Suborbital Lunar Hops

[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.