L1 jump point size

[Lagrange]

EDG's thread made me want to fully work out the size of L1 jump points.  Is it possible to guard a pirate point?  What's the timescale of an attack?  How many sensor platforms do you need to monitor?  Are collisions between jumpers likely?  The following is a calculation for the size for L1 jump points.  Maybe it's useful to anyone who wants super-crunchy battletech?  In any case, I'm happy to hear about any thoughts or corrections.

I'll use M_A and M_C to denote the mass of an ancillary body and M_P to denote the mass of the primary body, and we'll consider a 2-body system.  This is obviously an approximation, although it should be reasonably good in many situations.

The Center
The center of the L1 jump point (note: different from the L1 Lagrange point) is where the gravitational potentials of the two bodies cancel, thus:

M_A/d_A² = M_C/d_C²

where d_A is the distance to A and d_C is the distance to C.  Pictorially, we have:

C-------------------------------------------------------------------------------------------------------------------------J--A
where J is the jump point.

This can be rewritten as:

M_A/d_A² = M_C/(d_CA-d_A)²

where d_CA is the distance between C and A.  Using algebra we get:

(M_C-M_A)d_A²+2d_Ad_CAM_A-d_CA²M_A=0

Setting us up for the quadratic equation which has a solution at:

d_A=d_CA((M_AM_C)^0.5-M_A)/(M_C-M_A)

Notably, in the limit where M_C >> M_A, this simplifies as:

d_A~=d_CA(M_A/M_C)^0.5

Radial Boundary
The radial boundary applies to the direction between the primary and ancillary bodies.  Pictorially, it looks like:

C------------------------------------------------------------------------------------------------------------------------BJB-A
Where B is +/- radial boundary.

Calculating a boundary requires that we know the tolerance to non-flat spacetime for jumps.  This is given by the jump limit which in Sol is set at 10.2AU (per SO page 133).  Let's call this d_j.  Jump drives can tolerate

M_S/d_j²

nonflatness where M_S is the mass of Sol.

Using this, the equation for the radial boundary is:

M_A/d_A² = M_C/(d_CA-d_A)² +/- M_S/d_j²

An analytic solution is tricky here since this yields a quartic.  This could be solved numerically, or we can make analytic progress after approximating M_C >> M_A and d_CA >> d_A.  Under that approximation, we get:

M_Ad_CA² ~= d_A²(M_C +/- M_Sd_CA²/d_j²)

Implying

d_A~=d_CA(M_A/(M_C +/- M_Sd_CA²/d_j²))^0.5

Adding a further assumption that 1 >> M_Sd_CA²/(d_j²M_C) this gives:

d_A~=d_CA(M_A/M_C)^0.5(1 +/- M_S/M_C * d_CA²/(d_j²))^0.5

Which further approximates as:

d_A~=d_CA(M_A/M_C)^0.5(1 +/- 1/2*M_S/M_C * d_CA²/(d_j²))

Implying the radial dimension of the jump point is about:

d_r~=d_CA(M_A/M_C)^0.5M_S/M_C * d_CA²/(d_j²)

Perpendicular Boundary
The perpendicular boundary applies in all directions which are perpendicular to the radial direction.  As you leave the radial axis at the center of the jump point a small part of the gravitational potential adds together rather than canceling.  Pictorially, this looks like:
                                                                                                                                                                                   T
C------------------------------------------------------------------------------------------------------------------------BJB-A
                                                                                                                                                                                   T

In particular, the perpendicular distance to the boundary d_T satisfies:

M_A Tan(d_T/d_A)/d_A² + M_C Tan(d_T/d_C)/d_C² = M_S/d_j²

Solving this analytically is again tricky.  However, when M_C >> M_A the second term on the left hand side can be neglected.  Using this, we get:

d_T ~= d_A Tan^-1(M_S d_A² /(d_j²M_A))

Examples
Since solutions involve mass ratios and distance ratios, it's relatively easy to do various calculations.

Planet	MA/MC	dCA	dA	radial width (approx)	perpendicular width (approx)	Notes
Earth	3.0*10-6	150*106km	~260*103km	~2497km=139 hexes	~2497km = 139 hexes	Large enough that a single warship does not quite have fire control with direct fire weapons.  Since the moon orbits at 384*103 km it will substantially perturb the solution.
Mars	3.0*10-7	228*106km	~125*103km	~1200km = 67 hexes	~2772km = 154 hexes
Jupiter	0.000955	778*106km	~24.0*106km	~231*103km = 12.8*103hexes	~6.08*106km = 338*103hexes	Large enough to be beyond targeting range for a single large craft.
Ceres	4.8*10-10	414*106km	~9070km	~87km=5 hexes	~663km=37 hexes	Small enough for a traffic jam.

Looking at the trends, the L1 jump point gets larger the further from the center and the larger the ancillary.   The perpendicular over radial width apparently also grows with distance from the primary.

