Polishing a Gravity movie class- need math checks

[UH60guy]

I'm currently in the process of polishing a class I put together for cadets based on the movie Gravity. The class is mainly about the scenario of an astronaut trying to get from Hubble to the ISS, but I delve into chapters 23 and 24 from Journey of Flight to give the background info needed to frame the problem.

Unfortunately I had to teach the class before I was finished, so a) I'm slowly fixing it up so I can give it to y'all if you want a resource and b) I'm not in any hurry to do so since I already taught it. However, while the movie is in theaters is the perfect time to tie in some AE modules to it, and I think it's a great opportunity for other AEOs out there.

My issue is this: I wanted to tie in some real Rocket Science (tm) formulas so they can see if the scenario is realistic, plus trick them into enjoying or using math. Can I get some help checking my methodology and math? Surprisingly, when the math was tied to a real world scenario of trying to save our astronauts (not just numbers on a school math test) they actually enjoyed it- though I mostly showed how the formulas were used and just plugged in the numbers myself. Eventually I'd like to have a bunch of datasheets for the instructor so they could have tried different spaceships, engines, etc. I even got some fun questions, like just how much G force would George Clooney endure if he strapped the Space Shuttle Main Engines to his back to make the orbital transfer?


One more disclaimer- the math is not the main purpose of the class and most of this is hidden in notes pages for the instructor- but I want to make sure it's all correct, because Lord knows how many times I've been surprised by astute questions from some of these freaking genius cadets we have. Plus, they did think it was fun to not just answer the question, but explore some other scenarios.

Here are the equations I used for the class:

• Ve = Isp * g -> Exhaust velocity and Specific Impulse relationship (used because rocket datasheets have easy to find Isp around the internet, but the other formulas need exhaust velocity to work)
• mf = ma (e^(Δv/Ve)-1) -> mass of fuel required = mass you're trying to move times (e^(maneuver delta v divided by exhaust velocity) minus 1 )
• Δv = Ve * ln (m0/m1) -> maximum delta v equals exhaust velocity times natural log of (initial mass divided by mass after fuel is spent)

The content (I admit, much found on Wikipedia but I'd love better/more accurate datasheets to use as reference if I can):

First up, I found a NASA study somewhere indicating it takes a delta v of about 3100 m/s to reach ISS from the Hubble. (used this as a teaching tool to discusss apogee and non-coplanar changes)

A NASA space suit (Enhanced EMU shuttle model) is 124.7 kg. MMU is 148 kg. I estimate George Clooney weighs ~83 kg. Sandra Bullock weighs ~55 kg. Mass of two astronauts, two suits, and one MMU is 535 kg.

MMU exhaust velocity is not provided by datasheet, which *does* say its maximum Δv = 24.4 m/s (telling us the transfer is impossible as it is well short of 3100 m/s). Exhaust velocity can be calculated using the following information:
Mass of full system, including fuel: m0 = 148 kg
It is powered by two compressed nitrogen tanks of 5.9 kg each, for total propellant mass of 11.8 kg
Now that we know propellant mass, we calculate dry mass as 148-11.8 = m1 = 136.2 kg

Using the rocket equation and rearranging terms to get exhaust velocity, we get ve = Δv/(ln(m0/m1)) = 24.4/(ln(148/136.2)) = 294 m/s
You can verify this by plugging it all back into the required fuel equation, and using the terms to see how much fuel is needed to accelerate 136.2 kg by 24.4 m/s with a 294 m/s exhaust velocity. I think the answer checks out- you get about 11.8 kg of fuel (possibly some minor rounding errors).

To see if they could make the trip with just a little more gas, we calculated the amount of fuel the MMU and astronauts would need to reach the ISS from Hubble:
mf = 535 (e^(3100/294) - 1) = 20,306,460 kg
Fun fact: The Space Shuttle external tank carries 730,000 kg of fuel. They have to carry the equivalent mass of 28 external tanks in compressed nitrogen to reach the space station!

