Why send artists to the moon on BFR? #dearMoon


#1

This week we have an epic panel to talk about the recent #dearMoon announcement including Tim Dodd the Everyday Astronaut, Emory “VaxHeadroom” Stagmer, Dr. Niamh Shaw and Astronaut Nicole Stott. We focus on the impact of the recent announcement by using a trip around the moon and art to inspire humanity in a totally new way.

We went in to a more technical discussion of BFR in After Dark and have made that available immediately to everyone here: YouTube – If you like content like that consider contributing to the show at TMRO is creating Space and Science Webcasts | Patreon

Launch Minute:

  • Polar Satellite Launch Vehicle - NovaStar-1 and SSTL S1-4
  • Long March 3B - Beidou navigation satellites C33 and C34
  • HII-B - HTV-7

Space News:

  • Astronomers discover planet like Vulcan!
  • JPL Troubleshooting Data Glitch on Curiosity
  • Giant net catches Cube Sat

#2

“Why send artists to the moon on BFR?”
The Huygans Titan landing happened on 12:43 UTC SCET, January 14, 2005
It wasn’t till 05.04.06 that the data was turned into art:


The creativity and technical skill involved in rendering slivers of discrete pictures and raw numbers into a panoramic spectacle of sound and color is truly the act of artist. The ability to make science interesting to a wide audience is a special skill, and one that you guys at TMRO make use of every week. Keep it up.

#3

I wanted to follow up on what Emery was discussing on Saturday about delta V budgets and trajectories. I don’t know enough to process this information my self, but it seems like I’m seeing conflicting numbers from different sources. I started here with the Wiki:

It seems like Earth to Moon could be 3.1 km/s as stated, but it looks like Earth to infinity is 8.8 not 3.18. And Earth to Mars is somewhere in the 3-3.6 km/s range. Any help here?

Thanks!!


#4

Okay so first off.

The most important thing you should proably do to understand how orbits work, is to go buy and play Kerbal Space Program. Get yourself to a point where you can land on Duna and you should have a pretty good feel for how deltaV works. Even just a Mun landing would get you the basics.


Next, let’s talk about these numbers that are confusing you:

By far the most relevant number you need to look at is 11.2 km/s. This is the speed needed to escape Earth (aka “escape velocity” or the “second cosmic velocity”). If we subtract Earth’s orbital speed from this we have 11.2 - 7.9 = 3.3 km/s. Thus if we start at LEO, we would need around 3.3 km/s to leave the “Earth system”. If we go slower than this, we will not leave, but instead have a very long elliptical orbit which will eventually descend back toward Earth.

The moon’s orbit is pretty high up in Earth’s gravity well. So to get to the same altitude as the moon requires a very similar amount of deltaV. But if we want to stay at that altitude or pull into the orbit of the moon we’ll need to use a little more deltaV (so you might see velocities greater than 4 km/s to get to the orbit of the moon). For a fly-by mission, we do not stay in orbit so typically the second burn is not needed. Thus we have your deltaV to fly-by the moon of 3.1ish which is slightly-less-than Earth’s escape velocity.

Meanwhile if we had used exactly 3.3 km/s, we would come out of Earth’s sphere of influence and would pull into a solar orbit which looks very similar to Earth’s. In order to get to Mars, however, we would need to have a higher orbit. Thus we need just a little bit extra velocity, which is where you get your 3.6 km/s deltaV.

You also might see a smaller number to get to Mars such as your 3 km/s deltaV. This is probably because that mission started out in a higher orbit (with more energy), and thus required less energy to escape (i.e. they did not start at LEO as we have been assuming all this time).


Finally, there’s a number that’s confusing me:

Where are you getting 3.18 from? Oh, I think i see, you misplaced your decimal. You meant 31.8 km/s.

31.8 km/s is the “fourth cosmic velocity” in technical jargon. It’s the speed you would need to leave Earth and come to a standstill, cancelling out all of Earth’s orbital velocity (and then fall into the sun). If you want a full, math-filled discussion of this, then look at the google book: “Theory of Interplanetary Flights” By Grigor A. Gurzadyan (starting probably at chapter 6 on page 51).

The book even goes on to describe the next number 8.8 km/s. This is the “third cosmic velocity” (16.6 km/s: the velocity needed to escape from the sun when starting from LEO) minus the “first cosmic velocity” (7.9 km/s: the speed needed to orbit Earth at LEO). In other words, 8.8 km/s is the deltaV you would need to leave the solar system when starting from LEO.

The idea here being that you can get to the sun more easily by first leaving the sun then doing a small maneuver to cancel your remaining velocity and push you back toward the sun and letting its gravity pull you straight back.

That’s a very specific case of trying to get to the sun from Earth though; you don’t need to worry about any of that to understand deltaVs.


Hope that helps!


#5

I think that definitely helped. Unfortunately I am too busy for Kerbal. It looks awesome, but I spend enough time just thinking about these things. Maybe one day…

I think maybe what he was referring to was just what is needed to escape the sphere of influence that you mentioned. I was pretty sure he said 3.18. But I should rewatch to that part of the video.

Thanks for the help! Any other insight is welcome


#6

This is one of those things where you can spend all day thinking about it, but, in order to a real handle on it, there’s no substitute for seeing it visualized and manipulating the various outcomes. KSP’s great for that.
Edit:


#7

Ah, I see now. He misspoke. When he said “escape the solar system” he meant escape the “Earth system”. His “3.18” is our “3.3 km/s”, i.e. escape velocity. Not sure why the numbers are different, it’s probably to due to a lower orbit or different numbers used for the fundamental measurements used in the calculations. For instance using 6000 km for the radius of the earth instead of 6310 km.

Also there’s some significant variation in how high Mars is because its orbit is far more elliptical than Earth’s. So during some years it takes more/less deltaV than others.

Check out my post on Mars trajectories:

P.S. Can I just say… the crew’s poses at that time stamp in that Afterdark… reminds me of Da Vinchi’s Last Supper.


#8

Faulx, yes… I agree :grinning:


Btw… this was a great session; thank you TMRO for making this After Dark segment public available.