Project: Calculate the Lagrange points for an imaginary perfectly circular orbiting system where the 'Sun's' mass = 3, 'Earth's' mass = 1, and a Satellite has negligible mass.

Terms used in both the text and the equations below:
Sun: Our imaginary Sun (or the real one in context).
Earth: Our imaginary Earth (or the real one in context).
Satellite: Our imaginary satellite.

L: The Lagrange points: L1-L5.
B: The barycenter or center of mass between Sun and Earth or just one of them in context.

For all orbiting bodies their:
M: mass.
V: tangential velocity.
W: angular velocity (i.e. their 'RPM').
P: period (i.e. the 'year' of our system) is the reciprocal of W -- the faster things rotate, the shorter the year.
R: the orbital radius from B.
d: the distance between the barycenters of two objects ('d' and 'R' are often the same thing below).
C: centrifugal force outward from B.
G: gravitational pull (towards B or towards the sum of Sun and Earth calculated separately or to just Earth in context).

(Note, we use, eg: 'R1' and 'R2' where the radius of both bodies is considered in one equation, otherwise 'R' is the radius of whatever body is being discussed.)

All mathematical and graphic relationships hold true regardless of the units used so long as they are consistent with each other, so for simplicity this essay will use imaginary units normalized to '1' wherever possible such that many terms in the equations can be ignored, eg. Earth's orbital period and tangential velocity are both one unit so they can be ignored in multiplications.  There are only two given values: 3 is the mass of Sun and the radius of Earth, and 1 is the mass of Earth and the radius of Sun -- all other values derive from these. Also, this entire essay relies on just two equations: those for gravity and for centrfugal force, as well as on the definitions of the various physical units:

velocity = distance/time
acceleration = distance/(time squared)
energy = force times distance
energy = velocity squared

---------------------------------------------------------------------------
The Barycenter (B):

In any real orbital system, such as Sun and Earth or Earth and Moon, the smaller body is commonly considered to orbit around the larger body. (This is drawn as the dotted cyan 'pseudo' orbit in Diagram #4 and following.) In fact both bodies (and any other bodies added to the system) orbit around their common B, at which:

$M_1R_1 = M_2R_2$

... The mass times the radius of the first body equals the mass times the radius of the second body. This is exactly comparable to the the way that heavy kids and light kids balance on a teeter-totter -- the heavy kid has to sit further in. Whether the system is stationary or orbiting the principle is the same.

Referring to our imaginary Sun-Earth system, by definition L are those places in the system where Satellite will hold a fixed position relative to both Sun and Earth. That is, its orbital period around B will be the same as Earth and its distance from both Earth and Sun will not change. At these places C balances the combined G of Sun and Earth and so the orbit of Satellite is stable -- it's velocity and position don't change:

$C = G_E+G_S$

It turns out that there are five such points. Here they are:

Diagram #1: The Lagrange points:


[lagrange-diagram1.png]

If we think of this as a topographical map, 'down' would be the direction Satellite would 'roll' if placed somewhere. The 'lowest' spot is Sun, and Earth forms the secondary low spot to the right. Note, this is not a map of gravitational pull only, which would never slope 'outward', but of the difference between the outward C and the inward G which cancel exactly at any orbital point giving us a 'flat spot'. (Note that Diagram #1 has the blue orbital circle going around the Sun, NOT around B which is misleading.)

To elaborate, if we had a 'topographical map' of C only, it would look like a dome -- flat spot on the top (at B), and increasing slope as you move away. The steeper the slope, the greater the force and of course C increases as we move away from B (at constant P!). If our map were of G only, it would look like one of those illustrations of a gravity well -- quite flat around the periphery but falling more and more steeply into the central depression where the mass lurks. But Diagram #1 combines C and G forces, thus we have C 'winning' beyond a given distance, and G winning closer in, and Lagrange points where they balance.

But note that L1-L3 are unstable -- looking at L1 in Diagram #1, if Satellite were placed just to the left it would roll into a new orbit around Sun. If to the right, into a new orbit around Earth. But if it were placed slightly up or down, it would roll back to L1! Thus the red arrows indicate places where a body would tend to move toward the local L (stable) whereas blue arrows indicate movement away (unstable). This means that real satellites placed at L1-L3 must make micro-adjustments to their position to stay at the flat spot and avoid rolling away into some other orbit.  (However it does seem that there are 'Halo orbits' and 'Lissajous orbits' and 'Lyapunov orbits' that are almost stable in case you're interested in burning your brains out studying this further.)