Other rocket engines they can try strapping to the astronauts' backs or riding in the spacecraft:

• Space Shuttle Main Engines: Published Isp = 455s. I calculate: Ve = 4463 m/s. Use first equation. 535 (e^(3100/4463) - 1) = 537 kg of fuel required
• Centaur upper stage: Isp=510. I calculate Ve = 5003 m/s. Use first equation. 535 (e^(3100/5003) - 1) = 459 kg of fuel required
• Flying the Shuttle itself, using OMS and landing mass (Justification: SSMEs have no fuel after reaching orbit and dropping the tank, and we're calculating mass without fuel so landing should be near empty). ma = 104,328. Isp=316. I calculate Ve=4000 m/s. Use first equation. Answer:  104328 (e^(3100/4000) - 1) = 122,125 kg of fuel needed (Shuttle only carries 8174 kg OMS fuel).
• Apollo: Published delta V of Apollo is ~2800 m/s- but I'm not sure if that includes flying the mass of the LM. What about just the command module and service module? Command Module mass: 5560 kg. Service Module mass = 24,520 kg (fueled). Fuel mass = 18,410 kg. Use the second equation. Isp = 314s, I get Ve=3080 m/s. Δv = 3080 * ln(30,080/11,670) = 2916 m/s. Just a little shy of the required 3100 m/s to make it to the ISS from Hubble. I guess that's why they had an upper stage of the Saturn to get them to the moon!

At this point I was asked the G forces if they basically just rode a booster, Slim Pickens in Dr. Strangelove style. I didn't have the engine data sheets on me, but I could use some help here. I assume I just need the thrust rating of the engine, the mass of the astronauts, and Newton's second. F=ma -> published thrust in Newtons divided by mass equals acceleration (just instantaneous mass at a launch or burnout, I'm not getting into calculus or differential equations here). Divide by 9.81 to get number of G's. Would that be right? For example, Wikipedia says the SSME produces 2,279,000 N in thrust in vacuum. If George Clooney strapped it on before the fuel was exhausted, he'd accelerate at: F=ma -> F/m = a -> 2279000/535 = 4260 m/s^2, which if divided by 9.81 gives me 434 Gs. Sounds a little high, but I'd wager it's possible to liquefy someone with that thrust.

I closed out the class by showing them a delta V map of the solar system, and explained how they could use these formulas to determine delta V requirements and fuel requirements, and design a space ship to go anywhere. Unfortunately I couldn't get the math checked out ahead of time, but as I basically just plugged in these numbers for them and they weren't writing anything down, the point was more conceptual. However, before I would put this out as a resource for other AEOs, I'd want to make sure it's right.

[Al Sayre]

you can't use 9.81 m/s^2 in space for g

You would actually need to calculate  F(g)=G((m1*m2)/r^2) where G is the gravitational constant,  m1 is mass of person and m2 is mass of nearest large body and r is the distance between the centers of mass.  To be more accurate, you could use the vector sum of the forces of the nearest 3 or 4 bodies, but in earth orbit, the vectors considering the masses of the earth and moon would be "close enough for gov't work".   That's the reason we seem weightless in space.  Standing on earth, r is relatively small (r of earth) whereas in space r becomes significantly larger so as to drive g to approach 0.

We simulate gravity in space by substituting centripital force through rotational acceleration of the structure.

[UH60guy]

In this case I was under the impresion that the specific impulse formula uses gravity at sea level as a conversion factor- as it's defined as "impulse per unit weight-on-Earth of propellant," which is how the Isp value is actually the same in metric and standard (and measured in seconds oddly enough).

Example: For metric: 316 s * 9.81 m/s^2, the seconds cancel out and you're left with just meters per second (4000 m/s). Same if you use 316s * 32.2 ft/s^2, and you're left with exhaust velocity in feet per second (10,175 ft/s).

If we're talking equalizing the gravitational attraction between multiple bodies, that gets into the whole Lagrange Point discussion- while Hubble only orbits at about 550 kilometers up, the nearest Earth-Moon Lagrange Point (L1) is about 1.5 million km.

Though you're right, in low earth orbit, g isn't much far off from 9.81 anyway, and it doesn't enter into the follow-on equations.

Looking again at everything I see now I neglected the mass of the engines if strapped to a George Clooney- I might need to reword to say "what if his MMU produced as much thrust as..." rather than saying he's riding an engine.

[ol'fido]

You might contact the retired Canadian astronaut that got thrown out of a theater for refusing to stop heckling the movie.

