### Author Topic: Basic Rocket Science Q & A  (Read 346866 times)

#### ClaytonBirchenough

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##### Re: Basic Rocket Science Q & A
« Reply #820 on: 07/14/2013 02:32 PM »
Ok.
Now this would be quite light.
It's not pop corn which is green. Or even if it was green and wet, the cob would dry out [freeze dry] in space.
So, it's low density.

I would say impact on surface Earth would be somewhere around baseball pitcher's fast ball [100 mph].
But Mars would much faster.

So can I assume from your comments that the pop corn on cob, survives re-entry?
Or are saying *even if it somehow were to survived re-entry* it would hit the ground at very high velocity [both Earth and Mars]?

I'm starting to get confused...

IMO popcorn is a kernel that has popped, a kernel is a pod on a corn on the cob, and a corn on the cob is many kernels on the cob.

This is off topic and should not be here. If you want, you can open your own thread about popcorn reentry different atmospheres...
Clayton Birchenough
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#### ClaytonBirchenough

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##### Re: Basic Rocket Science Q & A
« Reply #821 on: 07/29/2013 07:13 PM »
Alright I got a few questions... (Keep in mind, I'm not a rocket scientist  )

How do you convert from km2s2 into km/s?

Also, how do you calculate the "slowing down" of a probe as it exits the solar system? (Like how the Voyager's velocity are slowing down relative to the Sun)

Thanks!
Clayton Birchenough
Astro. Engineer and Computational Mathematics @ ERAU

#### baldusi

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##### Re: Basic Rocket Science Q & A
« Reply #822 on: 07/29/2013 07:51 PM »
You can't, those are different concepts. Km²/s² is potential energy (and it's wrt a certain body). km/s is an impulse concept. Plus, you might need a lot of km/s to stay at the same km²/s². Say that you want to change your orbital plane, keeping everything else equal. You need km/s but km²/s² is the same.
As someone explained on a thread, you could thing of an orbit like a skater on a halfpipe. The higher that he gets the more km²/s² he has. But if he wants to go higher, lower or simply sideways, he'll need some impulse (km/s).

#### Robotbeat

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##### Re: Basic Rocket Science Q & A
« Reply #823 on: 07/29/2013 11:18 PM »
Sure you can convert it. Just take the square root of the C3 to get the asymptotic velocity (i.e. the velocity it would have at infinity, i.e. the excess above escape velocity). Take the square root of the whole quantity including the units to make sure it gives you the right units.
Chris  Whoever loves correction loves knowledge, but he who hates reproof is stupid.

To the maximum extent practicable, the Federal Government shall plan missions to accommodate the space transportation services capabilities of United States commercial providers. US law http://goo.gl/YZYNt0

#### R7

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##### Re: Basic Rocket Science Q & A
« Reply #824 on: 07/30/2013 06:20 AM »
What Robotbeat said. But make sure the value you got is C3, the same unit is also used for specific orbital energy (of which C3 is a special case) in which case you first need to multiply the value by two before taking the square root. Wiki explains these quite well;

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

(but buy the books )

#### ClaytonBirchenough

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##### Re: Basic Rocket Science Q & A
« Reply #825 on: 07/30/2013 03:40 PM »
(but buy the books )

What books?

So what I've been trying to figure out is, if a Delta IV Heavy has a C3 performance of 60 km²s−2: 2,521 kg, that would make the velocity relative to Earth...

60 * 2 = 120

Square root of 120 km2s2 = 10.95 km/s?

Correct?
« Last Edit: 07/30/2013 03:40 PM by ClaytonBirchenough »
Clayton Birchenough
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#### Robotbeat

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##### Re: Basic Rocket Science Q & A
« Reply #826 on: 07/30/2013 03:41 PM »
No, read carefully! Take the square root of the C3. sqrt(60km^2/s^2)=7.75km/s
Chris  Whoever loves correction loves knowledge, but he who hates reproof is stupid.

To the maximum extent practicable, the Federal Government shall plan missions to accommodate the space transportation services capabilities of United States commercial providers. US law http://goo.gl/YZYNt0

#### ClaytonBirchenough

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##### Re: Basic Rocket Science Q & A
« Reply #827 on: 07/30/2013 03:49 PM »
No, read carefully! Take the square root of the C3. sqrt(60km^2/s^2)=7.75km/s

Ah, I multiplied by 2 thinking it was specific orbital energy, but it wasn't.

So if the object is escaping Earth at 7.75 km/s, the object would have a relative velocity to the Sun of 26.75 km/s?
Clayton Birchenough
Astro. Engineer and Computational Mathematics @ ERAU

#### R7

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##### Re: Basic Rocket Science Q & A
« Reply #828 on: 07/30/2013 07:45 PM »
What books?

The one's recommended to you in the book thread.

So if the object is escaping Earth at 7.75 km/s, the object would have a relative velocity to the Sun of 26.75 km/s?