But it turns out that L4 and L5 are fully stable -- a satellite placed slightly off the exact point will actually orbit around the point even tho there's nothing there to orbit around! But the orbits will be chaotic, (see diagram #3 below). The explanation for these orbits has to do with the Coriolis force. (For reasons I still don't understand, this only works if Sun is at least 25 times heavier that Earth. Fun Fact: the value is: (25 + sqrt(621))/2)=24.9599...)

[ Aside: Real orbiting systems (in the Solar System anyway) almost always have very large mass ratios: The Sun is 333,000 times heavier than the Earth; the Earth is 81 times heavier than the Moon, etc. The only exception is Pluto and Charon, Pluto being only about 6 times heavier than its moon. That system's B is actually outside Pluto and some say that thus they should be called a binary planet. We are using a ratio of 3:1 simply because it makes it much easier to visualize what's going on in the diagrams. In fact the real Earth's L1 and L2 are roughly a million miles from the Earth (about 1% of the distance to the Sun) and drawing to that scale would make for very unsatisfactory diagrams.]

Could there be any other L points?  Intuitively, no. Try to find any more flat spots on the diagram and there just aren't any. We have the tops of two hills at L4 and L5, and three saddles at L1-L3, but that's all.

Here's another diagram of L but with a more 3D look to it:

Diagram #2: 3D view of the Lagrange points:


[lagrange-diagram2.png]

Think of this as looking like a very low volcano with Earth being the secondary caldera off to the right of the main Sun caldera. L3-L5 are along the rim of the volcano but L4 and L5 are slightly higher than L3. L1 is a saddle dipping down between L4 and L5 but still a high point between Sun and Earth. L2 is also a saddle but a bit higher than L1. We can easily see that L1 is the most unstable, L2 is a bit better, L3 much better, and L4, L5 are large, bean-shaped plateaus where stability is obviously going to be good.

Here's a computed orbit around the L4 point of Jupiter (astronomers note that asteroids get trapped there and call them 'Trojans'). The orbit is chaotic but will remain within the L4 'zone'.

Diagram #3: Chaotic orbits around L4 (or L5):


[lagrange-diagram3.png]

---------------------------------------------------------------------------
Simple systems:

Before we look at L, let's get the basics of orbital mechanics sorted out:

Diagram #4: The simplest case -- Earth and Sun:



[lagrange.ggb]

In our system we draw Sun one unit to the left of B, and Earth three units to the right so that, as explained above, mass times distance is equal for both bodies (3*1 = 1*3). B itself is at the algebraic/graphing Origin or center of the system (point (0,0)).

As mathematically normal, forces and distances to the right of the origin are positive and visa versa. Since G and C act in opposite directions, we have to make G negative so that our algebra and graphing behave properly.  Thus the gravitational force on Satellite when it is to the right of B are negative but centrifugal force is positive. 

The 'x' axis is R, the 'y' axis is the magnitude of the forces acting on the body (just Earth for now).

C = the green line.
G = the red line.
The dashed red line is the difference between C and G, so where that line crosses the x axis (zero), the two forces are balanced and we have a stable orbital point.


---------------------------------------------------------------------------
Gravity:

The universal law of gravitation (thanks Isaac):

$G=\frac{-gM_1M_2}{d^2}$

... The gravitational pull between two bodies equals the gravitational constant (g) times the product of the two masses divided by the square of the distance between them.  This is true whether or not the bodies are in orbit or even moving at all.

g = 6.67 × 10-11 N-m2/kg2 but we can ignore it here (or set it to '1') since its effect is simply to determine the P of our system. Stronger gravity would just require our system to rotate faster, but we've already normalized our P to '1' (say: 'one orbit per year') -- we don't care what the actual P is only that it is the same for Earth and Satellite (at L).

