° suborbital hop. Q: What is the optimal launch angle [the formula above is no good as it recommends launching into the ground...]?)

That said, this thread is an excellent resource for Moon studies, and so should be in the Moon section as it would be a shame for it to get buried in all the wonderful science fiction within this section! YMMV

ETA: Actually, looking at Hop's table again, I can see that for small increases in ΔV, the area able to be explored vastly increases. Probably the better solution would simply be to beef up your exploration lander. A ΔV of ~7 km/sec (which is certainly doable in theory) would give you round-trip access to all points on the Moon via suborbital hops. Still more ΔV than simply going from L2 and back, but then you've got to refuel your L2 depot, so the suborbital solution should be cheaper if your lander has enough total ΔV capability.