**Patching Conics**

Time and time again I've watched people patch conics by using straight addition and ignoring the Oberth benefit. It is a very easy mistake to make. I'll give an example of this error for an earth to Mars delta V budget.

**3 km/s**is needed to leave earth's 30 km/s heliocentric orbit and enter this transfer orbit. In a similar fashion it takes

**2.5 km/s**to leave Hohmann transfer and match velocities with Mars.

But before we go from one heliocentric orbit to another, we need to escape the planet's gravity well. Earth's surface escape velocity is about

**11 km/s**. Mars' surface escape velocity is about

**5 km/s**.

The novice will look at these 4 quantities and simply add them. 11 + 3 + 2.5 + 5 is 20.5. They'll tell you it takes about 20.5 km/s to get from earth's surface to Mars surface.

But to accurately patch conics you need to use the hyperbolic orbit that takes you out of the planet's sphere of influence to a heliocentric orbit. Hyperbolic orbit speed is sqrt(V

_{escape}

^{2}+ V

_{infinity}

^{2}). But what's V

_{infinity}? In this example it's the 3 km/s needed to go from earth's 30 km/s orbit to a Hohmann's 33 km/s. In Mars' neighborhood V

_{infinity}is the 2.5 km/s needed to exit Hohmann and match velocities with Mars.

If you remember high school math, sqrt( a

^{2}+ b

^{2 }) should look familiar. It's the hypotenuse in the good old Pythagorean theorem! And that's what I use to visualize hyperbola speed:

The novice will tell once you've achieved the 11 km/s to escape earth's gravity well, you need another 3 km/s. The informed will tell you only another

**.4 km/s**is needed.

For Mars to Earth the naive will tell you after you've reached 5 km/s to escape Mars then you need another 2.5 km/s to send the ship earthward. The savvy will tell you an additional

**.6 km/s**is needed.

.4 vs 3 and .6 vs 2.5. In this case the novice method results in a 4.5 km/s overestimation of the delta V budget.

**Erik Max Francis**

I used to call this the Erik Max Francis Error. Delta V budgets from one planet to another would often come up in space usenet groups. Erik would use his Python BOTEC and give an answer to ten decimal places. People would ooooh and ahhhhh. Wow! Accurate to 10 significant figures! In reality Erik's answers were accurate to

**significant figures. I finally prodded him to correct his error. Now his BOTEC is accurate to 1 or 2 significant figures. So far as I know, he still gives answers to 10 decimal places.**

*zero***Rune**

Later I called it the Rune error. The total Vinf can be roughly estimated by subtracting Mars 24 km/s from earth's 30 km/s. And this is what Rune uses on the New Mars forum when Louis asks how much delta V is needed after you get out of earth's gravity well.

Brute force" trajectories would take about as much delta-v as is the difference between the orbital speeds of mars and earth, so about 29.8km/s (for earth) - 24km/s (for mars) = 5.8km/sI have explained to the New Mars Forums many times that the speed of a hyperbola is sqrt(V

_{escape}

^{2}+ V

_{infinity}

^{2}). Will Rune ever learn it? I doubt it.

But no matter, Rune's a member of Zubrin's cult. People expect Zubrinistas to be innumerate, nobody takes them seriously. But sadly they are loud and high profile. John Q Public can be misled into thinking they speak for all space advocates.

It's more damaging when someone in authority commits this error. Now I'm talking about Dr. Tom Murphy.

**Professor Tom Murphy**

This is from a graphic from Tom Murphy's Stranded Resources:

Murphy writes:

For instance, we travel around the Sun at a velocity of 30 km/s, while Mars sails at a more sedate 24 km/s. So to meet up with Mars, we have 6 km/s of extra velocity to burn, helping us up the hill. We speak of this as a Δv (delta-vee) adjustment to trajectory.Same method as Rune for getting the total Vinf: Subtracting Mars' 24 km/s from earth's 30 km/s to get 6 km/s. An over estimation but not wildly inaccurate. And Murphy correctly shows earth's escape as 11 km/s and Mars escape as about 5 km/s.