The mass of Sun = 3 and Earth = 1, and the distance between them is 4, so the gravitational pull between Sun and Earth is:

$G = \frac{-gM_1M_2}{d^2} = \frac{-1*3*1}{4^2}$

Throwing out the useless 1's:

$G = \frac{-3}{16}$

(Note the vertical red line? As normal, we calculate gravity as tho the mass of the body in question were a single point. This means that, mathematically speaking, one can get infinitely close to this point at which time the gravity becomes infinite! This creates a mathematical asymptote -- a straight line that another line (in this case our gravity) gets infinitely close to, but never touches ... except that at infinity, it jumps over the asymptote and becomes negatively infinite. In the happy world of calculus this is called a Discontinuity. Of course in practice this never happens because all real objects have surfaces. But there's no reason to overrule the math -- the equations just do their thing and we let them rise to infinity as they desire.)
 
---------------------------------------------------------------------------
Centrifugal force:

$C = \frac{MV^2}{R}$

... Centrifugal force equals the mass of the body times its tangential velocity squared, divided by its radius.  At a constant V, C goes down in inverse proportion to R because the rate of curvature goes down which reduces the amount of acceleration needed to keep the body moving in a circle. Or:

$C = MW^2R$

... Centrifugal force equals the mass times the angular velocity (the 'RPM') squared times the radius. At a constant W, C goes up proportionately with R. This is intuitively true.

The mass of Earth = 1. And since the whole point of L is that everything rotates at the same period, let W = P = 1 orbit per year (or one year per orbit as you prefer):

$C = MW^2R = 1 * 1^2 * R$

... Centrifugal force equals 1 times 1 squared times the radius. Throwing out the useless 1's:

$C = R$

For any body in orbit C = -G (Or: 'C+G = 0') -- the forces cancel giving us:

$C=R=-G$

At R = 3, G = -3/16 (above) so the C line on the graph must be +3/16 at that radius given that Earth is in a stable orbit (where C = -G).  Thus we 'correct' the slope the C line like this:

$C(line)=\frac{R}{16}$

Note that the dashed red line shows how, at our given W, if Earth were placed further out, C would 'win' over G and visa versa. Obviously if W were to be increased (by spinning our system faster), C would become stronger than G and Earth would spiral outward.  Likewise, if Sun had more mass, increasing G, Earth would spiral inwards (unless we also increased W to match).

[Aside: there is always a stable orbit. The above presumes a constant W in keeping with the fact that we are studying Lagrange points. In fact if we accelerated Earth what would happen is that it would 'climb' to a higher orbit but loose speed in doing so such that eventually G and C would once more balance.  More about this below.]

---------------------------------------------------------------------------
Ok, now let's throw in Satellite:

The mass of Satellite is so small as to be irrelevant to the B of the system, but since C goes up exactly as G goes up if Satellite's mass increases, this cancels out and thus we can ignore the mass of Satellite (or set it to '1' which does the same thing).

If Satellite is put into orbit around our system at a great distance, the gravitational pull on it can be calculated as the combined mass of Sun and Earth coming from B. Thus Satellite obeys a different gravity equation. It isn't intuitive, but whereas Earth feels a gravity of 3, Satellite feels 4 (3+1). You could say that Earth is attracted to Sun, but not to itself whereas Satellite is attracted to both.  The gravity equation for Satellite in orbit around B is thus:

$G = \frac{-(3+1)}{R^2}$ not: $G = \frac{-3}{R^2}$

---------------------------------------------------------------------------
L1-L3:

The above is all fine if Satellite is at a great distance, but if it is closer in, the gravity of Sun and Earth should be calculated separately since at different points in the orbit, both bodies will be noticeably closer or further away from Satellite than the other. Also, angularity can exist between the gravity of Sun and Earth so that they don't pull in exactly the same direction. (At L1 they pull in opposite directions so instead of adding they subtract.) It is the interesting outcomes of these separate calculations that make the existence of L possible.

Diagram #5: Enter Lagrange Points 1,2 and 3:



[lagrange.ggb]

C = The green line.
G  = The purple lines.
The dashed purple line = C + G, that is, the difference between them. Orbits are stable where this line crosses zero.

