Author Topic: Basic Rocket Science Q & A  (Read 271329 times)

Offline gbaikie

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Re: Basic Rocket Science Q & A
« Reply #780 on: 05/21/2013 12:54 PM »
From Mars distance at average Mars orbital velocity of 24.1 km and doing standard Hohmann transfer to average Mercury distance [with average velocity of 47.9 km/sec] how much delta-v is needed. And how long does take to get to Mercury distance. And how long does it take to cross Earth's average orbital distance?

And is it true, that if instead you started from Mercury distance and did a standard Hohmann transfer to Mars, it takes the same amount of delta-v.
Or if wanted to circularize orbit at either Mercury or Mars distance is also same amount of delta-v?

For time of travel from Mars to Mercury can just add the length of each of their years and divide by 2. And divide by 2 again for duration of each leg?

Offline deltaV

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Re: Basic Rocket Science Q & A
« Reply #781 on: 05/21/2013 04:41 PM »

Offline IsaacKuo

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Re: Basic Rocket Science Q & A
« Reply #782 on: 05/21/2013 05:15 PM »
Also, it's not so simple because of the Oberth effect.  It takes much less delta-v to go between an interplanetary transfer orbit and orbit around a planet than it does to go between that transfer orbit and a point in open space.

Think of a planetary orbit as zooming around a racetrack.  It doesn't take as much delta-v to slow down to it or speed up from it because there's already some velocity just going around and around.  The true orbital mechanics are a bit more complicated than that, but it should give you an intuitive idea.

Offline gbaikie

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Re: Basic Rocket Science Q & A
« Reply #783 on: 05/21/2013 11:27 PM »
Also, it's not so simple because of the Oberth effect.  It takes much less delta-v to go between an interplanetary transfer orbit and orbit around a planet than it does to go between that transfer orbit and a point in open space.
It seems that the Oberth effect does complicate it. When one gets nearer to the sun one get more Oberth effect.

Mars and Mercury are similar planetary masses. The Oberth effect depends upon velocity. Because at Mercury distance one has a higher velocity, one gets less Oberth effect when one "involves" the planet Mercury.
Or because one has lower orbital velocity at Mars, by using the planet Mars one get more significant difference in the Oberth effect.

Quote
Think of a planetary orbit as zooming around a racetrack.  It doesn't take as much delta-v to slow down to it or speed up from it because there's already some velocity just going around and around.  The true orbital mechanics are a bit more complicated than that, but it should give you an intuitive idea.

My question is related to a fast trajectory from Earth to Mars. If you have a highly elliptical orbit, such as mercury distance at perihelion and Mars distance at aphelion. From Earth distance to Mars is fairly short travel time in terms distance and velocity. And you arrive at Mars distance at low velocity difference- between Mars distance orbital speed and the aphelion velocity of the highly elliptical orbit. So fast transit and least braking at Mars. And could have some kind of "free return" back to Earth.

And for a faster transit, one has the elliptical orbit have the perihelion closer to the sun than Mercury, and has slightly less total energy orbital energy if the aphelion of orbit is slightly less than Mars distance- but more importantly less travel distance in the Earth to Mars earth segment of trajectory [though requires much more delta-v to match this trajectory starting from Earth].
But generally it seems the more important variable is how close the perihelion of the elliptical orbit is to the Sun.
In terms of Mars cycler, perhaps the orbital period would a Earth year, some resonant orbit, or some precession orbit.
« Last Edit: 05/22/2013 01:00 PM by gbaikie »

Offline baldusi

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Re: Basic Rocket Science Q & A
« Reply #784 on: 05/22/2013 03:38 AM »
Remember plane change costs. Mercury's orbit is the most inclined of the planets.

Offline gbaikie

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Re: Basic Rocket Science Q & A
« Reply #785 on: 05/22/2013 06:39 AM »
Remember plane change costs. Mercury's orbit is the most inclined of the planets.

Hmm. That's kind of neat. It's an added element in terms of getting to Mercury.
But also seems like it gives more options in terms destination, which may be at inclination [a lot space rocks]. You can leave [or arrive] at Mercury at a zero inclination, but you leave [or arrive] at a 7 inclination to solar planetary disk.

Offline torzek

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Re: Basic Rocket Science Q & A
« Reply #786 on: 05/22/2013 11:20 AM »
I'm a student of Aerospace engineering currently working on my thesis.
Assumption is to put rocket of certain mass and payload into 600 km LEO . how to determine if 2 or 3 stages should be used? Is Tsiolkovsky equation enough? How to calculated velocities of each stage and total velocity taking into consideration earth rotation.