[ProdigalJim]

Ken, you know we're neighbors, right? I have an AEO in my squadron who actually IS a rocket scientist. PM me if you want to connect with him...?

[Panache]

While just skimming over this math made my eyes bleed, I didn't see that you mentioned deceleration anywhere.  So even if they have the thrust to get to the ISS, would they have the ability to slow down and stop at it, and not just helplessly shoot past?

Disclaimer: I have not seen the movie.

[UH60guy]

Yeah I'm not sure on that- I need to take another look at that nasa report if I can find it- I think the 3100 m/s includes the accel and decel (I think they're cumulative accelerations independent of direction) but I'm not sure. Definitely needs to be answered or that means the requirements are way off.

I started this stuff when I found a website who tried to calculate for just the MMU and used the 3100 value as the goal- but after I added in the other "options" to demonstrate how hard non-coplanar transfers are, I was off on my own. Thankfully most of this was hidden from the cadets- I just showed the formulas and that the results were big numbers. it worked in class, but if I wand to leave others a standing product I still need to double check things down to the footnotes.

[SarDragon]

While just skimming over this math made my eyes bleed, I didn't see that you mentioned deceleration anywhere.  So even if they have the thrust to get to the ISS, would they have the ability to slow down and stop at it, and not just helplessly shoot past?

Disclaimer: I have not seen the movie.

It's all about frames of reference.

Imagine that you are chasing a car in another car. From the road frame of reference (E), you are both traveling at some speed. We'll say initially 40 mph and 30 mph. From either car frame of reference (C), there is a 10 mph differential. As you speed up to catch the other car, your E speed is increasing, but your C speed is decreasing. When you catch the other car, and match speeds, similar to what happens in orbit, your E speed has increased, and the C speed has decreased to zero. No slowing at all.

This is very simplified, and in reality, some slight slowing is needed. This can be accomplished by maneuvering jets.

[MercFE]

Forget all the math...  All we need is Kerbal Space Program to experiment with!

[coudano]

Yah i'm not a rocket scientist (disclaimer)
but I don't think that's how orbitals work.

It takes x amount of thrust to get to a point x (in how much time)?
Also how far away is point x?

And as the other pointed, you can apply thrust to change your momentum in the right direction,
after which you can coast
but then you have to apply thrust again to slow down or line up with your target when you get there.
so you don't smash into it or coast right by it



I (think) it goes like this:


Consider a thought experiment.
Geostationary satellites are falling in sync with the earth's rotation, right?
Which should be approximately 25,000mph?
It is far away from the earth (well above things like the ISS)
If you could look up and see one, it would appear in a fixed point and wouldn't move.

Higher things have an absolute speed above escape velocity, and will gradually fall at further distances from the earth, and eventually head off into the solar system.  However, from the perspective of earth they are crossing the sky really slowly, and will just get smaller as they get further away, will be passed in the field of view by geostationaries (which will appear...    stationary)

Lower things are falling below escape velocity and are traveling slower, but they will appear to race across the sky faster, and pass geostationaries in your field of view (i.e. the ISS flying overhead streaks by) (whereas geostationaries would appear... well...  stationary)



Let's try a slightly less complicated problem for starters.
Suppose you are in a space suit, falling in the exact same orbit (at the exact same speed) as your space ship, which is 10km away from you (oops, not good) (we'll leave how you got into this pickle in the first place, for another time)

You would need to fire a retro rocket, exactly opposite of your momentum, tangent to the orbit (not off to either side)
This will slow you down, which will effectively drop you into a lower orbit (smaller circle)

Since you are now traveling around a smaller circle, closer to the earth, you will close the range on your ship which is falling in a higher orbit (bigger circle, further from the earth) in the same amount of time

When you get close, you'll need to fire a rocket in the direction of the orbit, to speed up, and climb back up to the higher orbit with your ship.  Since you are now falling on the same orbital height again, your absolute speeds will equalize.

The moral of the story is this,
you don't have to apply thrust continuously to power your way to your destination.
You just have to apply enough thrust to get to the orbit that puts you on an intercept trajectory, and then coast
until it's time to adjust






Now, to complicate matters, suppose you are falling/orbiting exactly along the prime meridian/international date line.
And your ship is falling/orbiting along the equator (90 degrees transverse)
In this case, you two might cross paths approximately every half orbit, right?

