Author Topic: Earth without atmosphere? (Read 17922 times)

jabe · « **on:** 09/24/2007 10:00 pm »

As a high school physics teacher I always tell the kids that friction makes the question "harder" which is why the questions always say...ignoring friction do the following...so..if the Earth didn't have an atmosphere how much easier would it be to get into orbit?
We will be getting into a discussion soon on putting air resistance into the question and how it could affect the answers...... so curious to see what the difference would be...
thoughts??
cheers
jb

khallow · « **Reply #1 on:** 09/24/2007 10:40 pm »

Depends how you do it. Rockets in Earth's atmosphere have two forces acting on them, gravity and air resistance. Accelerate too slowly and you spend too much of your propellant just overcoming gravity. Accelerate too fast while in atmosphere and you lose too much momentum to air resistance which at supersonic speeds scales portional as the cube of the rockets velocity and density of the atmosphere. It also depends on the drag of the vehicle. My impression is that optimal losses (with the right acceleration profile) are equivalent to around 1.5-2 km/s of extra delta v.

Some approaches (for example, high altitude plane or balloon launch) reduce this delta v by launching above most of the atmosphere. Space elevators move at a constant, subsonic speed (googling around, I've seen 120 MPH suggested as likely for a slow, prototype elevator). So air resistance losses would be considerable less than for a rocket. Rockets aren't a good model for a space elevator. No reaction mass is used and the momentum changes of the rising cargo are transfered to the cable and then to either the Earth or the counterweight on the other end.

jabe · « **Reply #2 on:** 09/24/2007 10:57 pm »

I'm thinking of it as a "typical" rocket launch. Maybe i could rephrase and ask..how much would extra mass could you lift from a given launch vehicle if no atmosphere was present.. just ball park figures would be great...
i.e. Falcon 1 can do ~750 kg to orbit if my quick search was right.. neglecting modifying engines due to no atmosphere at low altitude etc..how much more could it get to orbit energy wise since no energy will be lost due to friction?

AntiKev · « **Reply #3 on:** 09/25/2007 07:29 am »

Personally I would introduce this topic by changing your frame of reference a little bit. Start with stationary objects with a certain wind speed (e.g.: wind blowing around a flagpole). It will be easier for a high school kid (especially one not that interested in physics) to visualize. And then move up to moving objects. Remember if you want to do this intuitively it would be much easier to visualize something like a car or train (simplify the shape of course to a block or cylinder) because then you don't have to include anything about the pressure/density effects as you move higher in the atmosphere. If you have really advanced students then you can move on to the rocket launch scenario. Be warned, I know 3rd year engineering students (for that matter engineering graduates) that have a very difficult time grasping these concepts. But if you really want to know the math behind it, look for an entry-level fluid dynamics textbook.

On another note, my copy of Space Propulsion Analysis and Design gives the following delta-v losses for various rockets (I'm only going to give the gravity and air resistance losses for comparison):

Ariane A-44L: Gravity Loss: 1576 m/s Drag Loss: 135 m/s
Atlas I: Gravity Loss: 1395 m/s Drag Loss: 110 m/s
Delta 7925: Gravity Loss: 1150 m/s Drag Loss: 136 m/s
Shuttle: Gravity Loss: 1222 m/s Drag Loss: 107 m/s
Saturn V: Gravity Loss: 1534 m/s Drag Loss: 40 m/s (!!)
Titan IV/Centaur: Gravity Loss: 1442 m/s Drag Loss: 156 m/s

Now it's plain to see that gravity losses are an order of magnitude higher than drag losses, that is more a trajectory design issue than anything else. Both drag and gravity losses can be mitigated by changing trajectory (gravity losses are based upon the amount of time thrusting directly against gravity rather than orthogonal to it, drag is based on angle of attack). I know this is advance for high school physics, and I've highly simplified the idea. But you have to know the can of worms you're opening by including air resistance in these questions

.

meiza · « **Reply #4 on:** 09/25/2007 11:23 am »

...and there's the thing that atmosphere messes up with the nozzle flow so you get less ISP.
IE the nozzle is the device that accelerates the combustion products. In a one dimensional simplification, at the same time the temperature and pressure of the combustion products goes down. If the pressure goes much lower than the atmospheric pressure, the flow separates from the nozzle wall and that shakes it bad.
So if there is less atmospheric pressure you can have a bigger nozzle that accelerates the combustion products even more and you get higher exhaust velocity and thrust and ISP.
Atmospheric pressure also affects ISP and thrust more directly, but that's more complicated. Anyway, for both reasons, rockets work better in a vacuum. For a sea level rocket, ISP optimum is when nozzle exhaust is at ambient pressure. In space, it's an infinite size, which is of course not practical. You also hit diminishing returns.

