Author Topic: Basic Rocket Science Q & A  (Read 277934 times)

Offline Antares

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Re: Basic Rocket Science Q & A
« Reply #700 on: 02/10/2012 04:14 PM »
Agree with combustion stability and weight items.  Also a way to get some economy of scale in producing more of the same thing.
If I like something on NSF, it's probably because I know it to be accurate.  Every once in a while, it's just something I agree with.  Facts generally receive the former.

Offline Lee Jay

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Re: Basic Rocket Science Q & A
« Reply #701 on: 02/10/2012 09:02 PM »
Agree with combustion stability and weight items.  Also a way to get some economy of scale in producing more of the same thing.

Okay, much appreciated, all three of you!

Offline e of pi

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Re: Basic Rocket Science Q & A
« Reply #702 on: 02/10/2012 11:28 PM »
Is it possible to inject into a polar lunar orbit directly from Earth, as opposed to entering an equatorial LLO, then plane-changing into polar, and what would the associated delta-v be compared to going to an equatorial orbit? Additionally, if such an injection is possible, is it still able to offer a free-return trajectory? Additionally, in terms of offering total-surface access, would this be any better than staging from L1/L2, given the lower delta-v required of a descent stage going directly down from orbit instead of getting there from a Lagrange point? Thanks for any insight into this.

Offline Antares

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Re: Basic Rocket Science Q & A
« Reply #703 on: 02/11/2012 12:37 AM »
1st question: yes, look at Lunar Prospector and LRO.
If I like something on NSF, it's probably because I know it to be accurate.  Every once in a while, it's just something I agree with.  Facts generally receive the former.

Offline Proponent

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Re: Basic Rocket Science Q & A
« Reply #704 on: 02/11/2012 07:28 AM »
Getting into lunar polar orbit from earth is hardly any more difficult than getting into lunar equatorial orbit.

The moon is moving around the earth at about 1 km/s.  When the spacecraft arrives in the vicinity of the moon, it is near apogee of a very highly elliptical orbit and is moving very slowly with respect to the earth.  Looking at it from an earth-centered perspective, the spacecraft is nearly motionless and is swept up by the moon.

Entering polar orbit rather than equatorial orbit is principally a matter of arriving in the moon's vicinity above or below the plane of the moon's orbit by somewhat more than a lunar radius.  On arrival, the spacecraft will go into an orbit around the moon such that the earth will initially be in view at all times -- it won't loop behind the moon.  Thus, I don't think a free-return trajectory is possible, at least not without following a much higher-delta-V trajectory to the moon in the first place.

The delta-V requirements for getting to polar orbit directly from earth are basically similar to those for reaching equatorial orbit.  However, the orientation with respect to earth of a polar orbit, unlike that of an equatorial one, will change as the moon revolves around the earth.  If abort-anytime capability is desired, then a large delta-V reserve will be needed for aborts when the orbital orientation is unfavorable.

Offline nyrath

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Re: Basic Rocket Science Q & A
« Reply #705 on: 02/13/2012 08:03 PM »
Hi! This is a stupid newbie with a stupid newbie question, so feel free to throw popcorn at me.

I have a question about optimal acceleration profiles for a rocket lifting off from Earth's surface into low Earth orbit.

Say the rocket has a thrust-to-weight ratio way above 1 and an unreasonably high specific impulse of 20,000 seconds or so (NSWR, don't ask) so it is a single stage to orbit rocket. The engine can be throttled so the acceleration can be varied as the rocket loses mass.

Given that it is undesirable for the acceleration to rise above 4g or so in order to prevent damage to the astronauts.

Also given that every second spent in transit to orbit imposes a "gravity tax" of a 9.81 m/s reduction in speed so it is desirable to minimize the transit duration.

So the question is: with these givens, what sort of acceleration profile over the transit duration would be optimal?

Offline strangequark

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Re: Basic Rocket Science Q & A
« Reply #706 on: 02/13/2012 09:55 PM »
Hi! This is a stupid newbie with a stupid newbie question, so feel free to throw popcorn at me.

I have a question about optimal acceleration profiles for a rocket lifting off from Earth's surface into low Earth orbit.

Say the rocket has a thrust-to-weight ratio way above 1 and an unreasonably high specific impulse of 20,000 seconds or so (NSWR, don't ask) so it is a single stage to orbit rocket. The engine can be throttled so the acceleration can be varied as the rocket loses mass.

Given that it is undesirable for the acceleration to rise above 4g or so in order to prevent damage to the astronauts.

Also given that every second spent in transit to orbit imposes a "gravity tax" of a 9.81 m/s reduction in speed so it is desirable to minimize the transit duration.

So the question is: with these givens, what sort of acceleration profile over the transit duration would be optimal?

Well, for the very unreasonable case where you have an infinitely throttleable rocket, and a 4 g upper limit, you would do something like what’s in the attached. Basically, you keep your accel as high as you can, while staying under a maximum dynamic pressure limit (0.5*air density*velocity^2). Once you’re through the thick part of the atmosphere, density drops off, and you can punch it again. I chose 800 psf for the attached graph, because it is a reasonable upper limit on maximum dynamic pressure (or “Max Q”).
« Last Edit: 02/13/2012 09:57 PM by strangequark »
Don't flippantly discount the old rules of this industry. Behind each one lies a painful lesson learned from broken, twisted hardware. Learn those lessons, and respect the knowledge gained from them. Only then, see if you can write new rules that will meet those challenges.