In this case you would need to increase vector in the plane of the equator (lateral rcs)
while proportionally reducing vector in the plane of the meridian (retro rcs)
while maintaining the same net velocity (staying in the same orbital height)
eventually you will have arrested all of your momentum along the meridian
and established all of your momentum along the equator.

oh and you made sure to turn the right way so that you are falling in the same direction as the ship right?
else you will crash into each-other at equal velocities in opposite directions... (doh!)

--now you are in the same situation as the above.
both you and your target are flying along the equator in the same direction, with some amount of distance between you...
drop orbit to close range
then climb to a higher orbit when you get close, to intercept.

(or if you are in front of your target, climb to a higher orbit so your target can close on you, then drop back down to intercept)


(if you timed it right, with the right kind of math, when you pulled into the equatorial plane of motion you could arrive pretty darned close to an intercept with your ship without having to do much adjustment while completely in the equatorial plane)




Now consider that different space stations probably fly at different orbital heights, AND on different transverse vectors, in order to deconflict the chance of impact.  You are talking about changing your transverse momentum AND establishing the correct orbit for an intercept.  Probably not the sort of calculation you are doing in your head while tumbling through space in a space suit. (I haven't seen the movie yet)

Haven't touched on how LONG it will take you to get everything lined up (even with a computer) vs the life support duration on the suit.  But even the smallest changes in lateral momentum and total velocity (orbital height) will create a trajectory change that will -EVENTUALLY- result in a near collision or a collision.  The idea is to pick the most efficient path, that either gets you there in the least amount of time, or conserves the most amount of rcs fuel, or some optimal combination of both.



Anyhoo, you're better off inside the space ship, with the computer, and guidance inputs from the nerds in Houston. 

[Panache]

It's all about frames of reference.

And there's the rub.  This is assuming that the relative velocity and vector of both the Hubble and the ISS are more-or-less the same.  But even if it is, you're still going to have to factor in both acceleration to get to it and deceleration once you get there.  I imagine if the relative speeds and vectors are the same, you could keep your relative speed down to conserve fuel... assuming you had the oxygen to take your time.

Imagine that you are chasing a car in another car.

The second I read this, this image immediately popped into my head: 

Forget all the math...  All we need is Kerbal Space Program to experiment with!

My Kerbals' next of kin would disagree with that plan of action.

[coudano]

While just skimming over this math made my eyes bleed, I didn't see that you mentioned deceleration anywhere.  So even if they have the thrust to get to the ISS, would they have the ability to slow down and stop at it, and not just helplessly shoot past?

Disclaimer: I have not seen the movie.

It's all about frames of reference.

Imagine that you are chasing a car in another car. From the road frame of reference (E), you are both traveling at some speed. We'll say initially 40 mph and 30 mph. From either car frame of reference (C), there is a 10 mph differential. As you speed up to catch the other car, your E speed is increasing, but your C speed is decreasing. When you catch the other car, and match speeds, similar to what happens in orbit, your E speed has increased, and the C speed has decreased to zero. No slowing at all.

This is very simplified, and in reality, some slight slowing is needed. This can be accomplished by maneuvering jets.

Sort of, except that things work differently in space, than they do in the air.

(you thought you'd never use this again after jr high, right?)
If two jets are 20 miles apart,
And the front one is going 300mph
And the back one is going 340mph

Then the closure rate is 40mph
The back jet will catch the front jet in 1/2 an hour (30 minutes)
The back jet has to expend extra power, burn extra gas, to maintain its higher speed, and higher closure rate.

When the back jet catches the front jet, all he has to do is let off the gas,
Which will reduce his thrust relative to drag (thrust and drag were equal when he was stable at 340, right?)
And he will slow down to 300
And the intercept/join will be accomplished.


Same goes for cars running down the highway.


(we do this extensively, with practical application at VFWS, by the way...)