L1 = 1.44  (2.44 units from Sun, 1.56 from Earth).
L2 = 5.06  (6.06 from Sun, 2.06 from Earth)
L3 = -4.41 (3.41 from Sun, 7.41 from Earth)

Our new and improved gravity calculation for L1-L3, where all forces are in a straight line, looks like this:

$G=\frac{-3}{(R+1)^2} + \frac{-1}{(R-3)^2}$

3 = the mass of Sun, 1 = the mass of Earth. But the radius to each body must be corrected. Consider L2, to the right of Earth: Sun is one unit further away than B; Earth is three units closer, thus the equation above makes sense. Our new equation works perfectly for L1-L3 where pulls go positive or negative automagically. As in Diagram #4, the dashed line shows where C and G balance crossing the 'x' axis (value = zero) -- these are our Lagrange points:

---------------------------------------------------------------------------
The neutral point:

Where is the neutral point between Sun and Earth i.e, where gravity balances but the body is not in orbit and thus there is no C? It will be further from Sun than L1 where Satellite would be pulled into a circular orbit thus feeling C and so needing more G to balance it thus demanding a closer orbit. We could do more math but we don't need to: The neutral point (N) is where the purple line (gravity) crosses the x axis:
N = 1.54.

Diagram #6: The neutral point:


[lagrange.ggb]

To be honest, I'm not sure this neutral point is good for anything at all, but who knows.

---------------------------------------------------------------------------
L4 and L5:

And now for the fun part, the magic of L4 and L5. Algebraic demonstrations are of course possible, but the geometric demonstration is far more elegant.

Supposing that the mass of Sun and Earth really were concentrated at B, at what radius would Satellite orbit, keeping the same period as Earth? Easy, it would orbit at R = 4: Earth feels a pull of 3 from Sun at a distance of 4.  As discussed, Satellite will feel the pull of both Sun and Earth (4 units) at the same distance.  And at R = 4, C must also be 4, so there you have it: -G = R = C = 4 -- a stable orbit. 

Alas, in our actual system Satellite can't orbit there because the proximity of Earth at almost the same radius would pull it in.  In fact L4 is located such that the 'backward'pull of Earth, tending to slow Satellite down, is balanced by the 'forward' pull of Sun, tending to speed it up. But since Sun and Earth now pull in slightly different directions (giving us our parallelogram below), we must have a reduced G, thus a reduced C and thus a reduced R.
  Diagram #7: L4 and L5:



[lagrange.ggb]
 
Let 'B' be the barycenter, 'S' be Sun, 'E' be Earth:
- Construct equilateral triangle S,E,L4.
S,E = 4 (given), so S,L4 = 4 (equilateral).
- Construct equilateral triangle S,B,J.
S,B = 1 by definition, so S,J = 1 and J,L4 = 4-1 = 3.
- Construct parallelogram B,J,L4 (point K is created).
Since this is a parallelogram, K,L4 = B,J = 1 and similarly B,K = 3.
- Construct B,L4.
To determine the length of this line:
- Construct rectangle B,M,L4 (points M and O are created).
B,M = cos(30) * 1 = .8660.
B,O = 3 + sin(30) = 3.5.
Pythagoras: B,L4 = sqrt(.8660^2 + 3.5^2) = 3.6055, the resultant vector.

Now, it is easy to see that we have a parallelogram of forces here, with the inputs being J,L4 (3 units long) and K,L4 (one unit long), and the resultant being L4,B (3.6055 units long).

Is L4 really a Lagrange point?  We already know that: 'C = R = -G for any stable orbit. At L4, R = 3.6055 which must equal C, so G needs to be the same. L4 is equidistant from Sun and Earth so the gravity it feels from each will be in proportion to their masses -- the effect of radius cancels out.  L4 is at the same distance from Sun as is Earth (4), so it feels the same proportional pull, namely 3 units, which we see is the J,L4 input to our parallelogram. The logic is the same for L4,K: one unit of gravity pulling to Earth.  The resultant: -G = 3.6055 = R = C, and these forces resolve on B, so L4 is orbiting stably at the same W as the Earth and is therefore a Lagrange point.

Note that L4 and L5 are on both the real and pseudo orbit lines and that, whereas the other L will vary their relative positions between Sun and Earth if the ratio of masses changes, L4 and L5 are always at the apex of the perfect equilateral triangles we've seen -- it is geometrically necessary that they be there. Cool, eh?

===========================================================================


While we're on the topic of orbits, as we recall:

$C=MW^2R$

... Centrifugal force equals the mass times the angular velocity squared times the radius. And:

$G=\frac{-gM_1M_2}{d^2}$

... The gravitational pull between two bodies equals the gravitational constant times the product of the two masses divided by the square of the distance between them.