I would greatly appreciate your help as well as proper book suggestion.

Offline jabe

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Re: Basic Rocket Science Q & A
« Reply #787 on: 05/22/2013 12:06 PM »
I'm a student of Aerospace engineering currently working on my thesis.
Assumption is to put rocket of certain mass and payload into 600 km LEO . how to determine if 2 or 3 stages should be used? Is Tsiolkovsky equation enough? How to calculated velocities of each stage and total velocity taking into consideration earth rotation.

I would greatly appreciate your help as well as proper book suggestion.

wow..now that is an interesting question.. not so "basic" though. :)
Use whatever they have taught you in class and apply it... ;)
my newbie thoughts..pick a fuel, engine and play with numbers...Elon did that and got the Falcon series of rockets..
my 2 cents
jb

Offline ClaytonBirchenough

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Re: Basic Rocket Science Q & A
« Reply #788 on: 05/22/2013 12:37 PM »
I'm a student of Aerospace engineering currently working on my thesis.
Assumption is to put rocket of certain mass and payload into 600 km LEO . how to determine if 2 or 3 stages should be used? Is Tsiolkovsky equation enough? How to calculated velocities of each stage and total velocity taking into consideration earth rotation.

I would greatly appreciate your help as well as proper book suggestion.


Three stages will always have performance benefits over two stage rockets. However, cost and risk goes up with more stages. Check out the document below for optimal mass ratios for each stage.

Welcome to the forum!
« Last Edit: 05/22/2013 12:39 PM by ClaytonBirchenough »
Clayton Birchenough
Astro. Engineer and Computational Mathematics @ ERAU

Offline deltaV

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Re: Basic Rocket Science Q & A
« Reply #789 on: 05/22/2013 07:36 PM »
Remember plane change costs. Mercury's orbit is the most inclined of the planets.

NASA's GMAT software looks like it can be used to answer the question with high fidelity, but it isn't particularly easy to use.

Offline torzek

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Re: Basic Rocket Science Q & A
« Reply #790 on: 05/23/2013 05:41 PM »
thanks for these suggestions,  but still strugglingto find the answer...

Offline Remes

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Re: Basic Rocket Science Q & A
« Reply #791 on: 05/23/2013 09:53 PM »
The good news is, you are doing a 2D simulation, so no need for the Z-components.  ;)

In a first step you could ignore the orientation of the spacecraft. Simply treat the spacecraft as a mass point and apply the forces as you need them.

Typically vectors are used to do that sort of calculation.
https://en.wikipedia.org/wiki/Vector_%28mathematics_and_physics%29

You would have a vector for Earth and Mars position. They move on a elipsis independently. The spacecraft has a position and velocity vector vector.

Calculate the distance between the spacecraft and the earth/mars/sun (vector subtraction, then pythagoras) in order to calculate the gravitation forces. Calculate the forces you apply by means of rockets. Based on the sum of forces and the spacecraft mass calculate acceleration, from acceleration velocity, from velocity the new position. That would be one timestep. This procedure is repeated as long as you don't enter the perfect orbit or burn up in the mars atmosphere.

If you need an angle between two vectors you use the scalar vector product. If you need to rotate a vector, use the rotation matrix (https://en.wikipedia.org/wiki/Rotation_matrix).

Hope this helps.

Offline gbaikie

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Re: Basic Rocket Science Q & A
« Reply #792 on: 05/23/2013 09:54 PM »
thanks for these suggestions,  but still strugglingto find the answer...

Well, a one stage rocket could deliver a payload to 600 km orbit.

In terms of costs, a 2 stage rocket could be the cheapest and most reliable- and generally costs are the most important factor, in terms
of engineering anything. But if it's a larger payload, then one might consider 3 stages.
A two stage could considered the minimal because, rockets perform
differently at sea level pressure as compared to vacuum of space- so first stage can have rocket nozzle optimized for in atmosphere, and second stage made to function best in vacuum of space environment.
Also one needs a lot of thrust at the beginning of launch up to the space environment- and higher energy density per volume, such as kerosene and LOX is commonly used in first stage. And one can also use kerosene
with the second stage [see SpaceX]. One could also use solid rocket fuel
in first stage- in some aspect solids are better than Kerosene or one could use strap on solid rockets.
Generally, most efficient chemical rocket as second stage or in space environment is liquid hydrogen & LOX. If you had an existing rocket that used kerosense in both stages, you could probably increase the amount payload delivered to orbit by switching to LH&LOX for second stage.
Solids rockets can used in upper stages, though they not as efficient but are easily stored and are reliable- and many are available "off the shelf".
Etc.
« Last Edit: 05/23/2013 10:04 PM by gbaikie »

Offline clongton

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Re: Basic Rocket Science Q & A
« Reply #793 on: 05/23/2013 11:23 PM »
... rockets perform
differently at sea level pressure as compared to vacuum of space- so first stage can have rocket nozzle optimized for in atmosphere, and second stage made to function best in vacuum of space environment.