But then Murphy straight up adds 11+6+5:

Crudely speaking, we must have the means to accomplish all vertical traverses in order to make a trip. For instance, landing on Mars from Earth requires about 17 km/s of climb, followed by a controlled 5 km/s of deceleration for the descent. Thus it takes something like 20 km/s of capability to land on MarsSounds like he's being generous to the poor deluded space cadets by rounding 22 down to 20. But the distance from earth's C3=0 to Mars C3=0 is not 6 km/s. It's about 1 km/s. Here's Murphy graph corrected for the Oberth benefit:

"But wait!" a Murphy apologist might say. "Murphy didn't include the delta V needed to rise above earth's atmosphere. That's 1.5 to 2 km/s! That makes the budget more like 18 which can be rounded up to 20."

An atmosphere does indeed add to delta V for departing a planet. On the other hand, an atmosphere is a big help for planet arrival. Park in a capture orbit with periapsis velocity just a hair under escape. Position the periapsis in the planet's upper atmosphere. Each orbit at periapsis, atmospheric friction slows the ship. This is known as

**. Almost all of Mars' 5 km/s descent can dealt with via aerobraking. Here is Murphy's graph corrected for atmospheric influence:**

*aerobraking*Now it takes 13 or 14 km's to reach escape. But with descent taken care of with aerobraking it only takes another 1 km/s to reach Mars' surface. A more realistic delta V budget from earth surface to Mars surface is about 14 or 15 km/s.

And in fact numerous Mars landers and orbiters have used this method. I am stunned that Murphy, a self proclaimed space insider, has never heard of aerobraking.

Is a 6 km/s error a big deal? Since the

**exponent**of the rocket equation scales with delta V, it's a very big deal. Murphy himself would tell you exponential growth can be dramatic.

Above graph assumes hydrogen/oxygen bipropellent. Each 3 km/s added to the delta V budget about doubles propellent needed. Each 5 km/s added nearly triples the amount. Murphy's 20 km/s delta V budget would need a little more than

**triple**the propellent of the actual 15 km/s delta V budget.

In my opinion the limits to growth is the most important issue facing us. Can space resources raise the ceiling on our logistic growth? If so, it is worthwhile to invest in building space infra-structure. If not, expensive space infrastructure is a waste of money. We should look at the question seriously. That is why I get bent out of shape when a so-called authority makes common mistakes that would embarrass a freshman aerospace student.

Tom Murphy's arguments against space are often cited in discussions of limits to growth. When I find such a discussion, I will chime in that a bright high school student could tear apart Murphy's arguments. The most recent visit was at Mike Stasse's Damn The Matrix. As usual, the

*Do The Math*crowd response was insults, appeal to authority, but

**.**

*no math***do the Math. Take my word for it because I'm a Ph. D."**

*Don't*
## 11 comments:

Hi Hop,

interesting post. But given your figures, it still raises another question for me.

We'd still need to burn 30kg fuel for each kg payload? Is that right?

So if we assume a settled asteroid colony producing their own rocket fuel from the asteroids, and mining some kind of metal resource, isn't 30kg fuel per kg metal too expensive? What metal would break even on today's markets? I'm thinking we'll mine the sea floor before the asteroid belt.

Eclipse, that 30 kg is for a 15 km/s delta V budget to Mars.

Near earth asteroids are much closer in terms of delta V. There are some asteroids that could be parked in lunar orbit for as little as .2 km/s.

Also the propellent I looked at is oxygen/hydrogen bipropellent which has an exhaust velocity of around 4.4 km/s. The most popular asteroid retrieval plan at this time is the Keck proposal which calls for solar electric propulsion with xenon as the propellant. This would have an exhaust velocity of around 30 km/s.

With .2 km/s delta V budget and 30 km/s exhaust velocity, it'd take 1 kilogram of xenon to retrieve 150 kilograms of asteroid.

I do agree that we will probably be mining the sea floor before asteroids.

Great informative post.

However, I am still grappling with your intention regarding your corrected delta v graphs. It displays a flat region around Earth's orbit.

