\require{AMSmath} \require{eqn-number}

# Practical Solar System Travel

(last update:  09 Sep 2012)

Fundamentally the question is:

Is it possible to reduce the travel time between the planets of our solar system to a handful of days, rather than many years?

The reason why this is the question is because peoples lives and careers are in question. It can take a decade or more to develop a mission, and build a spacecraft. With that done, the folks who developed the mission, and the spacecraft may have to wait many years for a launch opportunity. After that it may require many years for the spacecraft to arrive at its destination.

So the mission design and execution cycle can span the careers of the people involved.

This is not efficient.

A great deal of time is spent hashing and re-hashing designs, since it takes so long (sometimes many years) to get to the destination. Often the design margins are too tight, and systems failures occur. All because of the notion that, "... we only have one shot at this..." Why? because it takes so long to get there, we can't just go when we feel like it.

Suppose we could "go when we feel like it".

What does that mean?

And is it even possible?

So... let's have a look...

## What might "practical" mean?

A destination which currently is not practical would be Pluto.

Having said that, there is a spacecraft on its way to Pluto. As of 09 Sep 2012 it has been traveling for 6.64 years, and has another 2.84 years to go. That's a total trip time of 9.48 years. Once it gets to Pluto it can't stop. It will streak by and head for the Kuiper Belt. It was not practical to provide a means of entering orbit at Pluto, in a similar manner to the Cassini mission to Saturn.

Waiting 9.48 years, plus mission development time, is quite a long wait. Especially for a quick fly-by.

That isn't what I would call practical. Ambitious, daring, maybe even desperate but not practical.

So how might such a journey be practical? By that I mean shorter in travel time. Much shorter.

By that I mean shorter in travel time. Much shorter.

But not absurdly so.

Surprisingly the amount of travel time to Pluto is much, much, much quicker if you don't coast, the way we do now. All you have to do is keep accelerating. But only at a modest 1g. Just like standing on the surface of the Earth or sitting in a chair, that's 1g. How would that effect travel time.

### Keeping it simple... at first.

Just to see what might be possible, let us start with an overly simplified case. But let us also make it ambitious.

How long would it take to get to the farthest point in Pluto's orbit, if you start with zero velocity at the Sun, and accelerate at 1 g for half the journey, and decelerate for half the journey?

In this simplistic view classical mechanics provides the law of linear motion:

\label{eqn:LawOfLinearMotion} d(t) = x + v t + \frac{1}{2} a t^2

The distance d from a starting point, as a function of time, is equal to some starting point x from d, plus the initial velocity v multiplied by time, plus one half the acceleration a multiplied by time.

We know the distance we want to travel d. That is the farthest distance Pluto can be from the Sun. What we want to know is how long it will take to get there if we accelerate for half the journey, and decelerate for the other half. This isn't realistic but it will serve to suggest what might be gained by a similar more practical scenario.

We will make some additional simplifications:

• Zero initial offset.
• Zero initial velocity.
• Acceleration of 1 g (9.8 \ \frac{m}{sec^2})

With those in hand equation\ref{eqn:LawOfLinearMotion} can be solved for time under acceleration over a distance:

\label{eqn:LinDistAccAlone} t(d,a) = \sqrt{\frac{2d}{a}}

Here d is the distance in question, and a is the acceleration.

The furthest point in Pluto's orbit (from the sun) is:

Pluto_{ap} = 7,375,927,931 \ km

or,

Pluto_{ap} = 49.304 \ AU

Recall an AU is an astronomical unit, which is the average distance from the Earth to the Sun, over the Earth's orbit. Using the AU makes the numbers less awkward.

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

## So you want to go to Jupiter...

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}

## No Free Lunch...

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 parking orbits: Orbital Rendezvous/Refueling

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.

## Let's make it simple... no Earth launch.

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.

## Houston... we have a problem...

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...

## Rethinking the problem.

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?

#### Things get complicated quickly...

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.

#### Technology limitations

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.

## Centaur/Juno Delta V Limit

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?

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

Earth mass:

m_e = 5.9736×10^{24} \ kg

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

Jupiter Solar Orbital velocity:

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