Only because of the nozzle design. Bell nozzles must be optimize for a specific barometric pressure. Aerospikes otoh are self regulating, dynamically changing their characteristics to match the changing air pressure as the vehicle ascends.
Chuck - DIRECT co-founder
I started my career on the Saturn-V F-1A engine

Offline gbaikie

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Re: Basic Rocket Science Q & A
« Reply #794 on: 05/24/2013 10:15 AM »
Quote
NASA's GMAT software looks like it can be used to answer the question with high fidelity, but it isn't particularly easy to use.

I think patience might be useful with GMAT, but I lack it.
This is simplier graphic in terms of a broad idea:
http://www.jgiesen.de/hohmann/index.html

So, using a hohmann transfer the planet which takes the shortest period of time to travel to from Earth is Mercury.
Next is Venus and then finally Mars.
Earth to Mercury: 105.5 days
Venus: 146 days
Mars: 259 days

If you are on Mercury, then shortest travel time is Venus, Earth, then Mars.
And from Mercury or Venus it faster to get to Mars than from Earth [Mercury has shortest travel time to Mars].

But if from Earth one could match the Hohmann trajectory going from Mercury to Mars [or Mars to Mercury] then you get to Mars quicker.
So if Mercury to Mars hohmann transfer was passing near Earth [say at distance 1 million km, so Earth gravity didn't affect the trajectory [by much] and launch rocket from Earth to catch up and dock with it. Then from Earth you get to Mars faster than from Mercury. And such a trajectory from Earth would not be a Hohmann trajectory. Though if you didn't stop at Mars you would return to Mercury's orbit [it's a Mercury to Mars Hohmann trajectory]. Such a non-Hohmann trajectory from Earth could get you to Mars in somewhere around 3 months [90 days] or less.

If use simple method [which isn't accurate] and half the years of Mercury and Earth.
So 365 + 88 = 226.5 and half again for 1/2 a year you get 113 days. As compared to 105 days.

So 88 day [Mercury year] and 687 day [Mars year] is 193.75 days for Hohmann trajectory [one leg to or from Mars].
So if using this rough reckoning and took 190 days [6 months] to do Hohmann trajectory from Mercury to Mars, part of the section of trajectory from Earth orbital distance to Mars orbital distance is a small portion of the trajectory, but one traveling at much lower velocity as compared most of the total distance which is traveled. So less than 1/2 but not a 1/4 or less of the total travel time.
Said differently the "year" of Mercury to Mars trajectory is about 387 days. At Mercury distance you going somewhere around +50 km/sec [Mercury average orbital speed is 47.9 km] and at Mars distance going less than Mars orbital average speed of 24.1 km/sec [if were going at Mars orbital speed you stay at Mars distance instead of falling back towards Mercury's orbit].

This fast way to Mars is not what most people expect or mention- they think if you go to Mars so one gets there in less than 3 month you will be going much faster than Mars orbital speed and at an sharp angle to Mars's orbit. And this trajectory you can go to Mars faster than any Hohmann type trajectory and one reach Mars at slower orbital speed than Mars and have no angle to Mars- going slower but exact same direction of Mars orbit.

Of course, we don't always take 8 1/2 month to reach Mars as would with a "true" Hohmann trajectory- you tweak it so it's a Hohmann trajectory with patched conic.

So if leaving from Mercury to Mars, you could also have Hohmann trajectory with patched conic and get to Mars quicker.
And from Earth one could match this faster trajectory- and so arrive at Mars faster and at small angle [which you do a course correction near Mars {patched conic}].

Doing what I am talking about requires more delta-v. But getting to Mars faster than 7 months require more delta-v.
Much is said about using nuclear propulsion to get to Mars fast and getting somewhere around say 20 km/sec of delta-v being applied from low Earth orbit.
So that is more delta-v and it's less efficient and takes longer- in comparison.
What talking about is done with chemical rockets [and can't be done with low thrust rockets] and uses much less total delta-v. And my rough guess it's instead about 10 km/sec of chemical high thrust delta-v from LEO or about 7 km/sec from high Earth orbit- a elliptical orbit with burn near Earth at perigee.