In space there is negligible drag.
So the back craft going 340 and the front craft at 300 continue to travel in that direction and speed
unless/until acted on by an outside force...
without exerting ANY thrust (zero thrust equals zero drag) that's unaccelerated "flight"
They can both just coast without expending any fuel, and they will close at a rate of 40
until they intercept and join
Then the back craft has to expend fuel opposite his momentum to match speeds with the other craft


--in orbit around the earth, there are other considerations, which I covered in my other post

[NIN]

Disclaimer: my eyes glazed over the instant I saw the delta symbol in the math. Done.

However, I do know that there are a whole BUNCH of dudes on this forum right here (www.nasaspaceflight.com) who probably have ALL the answers to these questions, and more. Matter of fact, they already have an entire thread (27 pages long, and counting!) devoted to Gravity. (http://forum.nasaspaceflight.com/index.php?topic=28092.0;topicseen)  If you dig into that thread, you'll find some honest-to-god rocket scientists who probably figured out these things (with a slip stick and a whiz wheel)

Or, they're itching to.

[THRAWN]

My cat's name is Mittens....

[UH60guy]

I appreciate all the discussion guys, this is great! Lots of times it seems we forget the "space" part of "Aerospace" Education Granted much of this is way over my head too, but I always enjoy trying to figure some of this out.

It takes x amount of thrust to get to a point x (in how much time)?
Also how far away is point x?

Right- though in this case, NASA did the work for me While the time is largely irrelevant (consider Deep Space 1 probe, thrust is equivalent to that exerted by a piece of paper in your hand on Earth, but over time it added up to a huge change in velocity), it's the accelerations- the push-and-coast if you will- that tell us how to change orbits to get where we're trying to go. For example, a delta-V map of the solar system may show that it takes 4.1 km/s of delta v (acceleration) to reach the moon- that's including the trans-lunar injection burn pointed in one direction, a long coast, and the lunar orbit insertion burn on the back side of the moon included and added together, but reduced to one number (4.1 km/s of velocity change) to just tell us the total fuel required for the mission.

That's why we can measure distance in space in terms of acceleration. Makes my brain hurt, but Journey of Flight chapter 23 explains it a little better. I think in this case, the 3100 m/s includes the apogee increasing burn (initial prograde burn), perigee change burn on the opposite side of the orbit (prograde), and the plane change maneuver to get the orbit from 51 degree inclination to 28 (North/South orientation, perpendicular to the orbit). Then again, maybe the actual rocket scientists found away to combine all into one maneuver, I don't know.

[JeffDG]

An interesting idea to look at when you're looking at Delta-Vs and such is the concept of "Transfer Orbits".

If you look at a circular orbit, say 150 miles (low) and geosync orbit (23,000 miles ish), you need to do a burn such that your perigee is 150 miles, and your apogee is 23,000 miles.  Then when you get to that apogee, you do another burn that pulls your perigee up from 150 miles to 23,000 miles.

Counter-intuitively, (and vastly simplified) a burn that you do at perigee will change your apogee, and conversely, a burn at apogee changes your perigee.  It works the same whether you're transferring from LEO to high orbit, to the moon, or (substitution perhelion and aphelion for periggee and apogee) to Mars or Pluto.

[NIN]

We can plug these numbers in your computer now, or even your phone, and get orbital maneuvers and burn durations that would have taken Apollo engineers down in "The Trench"  a half an hour or more to figure out with a slip stick.

Ain't technology amazing?

[a2capt]

There are still some who won't believe anything until they see it done with a slip stick.

[Panache]

We can plug these numbers in your computer now, or even your phone, and get orbital maneuvers and burn durations that would have taken Apollo engineers down in "The Trench"  a half an hour or more to figure out with a slip stick.

Ain't technology amazing?

Ask me this question again when we land somebody on Mars.  Or even the moon again.

[BHartman007]

I'm a bit late commenting on this, so I don't know if you're still looking for info or not, but what the heck.

The AE officer in my squadron is also the current Deputy Director of the Astronaut office (http://en.wikipedia.org/wiki/Eric_Boe). I was talking with him a while ago about that movie, and he said it's just not possible to go from Hubble to the ISS due to the different orbital planes. He said the orbital plane is fixed at launch, and has to do with the latitude from which the object was launched, and due to the physics of it, can't be changed later. Certainly not with a jetpack. I'm sure he would provide more info if I were to ask him.

He did say, though, that they did a very good job on portraying the "feel" and look of being in space.