Let's look at the orbits in a solar system with a few planets. We're not looking at Lagrange points now, just regular planetary orbits. Since they all orbit the same sun and since we've already discussed how the mass of a planet doesn't matter since C and G both go up at the same rate as a planet gains mass, we can simplify both equations (Note, in this context 'distance' is the same thing as 'radius'):

$C=W^2R$  and  $G=\frac{1}{R^2}$  so: $W^2R=\frac{1}{R^2}$

P (the period or 'year' of our system) is the reciprocal of W, that is, as the angular velocity increases P decreases. ( Obviously the faster you go, the less time it takes to compete one revolution.) So:

$W=\frac{1}{P}$

Substituting W with P:

$(\frac{1}{P})^2R=\frac{1}{R^2}$

... simple algebra gives us:

$$

... Kepler's third law: "The square of the period of orbit is proportional to the cube of the orbital radius."

More algebra gives us:

$V=\frac{1}{\sqrt{R}}$

... for example, double the radius and orbital velocity is divided by $\sqrt{2}$: 1.41.

Easy for us because we already know the law of gravity, but it hadn't been discovered when Kepler came up with his laws, all he had was the astronomical data of his mentor, friend, enemy and rival Tycho Brahe. Kepler's work was done by 1619 but it took another 68 years before Newton, using Kepler's laws, finally reverse-engineered the law of universal gravitation.

===========================================================================


... is the velocity an object needs to escape from a gravity well. It will keep moving away forever. It will be slowing down forever too, but it will never stop either. The curve of its deceleration is a hyperbola that will asymptotically approach zero velocity as it approaches infinite distance.

The Ve from Earth's surface is about 11 km/s (6.951 mi/s; 40,270 km/h; 36,700 ft/s; 25,020 mph, Mach 33). At 9,000 km altitude in space it is 7.1 km/s. Obviously the higher you go, the less velocity you need to break away because you've already done some of the work getting to the height you are at -- some of the original Ve has already been 'spent'. (We'll say 'height' from now on but this is exactly the same as 'radius'or 'distance'. Height is more intuitive in this context because we are 'climbing' out of a gravity well.)

[Aside: fascinatingly, if one is accelerating out of an orbit so as to reach Ve, one can accelerate tangentially or radially (horizontally or vertically or anywhere in between) and it makes no difference to attaining Ve.  So long as you are adding kinetic (tangential/horizontal) or potential (radial/vertical) energy, it works out the same in the end.]

Ve can also be described as the velocity at which the sum of the object's kinetic energy and its Gravitational Potential Energy (GPE) -- the energy to be had by droping the object from infinity to the launch surface -- which is always considered as a negative value, is equal to zero. If we launch with that much velocity/energy then we have enough velocity to 'return' to infinity, thus we have Ve. 

This is obvious: as the object rises, velocity is lost but exactly as much GPE is gained. Think of a perfectly elastic bouncing ball held at some height: it has 100% GPE but 0% kinetic energy because it isn't moving. Drop it. As it falls GPE is converted to kinetic energy but the sum (the total energy) does not change. As it hits the ground, kinetic energy is converted into elastic potential energy which is converted back into kinetic energy as it bounces upward, which is converted into GPE again and so on.

Diagram #8: Gravity, Energy and Escape velocity:


[HyperbolicExcess.ggb]

The equation for Ve (the purple line) is:

$V_e = \sqrt{\frac{2gM}{R_s}}$

where ...
Ve: escape velocity
g:  universal gravitational constant
M:  mass of Earth (or whatever gravity-well we are escaping).
Rs: radius from B of our 'launch surface' normally the surface of a planet, in this case Earth with Rs=1.

g * M is called the Standard Gravitational Parameter (SPG), that is, the actual gravitational field of some actual body.

But Ve is usually calculated from the surface of some body -- we launch all our satellites from the surface of the Earth don't we?  We know that gravity is proportional to 1/R^2, and if you throw in the SGP, we now have the Surface Gravity (s) -- the actual gravity at the surface of some actual planet (Earth's SG is about 9.8 m/s^2):

$s = \frac{gM}{R_s^2}$

Plugging 's' into our Ve equation above, simple algebra gives us:

$V_e = \sqrt{2sR_s}$

... the actual Ve from the surface of some actual planet.