Offline nyrath

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Re: Basic Rocket Science Q & A
« Reply #707 on: 02/14/2012 12:13 AM »
Thank you very much!

Offline IsaacKuo

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Re: Basic Rocket Science Q & A
« Reply #708 on: 02/14/2012 12:24 AM »
Also given that every second spent in transit to orbit imposes a "gravity tax" of a 9.81 m/s reduction in speed so it is desirable to minimize the transit duration.
This is not true.  The "gravity tax" is actually much less than that, depending on your thrust angle.

For example, if your acceleration is 4 gees, and you want horizontal acceleration, you only need to tilt your thrust upward by ~15 degrees in order to counteract a downward acceleration of 1 gee.  Your horizontal thrust is reduced, but only down to 3.87 gees.  In other words, the gravity drag is only 0.13 gees.

Offline nyrath

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Re: Basic Rocket Science Q & A
« Reply #709 on: 02/14/2012 02:04 PM »
Thank you, I was wondering about that.

Offline baldusi

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Re: Basic Rocket Science Q & A
« Reply #710 on: 02/14/2012 02:16 PM »
Isn't the gravity loss the integrations of g(distance to Earth).sin(gravity normal)? You'd have two components, one is the angle towards the center of the Earth, and the other would be the distance to the point source, right?

Offline Danny Dot

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Re: Basic Rocket Science Q & A
« Reply #711 on: 02/14/2012 02:47 PM »
There is no reason to limit to 4 gs.  NASA Standard 3000 allows for much higher g levels.
Danny Deger

Offline joertexas

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Re: Basic Rocket Science Q & A
« Reply #712 on: 02/24/2012 05:56 PM »
How is C3 calculated? I understand Dv, but C3 is beyond me..

Thanks,

JR

Offline deltaV

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Re: Basic Rocket Science Q & A
« Reply #713 on: 02/24/2012 06:32 PM »
How is C3 calculated? I understand Dv, but C3 is beyond me..

Thanks,

JR

Does http://en.wikipedia.org/wiki/Specific_orbital_energy help?

Offline IsaacKuo

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Re: Basic Rocket Science Q & A
« Reply #714 on: 02/24/2012 06:47 PM »
How is C3 calculated? I understand Dv, but C3 is beyond me..

One way to calculate it is:

v^2 - v_escape^2

Where v is your current velocity and v_escape is the relevant escape velocity.  Note that v_escape depends on your current altitude.

The utility of this value is that it remains constant if you don't perform any more thrusts, as you escape the local gravity well.  It tells you what your eventual velocity will slow down to as you escape, by taking the square root.

Does http://en.wikipedia.org/wiki/Specific_orbital_energy help?

It's important to realize that C3 is actually TWICE the specific orbital energy, since it is missing a factor of 1/2.

http://en.wikipedia.org/wiki/Characteristic_energy

Offline MP99

Re: Basic Rocket Science Q & A
« Reply #715 on: 02/24/2012 07:44 PM »
How is C3 calculated? I understand Dv, but C3 is beyond me..

I got a few answers re C3 here (and subsequent pages):-

http://forum.nasaspaceflight.com/index.php?topic=13543.msg521874#msg521874

cheers, Martin

Online DarkenedOne

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Re: Basic Rocket Science Q & A
« Reply #716 on: 02/25/2012 05:16 PM »
Hey this is just a quick question.

Why do rockets not use the same expanding nozzle technology used in fighter jet aircraft?

Offline Jim

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Re: Basic Rocket Science Q & A
« Reply #717 on: 02/25/2012 05:51 PM »
Hey this is just a quick question.

Why do rockets not use the same expanding nozzle technology used in fighter jet aircraft?

Exhaust is hotter and at higher pressure

Offline joertexas

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Re: Basic Rocket Science Q & A
« Reply #718 on: 02/27/2012 06:54 PM »
How is C3 calculated? I understand Dv, but C3 is beyond me..

I got a few answers re C3 here (and subsequent pages):-

http://forum.nasaspaceflight.com/index.php?topic=13543.msg521874#msg521874

cheers, Martin

My math skills just aren't up to this..

I needed to calculate what the approximate C3 would be for a translunar injection from LEO, say, 180km.

Thanks for the responses, though. I am trying to make sense of them.

JR

Offline strangequark

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Re: Basic Rocket Science Q & A
« Reply #719 on: 02/27/2012 07:08 PM »
Hey this is just a quick question.

Why do rockets not use the same expanding nozzle technology used in fighter jet aircraft?

Exhaust is hotter and at higher pressure

Jet nozzles are strictly converging as well (the throat is the exit), and the geometry of a converging nozzle doesn’t really matter to the flow. You have much more leeway because you won’t trigger shocks, and you can have unsmooth surfaces. Even neglecting the smoothness, if you tried to do that with a converging diverging rocket nozzle, you would probably be restricted to a conical nozzle (otherwise you would have to adjust the entire contour of the nozzle). With that, you either suffer huge cosine losses at high expansion ratio (which is the opposite of what you want), or your nozzle is obscenely long at low expansion ratio, which also makes it impractical.
Don't flippantly discount the old rules of this industry. Behind each one lies a painful lesson learned from broken, twisted hardware. Learn those lessons, and respect the knowledge gained from them. Only then, see if you can write new rules that will meet those challenges.