(last update: 09 Sep 2012)

I have often wondered how folks calculate what is required to get from point A to point B within our solar system. Not just what direction to head my space probe in but once I have the correct direction, how do I get it there?

Not just using a rocket or stage to zip my spacecraft there, but how big a stage do I need?

How long does it need to burn?

At what thrust setting can it operate?

What will happen when I get there or rather HOW DO I STOP?!

These are simple questions but the details they hide are called orbital mechanics. I have never had a proper coarse in orbital mechanics but I have had a physics class or two, and there are many sources of useful information. One of the first places to start is Wikipedia, where much useful information can be found on the planets, their orbits, masses, and all manner of other things. Another place is NASA's JPL or Jet Propultion Laboratory. More specifically the Basics of Space Flight web coarse. A discussion of how conventional rocket engines work can be found at the Rocket Propulsion website.

One of the best places to start is getting to know some mathematical modeling of the physical world. Why mathematics? Because it is a most efficient way to carry on a conversation about the world around us, without getting lost in all the jargon, and verbiage of spoken or written words. Language is great as a general tool but as soon as clarity, and consistency of thought is needed, it is not so good. Fortunately the mathematics which describe the motion of things in gravitational fields does not have to be particularly complicated, to get useful results.

A great place to start is classical mechanics, and Newton's law of gravitation.

\label{eqn:LawOfGravitation}
F = \frac{G \ m_{1} \ m_2}{r^2}

Newton came up with this equation around 1687.This equation says that F, the force (in Newtons "N") of a gravitational field between two masses, can be found using G, the universal gravitational constant, the masses of two objects m_1, and m_2, and the distance between the two masses r.

The gravitational constant is always the same, and has a published value something like the following:

\label{eqn:GravitationalConstant}
G = 6.67384 \cdot 10^{-11} \ N \frac{m^2}{kg^2}

Don't be confused by the m in equation\ref{eqn:GravitationalConstant}. In this context it stands for meters. A meter is about 3.281 feet. The N stands for Newtons, which is a unit of force. Often in the USA we talk about pounds of force. In the metric system the Newton is often used, although sometimes people will talk about kilograms (kg) of force. This does get confusing because the other unit in this constant is kg. In this case, and in just about every equation ever written, kg is a unit of mass, not force.

No matter the confusion, the gravitational constant in equation\ref{eqn:GravitationalConstant} is effectively a scaling factor. This scaling factor allows the equation to produce results consistent with the units used.

Equation\ref{eqn:LawOfGravitation} is all very well, and good but how is it going to help with getting from one place in the solar system to another???

All alone it can't. But there is more to classical mechanics...

There is another interesting mathematical model of the physical world. It is called Centripetal force. It looks like this:

\label{eqn:LawOfCentripetalForce}
F = \frac{m \cdot v^2}{r}

It was derived by another smart guy named Christiaan Huygens around 1659. Huygens was an older contemporary of Newton. What equation\ref{eqn:LawOfCentripetalForce} says is that a force is required to cause a moving body to rotate about a point. That force F is proportional to the mass m of the object, its velocity v, and how far away from the pivot point it is r.

Now equation\ref{eqn:LawOfCentripetalForce} holds for circular motion. Orbits are almost never circular. Of the eight major planets in our solar system none of the orbits is circular. All are ellipses, some are nearly circular, others less so. Keep in mind that a circle is simply a special form of ellipse, where the eccentricity goes to zero.

As an object's orbit departs from circular, equation\ref{eqn:LawOfCentripetalForce}'s results become less accurate. Since we will be presuming orbits are circular, we will have to accept some error in our calculations. This is done in an attempt to reduce the complexity of the calculations.

Mindful of the accepted error, we now have two expressions that find a force. So... let us combine them.

\label{eqn:CentripLoG}
\frac{m_2 \cdot v^2}{r} = \frac{G \ m_{1} \ m_2}{r^2}

What happened in equation\ref{eqn:CentripLoG}?