(Helpful fact: The escape velocity of an object in a circular orbit at any height is $\sqrt{2}$ times the orbital velocity at that height.)

As with Lagrange points above, things get messy when the B involves more than one body. Obviously when we are launching from Earth, at first we are only concerned with escaping from the gravity of Earth, but, for example in the case of some space probe like the Voyagers, as it moves away from Earth, the gravity well it is escaping from centers far more on the Sun. The math to sort this out is heavy duty. For everything below we assume the  launch of a 'Voyager' from Earth with no other gravity to worry about, this just keeps things simple.  Again, as in the discussion above, we can ignore 'g' and we can also ignore all masses since the relationships below are the same regardless.

---------------------------------------------------------------------------
Some calculus:

As we know (setting mass and the gravitational constant both to '1' so that they can be ignored), gravity is equal to the inverse square of the radius (or distance) from B:
$G=\frac{1}{R^2}$  (blue line)

and similarly simplified:

$V_e = \frac{1}{\sqrt{R_s}}$  (purple line)

... so simple algebra gives us:

$G=V_e^4$

Let's use calculus to prove the above equations correct:

An integral is the area under some curve on a graph, but what does that area represent?  Its unit will be the product of the unit of the 'y' axis times the unit of the 'x' axis.  This is exactly what we'd expect -- if our 'x' and 'y' axes both represented distance in meters, what would be found under any line we draw on that graph?  It would represent meters squared (area).  This is basically what a map is, a graph who's units on both axes are distance.

Now, gravity is an acceleration, so the 'y' unit for the gravity line (converting to using meters and seconds now for compactness) is the unit for acceleration: m/s^2.  The 'x' unit on our graph is meters, so the unit of the integral of the gravity line is:

$\frac{m}{s^2} * m = \frac{m^2}{s^2}$

What on earth is that?  Well, acceleration is a force, and a force thru a distance is work and work is one form of energy.  So the integral of the gravity line is a force times a distance: the energy expended in lifting out of the gravity-well.  So, as unintuitive as it might be, the unit for energy (or one way of expressing it) is: m^2/s^2.  That's the green line.

Now, what happens when we take the square root of our energy line?  Obviously we get m/s ... velocity! -- the purple line.  In summary:
$G=\frac{1}{R^$   (blue line)

$E_e = \int{G}$  (green line)

$V_e=\sqrt{E_e}$  or: $E_e=V_e^2$  (purple line)

...Calculus is our friend.

---------------------------------------------------------------------------
Hyperbolic Excess Velocity (HEV):

What if Voyager launches with more velocity than is needed to reach Ve?
This is called an Initial Excess Velocity (IEV or Vx) and Voyager will now asymptotically approach the Hyperbolic Excess Velocity (HEV or $V_\infty$).  As above, Voyager will be eternally slowing down, but its velocity will never drop below the HEV even at infinity.

Diagram #11: Hyperbolic Excess Velocity:


[HyperbolicExcess.ggb]

The equation for HEV is:

$V_\infty = \sqrt{V_x^2 - v_e^2}$ 

How does it work? We know that:

$V=\sqrt{E}$  or: $E=V^2$

So what the above equation does is first convert Vx (IEV) to energy by squaring, subtract the escape energy (because if we are voyaging to infinity, we must loose all that energy by definition) from the initial excess energy to find the residual Hyperbolic Excess Energy, then take the square root of that to find the HEV.  Simple and logical.

But the graph looks better if we add '1' to the IEV giving us the actual starting velocity not just the excess.  And our Ve is '1' so:

$V_\infty = \sqrt{(1+V_x)^2-1}$  (the tan line)

...the HEV obtained ('y' axis) with a given IEV ('x' axis). 

The dashed purple line is the asymptote of the HEV given a Vx of '0.5' -- the same equation but with Vx: = IEV (another 'slider', see above):

$V\infty=\sqrt{(1+IEV)^2 -1}$  (the dashed purple line)

The solid purple line shows the decline in Vx over distance given an IEV of 0.5.

