By Bryan Laubscher

**Introduction**

There is an old adage: “Getting there isn’t half the fun, it is all the fun!” I want to adapt it to spaceflight: “Getting there isn’t half the effort, it is all of the effort!” The reason for this is the physics of the rocket equation and the depth of Earth’s gravity well. In this missive I will perform a back of the envelope calculation for the cost per kilogram (kg) of transporting mass to the vicinity of Mars. This calculation is rough and certainly not the last word. It does not include the cost of rocket research, development or engineering. Moreover, the overhead of rocket structure, tanks and staging are not included. Thus the rocket equation represents the best case scenario – true rocket performance will be less. However, the resulting cost is stunning and creates a lot of questions around how much exploration, especially manned exploration, can we afford to do with rocket technology!

Note that in this discussion, the vicinity of Mars means that it is not in orbit nor on the surface – those two maneuvers cost more fuel and hence money. I chose the vicinity of Mars because at Mars aerobraking can be used to get into orbit or to the surface without fuel expenditure. Another reason for the choice is that the ΔV, (the amount of velocity change) to reach the vicinity of Mars is less that that required to “soft” land on the surface of the Moon. Thus all costs that I derive are greater for a lunar surface mission

**ΔV Considerations**

Earth’s gravity well is deep. It takes a ΔV of 9.7 kilometers per second (km/sec) to reach low Earth orbit (LEO) from Earth’s surface. Now, LEO is actually a range of altitudes each requiring slightly different energies but this is a rough calculation so I’m not sweating the details. This ΔV is achievable with chemical energy, albeit just barely. As an illustration, remember that the mighty Saturn V rocket configured for the Apollo mission expended 95% of its total mass to place the remaining 5% into LEO. As an astronomer, I wonder about our civilization had we evolved on a more massive world such that chemical energy was impractical to achieve orbit. In that case, we might have waited until nuclear propulsion was perfected. Where would we be at the present time in such a world?

For comparison, note that the ΔV from LEO to the vicinity of Mars is 3.8 km/sec and from LEO to the lunar surface is 5.5 km/sec. This is the origin of Heinlein’s declaration (paraphrased): “When you get to LEO, you are halfway to anywhere!”

**The Rocket Equation**

Consider the forms of the rocket equation:

ΔV = V_{p} ln(M_{i}/M_{f})

And

M_{i}/M_{f} = exp(ΔV/V_{p})

Where ΔV is the velocity change, V_{p} is the propellant velocity of the rocket, M_{i} is the initial rocket mass and M_{f} is the final rocket mass (payload). Note that the mass ratio, M_{i}/M_{f} depends exponentially on the ratio of velocities, ΔV/V_{p}. This means that if the required ΔV doubles, the initial mass must increase by a factor of around 7.8. If the ΔV increases by a factor of 5, the initial mass must increase by a factor of about 172. This is the reality of rocket technology.

One way to think about the reason for this “inefficiency” is that a rocket carries a tremendous amount of fuel to burn at higher altitudes. Lifting all that fuel takes additional fuel that ends up dominating the mass carried at most points in the trajectory. One can imagine getting around the rocket equation by burning all the fuel on the launch pad – an artillery shell. It is hard to imagine, however, containing an explosion of the enormity needed to throw a communications satellite to geosynchronous orbit. Also a satellite robust enough to survive the tremendous acceleration (thousands of times the acceleration of gravity) would be heavily built and quite massive requiring an even larger explosion and greater acceleration. Moreover, the air resistance encountered by a large payload achieving its maximum velocity at a hundred meters (or so) altitude after launch would be prodigious and require an even greater amount of initial energy and speed to achieve orbit above the atmosphere. Indeed, to minimize air resistance, modern rockets fly an “efficient trajectory” by throttling up only high in the atmosphere, purposefully carrying fuel to high altitudes.

As an aside, nuclear rockets, by virtue of relying on propellant are also subject to the rocket equation. The advantage of nuclear fission rockets is that V_{p} is 2 to 3 times higher than for chemical reactions. Fusion and anti-matter rockets are predicted to have much higher propellent velocities than fission rockets. Of course, some drawbacks exist to nuclear rockets although the negative impacts decrease outside the atmosphere.