This doesn't make any obvious sense to me. If you achieve v_infinity, then you will ultimately have zero relative velocity to Earth. You can not go anywhere, you will still be in Earth orbit.

I would imagine that Earth's gravity well should intersect with the blue line at v_infinity. Of course, you can do it however you like, but it's not clear what the definitions are as I'm reading it right now.

The Oberth effect should squish the blue line (representing Hohmann transfers about the sun) vertically, but I can't see how it would make it flat over any region.

"If you achieve v_infinity, then you will ultimately have zero relative velocity to Earth."

No, that's not right. Unless your orbit is parabolic in which case V_infinity is zero.

If you achieve 11.4 km/s at 300 km altitude, it's a hyperbolic orbit wrt earth with a Vinf of about 3 km/s which is sufficient to send you on your way to Mars. The top point is C3=0 (aka escape) which is about 11 km/s near the earth. Follow the flat line to Mars and you will see it go up a little, that's the .4 km/s. Once a capture orbit about Mars is achieved, the rest of the delta V can be achieved with aerobraking.

Regards the flat line, I just shifted Murphy's graph down and then drew a horizontal line to C3=0. That is confusing. I redrew the graphs with a more gradually sloping path from C3=0 to the 11.4 km/s velocity where V_infinity would be about 3 km/s.

I contacted Tom over this, and this is what he said:

Hi Mike (hey--DtM is both Do the Math and Damn the Matrix),

I don't dispute the more careful approach used on hopsblog. I put pieces together very simply, which may not represent more clever ways to manage interplanetary trajectories. That said, I stated clearly what I was doing, so that it's an easy job to pick it apart. I'm fine with that. I hope I never appealed to my authority as an orbital mechanics expert, because I am not. I just try to put scales on things and sort out roughly how hard things are. At the pace of a post a week (during that time)--on top of a busy job--I could not spend time polishing. 15 km/s (allowing a bit of rounding) is still frikin' hard, so my main point is barely scratched. I don't think resource extraction from space is likely to extend our growth phase on earth. It's easier to get there with our imaginations (and math, for that matter) than to actually pull it off.

This is a case of making a mountain out of a molehill, I'm my mind. Lots of strawman-flavored bits in his post.

Mike, 15 km/s is the delta V budget it takes to put a com sat in geosynchrous orbit. If it's so frikin' hard, why are there so many GEO com sats? 15 km/s isn't even close to a 20 km/s delta V budget. His point isn't merely scratched, it's gored.

You quote Tom Murphy:

I hope I never appealed to my authority as an orbital mechanics expert, because I am not.Which is exactly the point of this post. He makes common errors a freshman aerospace student wouldn't commit. I'm happy to hear him admit this because Murphy seems to puff himself up when he proclaims he's a space insider. And his fans most certainly commit the Appeal to Authority fallacy. See the screen capture above of you and Maponos.

I am happy you have the ear of Tom Murphy. His novice blunder pointed to here is only the start of a series of mistakes stemming from ignorance or dishonesty.

Ask him if he's read the Keck report,.

In

Murphy suggests it'd take 5 km/s to bring an asteroid to our neighborhood. Then he notes using oxygen and methane (exhaust velocity about 3 km/s) it'd take more than an asteroid's mass in propellent to retrieve it.Grab That AsteroidBut the Keck authors note there are asteroids that can be parked in our neighborhood for as little .17 km/s. They propose to use xenon as a propellent with an exhaust velocity of 30 km/s.

I know you and Maponos can't plug those numbers into the rocket equation. But maybe Murphy can. Ask him if the propellent mass needs to be greater than the mass of the asteroid.

Hi again Hop,

Good news, so thanks for the reply.

Now, they've just started this conversation over at the nuclear-green blog by Barry Brook, head of the Climate Department at Adelaide Uni. He's talking about Zubrin, and so you might have a few cautionary tales to tell, but he's definitely a fan of space.

http://bravenewclimate.com/2014/03/04/entering-space-energy-resources/

In the Hohmann diagram, it looks like the arrow lengths and the numbers at Mars are swapped.

-Melcon37

Melcon37, you're right. I will try to correct that diagram when I have time.

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