Depends on to what direction the object escapes.

#### MP99

##### Re: Basic Rocket Science Q & A
« Reply #829 on: 07/30/2013 08:35 PM »
What Robotbeat said. But make sure the value you got is C3, the same unit is also used for specific orbital energy (of which C3 is a special case) in which case you first need to multiply the value by two before taking the square root. Wiki explains these quite well;

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

(but buy the books )

I'd guess x2 applies for a circular orbit?

Cheers, Martin

#### R7

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##### Re: Basic Rocket Science Q & A
« Reply #830 on: 07/31/2013 09:42 AM »
Circular orbit simplifies calculations a lot.

r = radius of circular orbit
μ = gravitational parameter of the celestial body you are orbiting

specific orbital energy e = -μ/2*r
characteristic energy C3 = 2*e = -μ/r
orbital velocity v = sqrt(C3) = sqrt(μ/r)

the minuses may seem confusing but it's because potential energy at infinite distance is agreed to be zero to simplify rest of the math. The result is that objects on circular/elliptical orbits always have negative specific orbital energy (thus C3 too).

#### MP99

##### Re: Basic Rocket Science Q & A
« Reply #831 on: 07/31/2013 09:40 PM »
Thanks.

I believe I was confusing specific orbital energy with it's gravitational potential energy component.

cheers, Martin

#### Robotbeat

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##### Re: Basic Rocket Science Q & A
« Reply #832 on: 08/01/2013 02:42 AM »
No, read carefully! Take the square root of the C3. sqrt(60km^2/s^2)=7.75km/s

Ah, I multiplied by 2 thinking it was specific orbital energy, but it wasn't.
easy mistake, but that's why we use ye ole fudge factor, or fixer coefficient.
Chris  Whoever loves correction loves knowledge, but he who hates reproof is stupid.

To the maximum extent practicable, the Federal Government shall plan missions to accommodate the space transportation services capabilities of United States commercial providers. US law http://goo.gl/YZYNt0

#### Hershey

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##### Re: Basic Rocket Science Q & A
« Reply #833 on: 09/17/2013 01:18 AM »
Whenever I read about bimodal NTR's, the brayton power cycle always uses a helium-xenon working fluid. I understand He, but why is Xe used?

#### R7

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##### Re: Basic Rocket Science Q & A
« Reply #834 on: 09/18/2013 06:56 AM »
I understand He, but why is Xe used?

http://proceedings.aip.org/resource/2/apcpcs/880/1/559_1?isAuthorized=no

Quote
The selected configuration for the Project Prometheus Space Nuclear Power Plant was a direct coupling of Brayton energy conversion loop(s) to a single reactor heat source through the gas coolant/working fluid. A mixture of helium (He) and xenon (Xe) gas was assumed as the coolant/working fluid. Helium has superior thermal conductivity while xenon is added to increase the gas atomic weight to benefit turbomachinery design.

Heavier gas = smaller turbines. Hmm, if turbines were replaced with piston driven generator there would be no need for xenon?

#### M129K

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##### Re: Basic Rocket Science Q & A
« Reply #835 on: 09/29/2013 01:45 PM »
I've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.
How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.

#### baldusi

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##### Re: Basic Rocket Science Q & A
« Reply #836 on: 09/29/2013 06:06 PM »
I've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.
How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.
You mean gravity losses? That's the integral of  g(h(t)).sin(rho(t)), where rho is the angle to the normal of the gravity source (or cos() if you take to the tangent to the Earth's surface) and h it the distance to the gravity source. You can assume that in Earth's case, at 6,371km distance the gravity is 9.8m/s.

#### M129K

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##### Re: Basic Rocket Science Q & A
« Reply #837 on: 09/29/2013 06:58 PM »
I've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.
How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.
You mean gravity losses? That's the integral of  g(h(t)).sin(rho(t)), where rho is the angle to the normal of the gravity source (or cos() if you take to the tangent to the Earth's surface) and h it the distance to the gravity source. You can assume that in Earth's case, at 6,371km distance the gravity is 9.8m/s.
If I understand it correctly, t is total time? If so, thank you! Seems very accurate.
« Last Edit: 09/29/2013 07:00 PM by M129K »

#### baldusi

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##### Re: Basic Rocket Science Q & A
« Reply #838 on: 09/30/2013 01:05 AM »
I've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.
How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.
You mean gravity losses? That's the integral of  g(h(t)).sin(rho(t)), where rho is the angle to the normal of the gravity source (or cos() if you take to the tangent to the Earth's surface) and h it the distance to the gravity source. You can assume that in Earth's case, at 6,371km distance the gravity is 9.8m/s.
If I understand it correctly, t is total time? If so, thank you! Seems very accurate.
Nope, h(t) and rho(t) are functions of t. Thus, you have to integrate from t=0 to T  the function over dt.