I already stated that the Saturn V, the most powerful rocket ever flown, placed 5% of is mass into LEO. Therefore, let me start with that information. This fact means that for every kilogram of payload lifted to LEO, 20 kilograms of fuel was required. Using 20 as the mass ratio corresponding to the ΔV to LEO I will use the rocket equation to scale the mass ratio from LEO to Mars vicinity. I get the result that 2.39 kg of fuel are required to transport each kg of payload in LEO to the vicinity of Mars. Thus the total mass lifted to LEO is 3.39 kilograms, 1.0 kg of payload and 2.39 kg of fuel to take that payload to the vicinity of Mars. Because this is a back of the envelope calculation, I’ll simplify this to 3.4 kilograms.

**Rocket Launch Costs**

The energy required to launch one kilogram to LEO is well known – it is 17.2 kiloWatt hours (kWh). Home electricity prices are less than $0.10/kWh so using that value the corresponding cost for the energy to LEO is $1.72/kg!

Rockets are expensive and complicated. They are mechanical machines that operate at the limits of materials science and chemistry. They also occasionally fail. Decades of experience with rockets points to more than $10,000 per kilogram to LEO. Attempts to radically reduce this number have failed and the claims for the greatest reduction hover around a factor of 2. (See When Physics, Economics and Reality Collide: The Challenge of Cheap Orbital Access by Jurist, Dinkins and Livingston, 2005, for a well thought out treatise.)

This is our current reality and the rocket equation, coupled with our deep gravity well, combine to ensure that there will not be much movement in this number. Rocket technology is surprisingly mature. The issues have been thoroughly understood for many decades. As an example, the Saturn V first stage possessed a mass ratio of 94%. That means that 6% of the first stage’s mass was structure, tanks, engines and controls – the remainder was fuel.

**Cost per Kilogram to Mars Vicinity**

Using the value of $10,000/kg, the cost of moving one kilogram of payload from the surface of Earth to the vicinity of Mars is $34,000. The cost to the lunar surface is greater. I claim that this is good to at least a factor of two barring a glacial bureaucracy that would drive the cost higher than $68,000/kg. I do not foresee the lower limit of $17,000 per kilogram being realized but most people would agree it is a lower limit.

NASA’s Space Exploration Initiative called for a Mars rocket with a mass of 1000 metric tons which corresponds to $340 billion launch cost to Mars! The innovative Mars Direct plans called for a Mars spacecraft of 87 tons implying $2.9 billion launch cost to Mars! These costs do not include research, development, fabrication, construction or test flights. Also, these costs are not rigorous since these ships were to be constructed and launched from LEO so the mass may include their fuel to Mars – my source did not break down the mass. If the fuel to Mars is included, then the launch costs change to $100 billion and $853 million, respectively. In either case, the magnitude of these numbers is useful to realize what we are looking at in terms of launch costs.

A manned outpost or colony would require many, many tons of shelter, equipment, food, water etc. to be sent to Mars over a long period of time. If the plan is to “live off the land”, initial missions will still require tremendous amounts of logistical support. The moon requires even more in-situ support since it lacks the inherent resources and advantages of Mars. Of course many fewer resources are required for the 3-day trip to the moon versus the many months travel time to Mars.

**Conclusion**

My question is: How much exploration, especially manned exploration, of the moon and Mars will we be doing at $34,000 per kilogram? My guess is that we’ll do pretty much what we’ve done over the last 35 years since the last Apollo mission.

If you agree, then consider that the Space Elevator is the answer. The cost per kilogram for the first Space Elevator launch is expected to be around $1000 - $3000/kg and it can launch the payload to Mars with only a small rocket engine for a plane change maneuver. The potential for savings is stunning! How much exploration could we do with a launch infrastructure that cost $1000 per kilogram to LEO or GEO or to Mars? Moreover, Space Elevator technology is subject to the economy of scale. As an infrastructure of high-capacity Space Elevators is constructed, the cost will fall dramatically from $1000/kg to $50/kg. No matter how difficult it may seem to develop Space Elevator technology, it is the missing piece to open space as a place to solve problems here on Earth.

I want to hear thoughtful responses from the community. What do you think? What is your estimate of the cost? What is a realistic estimate for human missions to the moon and Mars? What conclusion do you draw from these numbers?

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