I substituted the F, on the left side of equation\ref{eqn:LawOfGravitation}, with the stuff on the right side of equation\ref{eqn:LawOfCentripetalForce}. I also chose to set the m of equation\ref{eqn:LawOfCentripetalForce} equal to m_2 of equation\ref{eqn:LawOfGravitation}. What is being said here is that there are two expressions for a force. If I insist that both forces are the same, I can swap either expression, for the force in question. However, the contexts are quite different. In other words, a force, is a force, is a force, whether I'm talking about gravitational attraction or objects moving around a point.

But... what about the mass?

I'm saying that a particular mass m_2 is common to both equations. In one it is attracted to another mass m_1, and in the other it is moving around some point in space. I'm not specifying where. I'm just saying that it is doing that.

Now we can do some math... divide both sides by m_2...

\label{eqn:CentripLoG-01}
\frac{v^2}{r} = \frac{G \ m_{1} \ m_2}{r^2 \ m_2}

... but the m_2 on the right side cancels...

\label{eqn:CentripLoG-02}
\frac{v^2}{r} = \frac{G \ m_{1}}{r^2}

we can multiply both sides by r, which removes the squared r on the right as well...

\label{eqn:CentripLoG-03}
v^2 = \frac{G \ m_{1}}{r}

now take the square root of both sides,

\label{eqn:OrbitalVelocity}
v = \sqrt{\frac{G \ m}{r}}

What we now have in equation\ref{eqn:OrbitalVelocity} is a general expression for the velocity of an object's orbit about another mass m when it is a distance r away from m.

In reality it is a bit more complicated than that. However, if the mass around which the orbit is centered (the Earth) is much larger than the orbiting mass (a satellite or space ship), the previous statement holds. Equation\ref{eqn:OrbitalVelocity} holds no matter what but things can get complicated when both the objects have a similar mass. For the purposes of simple orbital mechanics, where one mass is much larger than the other, it is fair to say that equation\ref{eqn:OrbitalVelocity} is a general expression for the velocity of an object's orbit about something like the Earth, the Moon or the Sun.

If you plug some numbers in, you will find that this simple equation is within 0.353% for the Moon's average velocity around the Earth. Using numbers for the Earth and Sun, the error is about 0.045%.

Bear in mind that the moon's orbital velocity changes by almost 3%, over its average velocity, depending on where it is in its orbit, so 0.353% isn't doing too bad, for rough calculations. In the case of the Earth's orbit around the Sun the measured speed variation is less than 0.8% the average value. But our calculation is only 0.045% the average value, so again our calculation is doing OK for our purposes. In these cases, and many others, one of the things causing error is the difference between a circle and an ellipse. Regardless of the causes of the error, our results are working out pretty well.

Where something ends up in the solar system depends on its energy with respect to the sun, provided it does not get too close to or hit something else. An object like the sun contains 99% of the mass of the solar system. With that much mass, it dictates what happens to all the other mass in the solar system. The only way other things, like planets, and spacecraft, can have any say in their fate is to have some energy. But not just any energy. They need kinetic energy or energy of motion. The direction of that motion is also important.

If you are the planet Earth or a spec of dust equally far away from the sun, and you don't have enough kinetic energy, orthogonal (that means perpendicular to the gravitational force vector - in this case) to the sun's gravitational pull, you are going to end up in the sun. Perhaps sooner rather than later. It doesn't get much simpler than that. Why? Because even objects in the solar system with kinetic energy directly away from the sun can still fall back into it.

As a simple example suppose we had a stationary gun pointed away from the sun, located as far away as the earth is from the sun. Using some simple approximations the bullet fired from that gun would travel away from the sun for about 14.76 hours. At that time it would stop and begin to fall back toward the sun, eventually to fall into the sun itself. Long before falling into the sun it would likely be vaporized, and the atomic nuclei of the bullet would be blown back out on the solar wind. The point is that common every day notions about motion can be misleading, when one wants to calculate orbits.

I have mentioned some "simple" approximations in the previous example. The reason for them is that calculating the effects of gravity on any object can get complicated, even with the simple bullet example. Why? Because gravity does not stay constant, and all non zero masses have gravity. This means that any given object is under the influence of many others. The force acting on any one object is the result of all the forces of the others, which is proportional to how far away a given object is from any other. Adding all those forces up, especially as they change with distance and time, can be a daunting task. Even for computers.