Those are also the reasons why the DC-X used RL-10 engines with a truncated nozzle. And why the Merlin nozzle will have an extension when it's used for upper stage flight.

renclod · « **Reply #5 on:** 09/25/2007 03:31 pm »

If you input the AntiKev's drag loss examples (say, 100 to 150 m/s) into the rocket equation, with engines of 300 to 450 Isp, you get something like this:
the shorter burn of the last (or only !) stage of the rocket will save 2...5% of the mass.
(100...to...150) m/s = (300...to...450) s x 9.81 m/s/s/ x ln (mass_fraction).

One thing: do not let yourself surprised off guard... the absence of the atmosphere on a planet like ours is of much, much greater importance with regard to... descending (re-enter) FROM orbit.

Absent the atmospheric friction, the prefered method of losing energy, a spacecraft would need to do a powered descent. Result: only less than 0.1 of the mass in orbit will touch down.

khallow · « **Reply #6 on:** 09/25/2007 05:16 pm »

Ouch, I was deeply in error there. AntiKev, thanks for the correction.

hmh33 · « **Reply #7 on:** 09/25/2007 07:44 pm »

Drag losses are much more significant for small vehicles (which is part of the reason why the Saturn V has only 40m/s listed). For small suborbital rockets, they can be a substantial proportion of the total mission delta V. I'm working on a project to launch a rocket from a helium balloon at 30km altitude, with the design goal of reaching 100km. The same rocket launched from the ground would only reach about 3km.

jabe · « **Reply #8 on:** 09/25/2007 07:56 pm »

Great input guys!!
I personally thought that air resistance was a much larger delta V loss...
and only 40 m/s for saturn??WOW!!! whats with that low number

We do a short but neat lab for terminal velocity by timing various bundled coffee filters falling and plot the air resistance versus terminal velocity and plot on log paper. The relationship usually works out to squared if they take care with the recording of data.
We usually talk about air density etc..Being a space "nut" we usually talk about Max Q..why there are airstart solids and not all fired at once the surface (I think i know why they do that

)
I talk about ISP, nozzle design etc..but never really knew what the loss was due to air resistance.
The whole landing issue I talk about as well..but never realized until recently that since mars atmosphere is so thin that landing masses over 2 tonnes is a real pain. It would be better to have no atmosphere at all! I haven't checked if has been talked about here or not, but article is here

keep input coming..
cheers
jb

edit: thanks hmh33 for clarification for the saturn and rocket size issue...

meiza · « **Reply #9 on:** 09/26/2007 11:45 am »

Saturn V had a very slow liftoff, meaning it was higher up when it started going at high speeds.
You can see it's gravity losses are very high too.
I'd also hunch the shape and size, leading to small frontal area compared to mass reduced the drag losses calculated in delta vee. If you calculate the drag loss energy in Joules, you'd end up with a huge number since the launcher was so huge. So smaller rockets have bigger drag losses (proportionally, ie in delta vee) because of that. That's also why small launchers tend to be long and thin and bigger launchers can be stocky. In a sense, one ballistic number is mass per frontal area, meaning rockets should only grow in width to keep the ballistic number the same.

That's also the reason why a golf ball decays from orbit faster than ISS, although the ISS looks thin and spindly - the golf ball still has much less mass per frontal area - or depth. It's very unintuitive, but results from th efact that when keeping the shape and density similar, drag scales with the square of size while mass scales with the cube of size, meaning bigger things have proportionally less drag.

hmh33 · « **Reply #10 on:** 09/28/2007 05:49 pm »

"Absent the atmospheric friction, the prefered method of losing energy, a spacecraft would need to do a powered descent. Result: only less than 0.1 of the mass in orbit will touch down."

There is always lithobraking...