[Daryk]

Nice work, but for true crunchiness you need quotes around "L1".  The actual L1 point includes factors for the centripetal force at the orbit.

[Lagrange]

Nice work, but for true crunchiness you need quotes around "L1".  The actual L1 point includes factors for the centripetal force at the orbit.

Agreed about the difference.  I added a note and made sure to always refer to the "L1 jump point" to disambiguate.

I reread the SO rules w.r.t. jumps.  Since objects in adjacent space hexes take damage during a jump, you plausibly need a 3-space-hex diameter to support jumps without taking damage.  So Ceres, the largest asteroid, barely generates a sufficient L1 jump point.  Looking through the list here, there are a couple dozen relevant objects inside the jump limit of Sol.

[Daryk]

Gútegow Belta!  8)

[Andras]

This brings up a question, is it possible that, since L1 point moves with the orbit of the secondary body, the jumpship drifts out of the acceptable jump-limits?

The JS doesn't arrive with momentum so it doesn't have the velocity to keep up with the L1 movement right?

[mikecj]

Thanks for laying it out that way!

[Daryk]

This brings up a question, is it possible that, since L1 point moves with the orbit of the secondary body, the jumpship drifts out of the acceptable jump-limits?

The JS doesn't arrive with momentum so it doesn't have the velocity to keep up with the L1 movement right?

That's how I understand it.  But even "station keeping" thrust is enough to get up to orbital velocity over time.

[Lagrange]

This brings up a question, is it possible that, since L1 point moves with the orbit of the secondary body, the jumpship drifts out of the acceptable jump-limits?

The JS doesn't arrive with momentum so it doesn't have the velocity to keep up with the L1 movement right?

JS drives are pretty monstrous compared to anything that we can do, since they can sustain .1g for weeks/months.  (In fact, they are aphysical.)

The 'gravity' of the sun at earth's orbit is apparently about .0006g (= mass of sun / mass of earth * earth radius^2 / AU^2)(*).  Even at Mercury's orbit, that's only increased by a factor of 10 or so.  This is also the centrifugal acceleration since these balance.  Hence, staying 'on' an L1 point is pretty easy according to the rules.   

SO page 89 says "Jumpships always arrive stationary with respect to the destination jump point."  The exact meaning of this is a little bit ambiguous since you could apply a rest frame with a velocity matching a L1 jump point's velocity or matching the primary's velocity.  These coincide at the zenith/nadir but differ for an L1 jump point.  Overall, I'd lean towards the former, since (a) the jump point is specifically mentioned rather than the primary and (b) arriving with a significant velocity relative to the jump point is nonintuitive.

If in some game it's the latter, the orbital speed of Earth is ~30 km/s, so catching up at roughly 1 m/s/s requires 30k seconds = a little over 8 hours at .1g.  Some futzing to avoid overshoot suggests it might take a day to get on station.

[Daryk]

I thought the consensus we reached in that thread was that 0.1g was in the realm of the possible, but that 1g (or more) was right out? ???

[Lagrange]

I thought the consensus we reached in that thread was that 0.1g was in the realm of the possible, but that 1g (or more) was right out? ???

There isn't really a physical limit on instantaneous thrust (a fusion nuke can give lots of thrust...).  The physical limit is on fuel consumption---you need about .1% of mass for 50 fuel points.

[Daryk]

Right, and at 0.1g, that fuel fraction is workable.

[Lagrange]

Right, and at 0.1g, that fuel fraction is workable.

As an example, with 100 tons of fuel on a 100K ton ship, you can thrust at 1g for 25 minutes or .1g for 250 minutes.  Station keeping though requires much less in general---you could thrust at .0006g for about 29 days.  And station keeping at the Zenith or Nadir would require x100 less.

[Daryk]

You calculated 17% as the necessary fuel fraction... I thought we concluded that was workable? ???

[Lagrange]

You calculated 17% as the necessary fuel fraction... I thought we concluded that was workable? ???

For a Dropship or Warship a 17% fuel fraction on a mission seems fine.  For a Jumpship, less so.

[Daryk]

True... using your math in this thread, 200 tons of fuel on a 100K JumpShip should be enough to get up to oribital speed.

[Lagrange]

True... using your math in this thread, 200 tons of fuel on a 100K JumpShip should be enough to get up to oribital speed.

Right. 

I remain skeptical about the need to do so though---stationary in the instantaneous rest frame of the L1 jump point at the moment of transition makes more sense.  If that's so, then you don't have any need to catch up to orbital speed and the fuel on a typical jumpship is sufficient to stay on a L1 jump point for weeks or a Zenith/Nadir for months.  For example, a Scout with 46 tons of fuel could stay on an L1 jump point for 2 weeks.  This could be extended via orbital insertion or weekly planetary resupply.

[Daryk]

The more I think about it, the more I like the "stationary with regard to the moving reference frame", since that's exactly what a ship is doing in a jump anyway.