Let's look at the bullet example in a bit more detail. Newton's law of gravitation is:

F = \frac{G \ m_{sol} \ m_b}{r^2}

Where the mass of the bullet is m_b, the mass of the sun is m_{sol}, the distance between the two masses is r (often a radius), and G is the gravitational constant. Also from classical mechanics we have the relation:

F = m \ a

If we swap in m_b, for m, and substitute the expression for F for the m\ a, we can get an equation for the acceleration of the bullet we fired away from the sun:

a\cdot m_b = \frac{G \ m_{sol} \ m_b}{r^2}

dividing by m_b on both sides:

a = \frac{G \ m_{sol}}{r^2}

This expression seems pretty simple except if one tries to use it to calculate the bullet's position over time. It soon becomes apparent that each time you try to calculate the level of acceleration it depends on where the bullet was before you made the calculation, which depends on where the bullet was before that calculation, which depends... you get the idea.

But we can use classical mechanics to help out...

s(t) = \frac{1}{2} a t^2

where s(t) is the position of an object at time t, and a is the acceleration. If we rename some of the variables, and do some substitution we get,

r(t) = \frac{1}{2} \frac{G \ m_{sol}}{r^2} t^2

but r^2 is a functin of time, and can be moved to the left side as in,

r(t)^3 = \frac{1}{2} G \ m_{sol} \ t^2

which can yield a simple function in time for the radius,

r(t) = \left( \frac{1}{2} G \ m_{sol} \ t^2 \right)^{\frac{1}{3}}

but there is one other thing to add, and that is the offset of the initial radius r_i.

r(t) = \left( \frac{1}{2} G \ m_{sol} \ t^2 \right)^{\frac{1}{3}} + r_i

Let's form an estimate for the amount of energy we need to put into the Juno spacecraft to go from the Earth to Jupiter. To make things simple let's use a Hohmann transfer orbit. A Hohmann transfer orbit isn't the fastest way to get to Jupiter but it consumes the least amount of energy. The Hohmann transfer orbit matches one elliptical orbit to another, using yet another ellipse. Remember that a circular orbit is simply a special case of an elliptical orbit. Both Earth and Jupiter have elliptical orbits.

So our job will be to calculate a "transfer orbit" to get the Juno spacecraft from the Earth to Jupiter. We can use the Obital Mechanics section of the Rocket Propulsion or the Orbit Altitude Changes sections and websites as starting points.

First we will need the semi major axis of the Hohmann transfer orbit:

\label{eqn:SemiMjrAxisHohmanTxfrOrbit}
a_{tx} = \frac{r_A + r_B}{2} = \frac{E_{or} + J_{or}}{2} = 4.843x10^{8} \ km

We will also need to match the orbital velocities at the start and the end of the Hohmann transfer orbit. These two velocities are simply the Earth's orbital velocity, and Jupiter's orbital velocities. The \(G\) in the equations is the gravitational constant. The dominant mass is the sun's mass \(m_{sol}\) sometimes shown as \(M\).

\label{eqn:InnerOrbitalVel}
V_{iA} = \sqrt{\frac{G \cdot m_{sol}}{r_A}}

for the initial velocity at the Earth's orbital radius, and

\label{eqn:OuterOrbitalVel}
V_{fB} = \sqrt{\frac{G \cdot m_{sol}}{r_B}}

which will be the orbital velocity at Jupiter's radius.

The idea here is to simply match the speed of our spacecraft to the speed of Jupiter's orbit. This does not include any additional velocity for an orbit around Jupiter. It is supposed that we can park our spacecraft far enough away from Jupiter so that it will simply follow along the same orbital path.