$V_x = \sqrt{(1+IEV)^2 -1+\frac{1}{h}}$  (the purple line)

So, as always we first convert velocity to energy by squaring, subtract the escape energy ('1') but restore the escape energy at that height ('1/h'), then take the square root of that to find the HEV at that particular height.

Note that you get more than you pay for!  For example, in our system, at a radius of one unit, adding 0.5 units of IEV over Ve (going from 1 unit to 1.5 units) yields a HEV of 1.118 units. Even at infinity Voyager never drops below that velocity (the dashed purple asymptote). (Fun Fact: Voyager I has an HEV of 16.6 km/s and is expected to be 1/3 of a light year from the Sun in the year 7987.)
---------------------------------------------------------------------------
Hyperbolic Deficit Velocity:

Obviously if Voyager is moving at less than Ve it will not escape Earth's gravity-well but will reach some height and then start falling back (into an elliptical orbit in fact unless it crashes into something). I'm calling this Hyperbolic Deficit Velocity (Vd 'Velocity(deficit)') for lack of a better term. Here's the equation for the height obtained for a given Vd:
$h = R_s\frac{x^2}{1 - x^2}$

... where x = Vd/Ve : the ratio of the initial velocity to the Ve at Rs.
(Unlike escape velocity, the direction (vertically up) is important to achieve maximum height because only 'vertical' velocity counts.)

Diagram #10: Hyperbolic Deficit:


[LagrangeHyperbolicDeficit.ggb]

Conversely the Velocity Required (Vr) to reach a given height (h) is:

$V_r = V_e\sqrt{\frac{h}{R_s+h}}$

... where:
Vr: y-axis: the 'deficit'velocity required to reach a height of 'h'.
Rs: the radius of the surface of Earth.
Ve: escape velocity, taken at Rs.
h:  x-axis: the height reached above the surface.

But our Ve is just '1', Rs is also '1', and for ease of reading the graph (so that x-axis values are 'h' values), we'll measure 'h' from B, not from the surface, giving:

$V_r = \sqrt{\frac{h-1}{h}}$(the dashed red line)

... to achieve a height of '1' (that is, to just sit there on the surface) takes zero Vr. At infinity the fraction becomes infintely close to '1' so the velocity needed is '1' -- the escape velocity.

We know that E is simply the reciprocal of height, and velocity is the square root of E, so the above just finds the amount of E needed to reach 'h' as a simple fraction, and then takes the square root of that.

Note how most of the 'work' is done low down on the curve -- as you get closer to escape velocity, extra height becomes much easier to achieve until, finally, at a velocity of '1' you are at Ve.  The graph shows a height of '3' being reached with a velocity of 81.6% of Ve.  No wonder rocket scientists struggle for even the smallest increases in power -- a little bit makes a huge difference.

The dashed purple line shows the corresponding decline of velocity with increasing height, from 0.8165 units at launch down to zero at a height of '3'.  (Note that the curve continues below the x axis as the object regains its speed falling back down again -- negative velocity.)

How do we construct the graph of the dashed purple line?  Let's start with energy: Point 'H', (where our 'HeightGiven'= 'h', the x-axis value) is the point at which we run out of velocity; that's because it is the point at which we run out of energy.

[Aside: GeoGebra 'sliders' let us modify some variable by dragging the 'dot'you see just below 'HeightGiven' in the diagram above.  In this case, we can modify the target height value and the whole diagram redraws itself to suit that situation.]

Lifting a given mass from point 'a' to point 'b' in a given gravitational field must require a fixed amount of energy because force times distance equals energy. This is true irrespective of velocity or time.  Thus, the graph of the energy in a Hyperbolic Deficit situation ('Ed', read: 'Energy(deficit)') is merely the graph of the Energy (escape)(Ee) (the green line), 'dropped' such that it hits zero at point 'H', giving us the dashed green line.  In other words, the energy lost by Voyager lifting from point 'a' to point 'b' is constant regardless of how much we had to start with.  Where we have insufficient velocty to achieve escape, we run out of energy (thus velocity) at point 'H'. We 'drop' the green line like this:

$$ (the dashed green line)

And since we know that velocity is the square root of energy:

$V_d=\sqrt{E_d}$  (the dashed purple line)

---------------------------------------------------------------------------
Not calculus but interesting:

The relationship between the dashed red and dashed purple lines in interesting.  Boiling them both down to basics:

$V_r = \sqrt{1-\frac{1}{h}}$ (the dashed red line)

$V_d=\sqrt{\frac{1}{h}-\frac{1}{H}}$  (the dashed purple line)

... where:
h: x-axis: height going to infinity.
H: HeightGiven: a specific target height.