To do this matching of velocity at Jupiter's orbital radius, the velocities of the initial transfer orbit, and final transfer orbit must be calculated. Toward that end the inner orbital velocity (at Earth's radius) will be,

\label{eqn:InnerOrbitalTxfVel}
V_{txA} = \sqrt{G \cdot m_{sol} \bigg( \frac{2}{r_A} - \frac{1}{a_{tx}}\bigg)}

and the outer orbital velocity (at Jupiters radius) will be,

\label{eqn:OuterOrbitalTxfVel}
V_{txB} = \sqrt{G \cdot m_{sol} \bigg( \frac{2}{r_B} - \frac{1}{a_{tx}}\bigg)}

Equations \ref{eqn:SemiMjrAxisHohmanTxfrOrbit}, \ref{eqn:InnerOrbitalVel}, \ref{eqn:OuterOrbitalVel}, \ref{eqn:InnerOrbitalTxfVel}, and \ref{eqn:OuterOrbitalTxfVel} can then be used to calculate the distances, and velocities which will be required.

We already have a value for \ref{eqn:SemiMjrAxisHohmanTxfrOrbit}, of \( ~ 4.843x10^{8} \ km \).

Gravitational Constant :

\label{eqn:GravConstant}
G = 6.67259 \cdot 10^{-11} \ \frac{N \cdot m^2}{kg^2}

Combining these with the solar mass.

Solar mass:

\label{eqn:MassOfSol}
m_{sol} = 1.9891 \cdot 10^{30} \ kg

Will provide the two velocities that we will need to generate and/or match at the beginning and end of our Hohmann transfer orbit. The first is the velocity we will need when leaving the orbit of the Earth around the sun:

\label{eqn:HomanXfrOrbitFromEarthToJup}
V_{txE} = \sqrt{G \cdot m_{sol}\left( \frac{2}{E_r}-\frac{1}{a_{tx}} \right) } = 38.356 \ \frac{km}{sec}

Which implies a change in velocity or "delta V" of:

\label{eqn:DeltaV_HomanXfrOrbitFromEarthToJup}
\Delta V_{txE} = V_{txE} - V_{Eaov} = 8.576 \ \frac{km}{sec}

where the average Earth solar orbital velocity is:

\label{eqn:AvgOrbVelEarth}
V_{Eaov} = 29.78 \ \frac{km}{sec}

This delta V of 8.576 km/sec is how much additional velocity will be needed at the "lowest" point of Earth's orbital ellipse about the sun, in order to begin a Hohmann transfer orbit to Jupiter.

Note that any object in Low Earth Orbit (LEO) is traveling about 7.611 km/sec already. That orbital path about the Earth will need to be broken, such that the spacecraft will not fall back into the Earth. To do this the object in orbit needs to achieve "escape velocity". For earth escape velocity is:

\label{eqn:EscVelEarth}
V_{Eesc} = 11.186 \ \frac{km}{sec}

Note that escape velocity is dependent on the distance between the two objects. An object orbiting far away from another object can have an orbital velocity which is quite low. The velocity of equation \ref{eqn:EscVelEarth} is based on an earth surface orbit. This is not realistic. A more realistic escape velocity would be that from a "parking orbit", which is often a modest LEO. The Ranger program of the early 1960's used a parking orbit of 185 km (115 miles) above the surface. Such a low parking orbit reduced the escape velocity only slightly:

\label{eqn:EscapeVelocity}
V_{esc} = \sqrt{\frac{2 \cdot G \cdot m_{sol}}{r}}

so,

\label{eqn:EscVelRangerParkOrb}
V_{Eesc} = 11.0284 \ \frac{km}{sec}

In the case of the Ranger program, as equation \ref{eqn:EscapeVelocity}, and \ref{eqn:EscVelRangerParkOrb} suggest, the parking orbit didn't provide much in the way of delta V savings. However it does provide time to check out the space craft, and confirm initial velocity parameters before committing to breaking the orbit.

If a higher parking orbit could be used, perhaps 400 km, equation \ref{eqn:EscVel_400km_ParkOrb} suggests a good deal of delta V savings may be had.

\label{eqn:EscVel_400km_ParkOrb}
V_{Eesc400} = 7.9739 \ \frac{km}{sec}

There is no free lunch here. Whether a high LEO or a low LEO is chosen as a parking orbit, no "savings" in energy is to be had. In order to gain altitude, energy will have to be spent. The choice to use or not to use a parking orbit is more an operational choice. The one caveat to that is that with lower orbits there is more drag on the spacecraft, since there is more atmospheric gas present. This will cause the orbit to decay faster than higher LEOs.

The real value of a parking orbit would be realized if refueling could be done. Another alternative would be orbital rendezvous with a fully fueled escape stage. In this way more modest escape stages could be used or heavier spacecraft could be given enough delta V to achieve escape velocity, as well as higher transfer obit velocities.

Higher transfer orbit velocities are a good thing because the higher the transfer orbit velocity, the less time it takes to get where you want to go. This can be important when traveling to Mars, Jupiter, Saturn, and beyond.

Unfortunately escape stages capable of, refueling, orbital rendezvous, and firing multiple times in zero g, are more difficult to design, and expensive to build. However, once the technology is in place, the operational benefits are so great that they would be measured in orders of magnitude.

To get a feel for how important delta V is, with respect to a transfer orbit, the problem can be simplified. The simplification comes by neglecting the Earth's gravity well. We can simply pretend it isn't there.

For the time being we will presume that the spacecraft is not in LEO, and is simply orbiting the sun at the Earth's solar orbital radii but is not under the influence of Earth's gravitational field. In this way there is no energy requirement to get up out of the Earth's gravitational well. Somehow the spacecraft is just there, ready to go.

This reduces the problem to one of spacecraft mass, and the amount of energy in the escape stage, that can be devoted to delta V. It is delta V that dictates the amount of time required in the transfer orbit. Keep in mind that Hohmann transfer orbits are all that are being considered for the moment.

This presents us with a problem because one of the most powerful upper stage boosters used to propel spacecraft into Earth orbit, and destinations beyond earth orbit, is the Centaur. Unfortunately the Centaur can only generate 6.688 km/sec of delta V, when working with Juno spacecraft type masses. However, we know from equation \ref{eqn:DeltaV_HomanXfrOrbitFromEarthToJup}, that we need,

\label{eqn:DeltaV_HomanXfrOrbitFromEarthToJup_simple}
\Delta V_{txE} = 8.576 \ \frac{km}{sec}

which means our spacecrft is 1.696 km/s short of the required delta V to get to Jupiter.

Our equations are telling us that it is impossible to get to Jupiter, using the Centaur, with the Juno spacecraft attached. There is just too much mass for the energy available. This presumes that there is no velocity/energy available, other than the solar orbital velocity at the radii of the Earth's orbit.

The analysis is telling us that our Juno spacecraft "can't get there from here...", at least not in this simplistic view of the problem. Perhaps, in our analysis, things are too simple...

More delta V is needed. This means more energy is needed. Where will the energy come from?

Make a bigger Centaur? A Centaur Stretch?

... wait a minute... didn't we have some velocity when in LEO?

Can that energy/velocity be leveraged into transfer orbit velocity?

Is it possible? We have the Earth's solar orbital velocity plus some LEO velocity. Is there a way to leverage some of the LEO velocity, so as to reduce the amount of delta V needed in equation \ref{eqn:DeltaV_HomanXfrOrbitFromEarthToJup_simple}?

But we would need to keep the escape velocity requirement as low as possible, otherwise we would spend our Centaur energy (delta V) just getting free of Earth's gravity well.

But can that even be avoided?

These are the sorts of questions that start popping up, as more complexity is added to the problem. Some of the answers help solve our energy budget problem, some of the answers hinder the overall energy budget.

Taken together, the issues all boil down to the problem of climbing up out of the Earth's gravity well. The more mass one is trying to lift out of the well, the more fuel (stored energy) one needs to do so. The more fuel one needs (additional spacecraft mass), the more fuel one needs to lift the fuel. The law of diminishing returns is creeping in here.

A given rocket technology can provide only so much lift, before the required fuel, and structural mass to lift the fuel, start to seriously eat into the useful payload. Eventually the entire fuel, and structure budget (mass, energy, structural design margin) is spent on the fuel and structure, leaving nothing left for the useful payload.