I had suspected that the relationship was a calculus sort of thing -- integral vs. derivative, but no, it's just that Purple (the dashed purple line) is built around a specific 'H'and Red (the dashed red line) is not.  Thus Red is 'general' and Purple is 'specific'.  Consider the diagram above where H = 3. Instead of a curve, we request a specific point on that curve:
$V_r = \sqrt{1-\frac{1}{H}} = \sqrt{1-\frac{1}{3}} = \sqrt{\frac{2}{3}} = 0.8165$ 

Consider Purple when 'h' = 'H': plug in the numbers and we quickly see that the velocity is zero.  But when 'h' is '1' (at the launch surface):

$V_d=\sqrt{\frac{1}{h}-\frac{1}{H}}=\sqrt{\frac{1}{1}-\frac{1}{3}}=\sqrt{\frac{2}{3}}=0.8165$ 

... Purple 'starts'(at the surface), at the velocity given by the red line and 'ends' at the 'H' specified when velocity drops to zero.  The dotted black lines show how Red 'controls' the starting and stopping points of Purple. No calculus involved.

---------------------------------------------------------------------------
Next question: Space And Time (not bothering Einstein tho):

It bothers me that acceleration (like gravity) is defined as 'm/s^2' -- that is, as a change in velocity over time, but the equation for gravitational fields '1/r^2' is defined in terms of change in acceleration over distance.  So, in the above Hyperbolic Deficit situation, what does the graph of velocity over time look like?  Obviously after launch, the rate at which we climb slows down as we loose velocity (duh), so at first the drop in velocity per unit of time will be steeper than the drop over distance, but then becomes more shallow. However this depends on the strength of gravity (since the sliders let us play with that) -- the lower the strength of gravity, the longer the VOT line becomes.

Diagram #11: Hyperbolic VOD to VOT:


[LagrangeHyperbolicDefDistanceToTime.ggb]

[
Acronyms below are constructed from:
V = velocity
E = energy
O = 'over'
D = distance
DD = deficit distance
T = time
X = the distance from 1 to H, that is the actual 'x' distance traveled
]

The logic chain is long, but I think the dashed red line is VOT. We've already seen how our EODD (green) and VODD (blue) curves are derived.  The next thing is to derive the TOD line: 

Take any unit of distance (x-axis). If velocity is, say, 3m/s then it will take the reciprocal: 1/3 second, to traverse that unit of distance. Thus the reciprocal of m/s:m is obviously s/m:m. Basic calculus shows us that it doesn't matter how small the units of distance are, even down to the infinitesimal. And the integral of the above must be: s/m:m * m = s:m = TOD.Diagram #12: Hyperbolic VOD to VOT demonstration:


[LagrangeHyperbolicDefDistanceToTime.ggb]

And now the tricky bit:  Point B in the diagram above is on the VOD line and point A is on the TOD line, that means that the 'y' value of point B (.6528) is the velocity and the 'y' value of point A (1.699) is the time, both of them at the same distance (2.2).  Point C borrows the 'y'value of B (the velocity) as its own 'x' value, and the 'y' value of A (the time) as its own 'y' value thus creating a point at the same velocity as point B, but now its 'x' axis is time, not distance: VOT.  Pretty clever if I don't say so myself.  But there's probably an easier way. Also, it's probably at bit unhygenic to have two graphs on the same grid even tho their axes represent different things.  But I'm a bad boy.

One more thing:  The all the lines we see are mathematically 'proper'-- they curve up to infinity** at the barycenter (the 'y' axis) so at launch we are sorta already at time=.5708 in about the same way that we are already at height=1. Obviously in any real situation, the 'clock' would be started at zero at launch.  What is interesting is that at infinity, all these lines can still tell each other appart, thus there are many different infinities.

(** The excption is the TOD line since time, obviously, starts at zero, not infinity.)

[The graphs in this paper were created using GeoGebra software, the active files are available on request. It is hugely fun to play with the sliders and see how the graphs change as the input values change.]