Space X claims a mass ratio of about 30:1 with their unproven Falcon 9 Heavy booster stages, with a competitor rocket, the Delta IV of United Launch Alliance (ULA), capable of only about 10:1. The validity of these claims is less important here, than the idea of mass ratio. Mass ratio is the ratio of propellant (fuel and oxidizer) to rocket structure. The larger the ratio, with equivalent thrust, the better.

Mass used for anything other than payload should be kept as low as possible, without sacrificing reliability, and thrust. In this way the most is made from the available energy (propellant). Mass used in rocket structure subtracts either useful propellant or payload. However, what really matters is the mass to thrust ratio. It is this ratio that dictates how high, and how fast the spacecraft will fly, after the boost to orbit.

Here's the problem. If a rocket has a 30:1 mass ratio but produces 1/3 the thrust, it will lift the same spacecraft, to the same orbit as a rocket with a mass ratio of 10:1, which delivers 3 times the thrust. Both rockets provide similar service to orbit, although their marketing numbers are quite different.

There are many technologies. VASMIR (Ad Astra), Solar Sail, Project Orion (pulsed nuclear), Nuclear Thermal Rocket and NERVA, ion propulsion... the list can get quite lengthy.

In one way or another most of these are reaction engines. Such engines accelerate a mass, and take the energy of doing that as a force to accelerate the engine itself, and anything attached (the payload, fuel... etc.) to it. All of the details of such an arrangement is simplified in the so called "rocket equation":

\label{eqn:Rocket_Equation}
\Delta v = V_e \ ln \ \frac {m_0}{m_1}

where V_e is the velocity of the rocket exhaust stream, m_0 is the initial mass of the rocket with fuel, m_1 is the mass with the fuel spent, and \Delta v being the Delta V we have been talking about, the change in velocity of the rocket. Note that ln is the natural log function.

The term V_e is a bit complicated. Not all of the exhaust stream travels at the same velocity. So the term is better thought of as the aggregate exhaust stream velocity. This would be the equivalent uniform velocity or the velocity integrated over the span of the variations with respect to the mass of the exhaust stream. A portion of the exhaust mass will be flowing at one velocity, and another portion at another velocity, and so on, until all the variations are accounted for. A simpler way to think about such a complicated sum, is to pick a uniform velocity (or a smaller mass) which has the same velocity multiplied by mass product.

Note that hidden in the V_e term is the efficiency of a given rocket engine's ability to convert its propellants to V_e, or thrust and \Delta v.

There are competing needs here as well, which are often referred to as design trades or design trade-offs. A structure, rocket or otherwise, floating in space in micro gravity can have much different requirements, than one sitting on the launch pad on the surface of the Earth.

A rocket leaving the surface of the earth needs to support itself under launch thrust, aerodynamic loads, and a host of other conditions, especially launch vibration. In orbit, well above the surface of the earth or in interplanetary space, the spacecraft has far lower loads. The overall environment is harsh but the loads on the spacecraft are far lower. The only significant loads, with some exceptions, are loads imparted by attitude control thrusters, or larger higher thrust orbital maneuvering engines.

If a spacecraft is designed solely for interplanetary flight loads, it will likely fall apart under launch loads and vibration. Thus most spacecraft are structurally over built. This means they are heavier than they need to be, and therefore require more energy to launch them successfully.

This is one example of a design trade. Add some structure (increased mass at launch) to insure that the spacecraft survives the launch. Suffer with unnecessary additional orbital flight mass, in order to be sure the spacecraft is not broken by the time it reaches orbit. Even though the additional mass serves no purpose, and may be a liability, during the orbital flight and mission.

OK... so where did the 6.688 km/sec delta V limit come from?

And... suppose a full 8.576 km/sec of delta V could be provided, how long would it take to get to Jupiter's orbit?

Earth Solar Orbital Radius:

E_{or} = 152,098,232 \ km

Earth mass:

m_e = 5.9736×10^{24} \ kg

Jupiter Solar Orbital Radius:

J_{or} = 816,520,800 \ km

Jupiter Solar Orbital velocity:

V_{Jaov} = 13.07 \ \frac{km}{sec}