This diagram of space rock is vaguely similar trajectory:
http://ssd.jpl.nasa.gov/sbdb.cgi?sstr=2013+JL22&orb=1

This rock would be harder to get to from Earth than what I am talking about. Though if one wanted to get to it with robotic mission, instead going directly from Earth to Rock, one probably go to inner planets and use gravity assist from Venus or Mercury [or both].
But assuming the orbit reached instead nearer to Mars, and assuming one wanted to go directly from Earth to rock, how delta-v would it take?
« Last Edit: 05/25/2013 02:51 AM by gbaikie »

Offline torzek

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Re: Basic Rocket Science Q & A
« Reply #795 on: 05/25/2013 02:25 PM »
Fisrt of all, my question was about earth to LEO transfer.
Secondly, i found that calculating velocieties of various stages is possible by using Tsiolkovski formula but how do i calculate the component of velocity comming from earth rotation????

I would realy appreciate if someone  could post some real calculations insted of basic theoretical description.

Offline R7

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Re: Basic Rocket Science Q & A
« Reply #796 on: 05/25/2013 02:40 PM »
how do i calculate the component of velocity comming from earth rotation????

465m/s * cosine ( latitude )


Quote
I would realy appreciate if someone  could post some real calculations insted of basic theoretical description.

Buy a book, a bargain for 11 bucks.

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Offline torzek

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Re: Basic Rocket Science Q & A
« Reply #797 on: 05/29/2013 04:34 PM »
Thanks a lot for all answers. Qeastion solved

Offline gbaikie

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Re: Basic Rocket Science Q & A
« Reply #798 on: 05/30/2013 05:30 AM »
Remember plane change costs. Mercury's orbit is the most inclined of the planets.

Hmm. That's kind of neat. It's an added element in terms of getting to Mercury.
But also seems like it gives more options in terms destination, which may be at inclination [a lot space rocks]. You can leave [or arrive] at Mercury at a zero inclination, but you leave [or arrive] at a 7 inclination to solar planetary disk.

I was looking for trajectories to space rocks to see if any trajectory to a rock would be similar to fast Mars. And this rock:
http://ssd.jpl.nasa.gov/sbdb.cgi?sstr=2002+NW16&orb=1
costs 10.995 delta-v and  346 days to get to:(2002 NW16)
http://neo.jpl.nasa.gov/cgi-bin/nhats?dv=11&dur=360&stay=8&launch=2015-2040&H=18&occ=7&sort=n_via_traj&sdir=DESC&action=Display+Table#top
[edit- link appears to work]
That link probably will not work, so
http://neo.jpl.nasa.gov/cgi-bin/nhats
And put in variable of a high delta-v [<=11 km/sec] and H<=18
Anyhow, it reminded me of high delta-v costs to change orbital plane of inclination.
And basically if you want a lower cost to make larger change inclination, go out to Jupiter. But for same reason Mercury is bad place to change inclination.
Anyways got me wondering how they get anything to Mercury. So upshot is I think they wouldn't arrive at Mercury when it's orbit crosses
the solar plane, but change the inclination prior to getting to Mercury.

So changing inclination at Earth requires a lot of delta-v and changing it at Mercury requires even more delta-v.
And I should check Mercury Messenger to see how they did this exactly.*

But anyhow, changing inclination is similar to getting a fast transit to Mars from Earth. The faster the orbital velocity the more delta-v is required to alter it
And I believe both are better done when in a gravity well.

And the difficulty of getting to Mercury has a lot to do with inclination rather than the nearness to the sun.
Of course, fast transit to Mars doesn't need to be same inclination as Mercury. Or it requires less delta-v to get to Mercury distance without the need to change inclination. In other words, if  you send a spacecraft to Mercury's distance from the sun and do Hohmann trajectory to Mars and from the point of Mercury distance, it gets to Mars faster than Hohmann trajectory from Earth distance.
And finally I am not suggesting one would send crew to Mercury distance to get to Mars, rather start from Earth distance and match the trajectory of a Mars to Mercury Hohmann trajectory. [Mercury's distance from the sun- not a trajectory with Mercury's inclination]

And the delta-v to do this, is similar to making a inclination change at Earth orbital distance.

A question is how much inclination change is it similar to?
So, roughly, is it like a 5 percent inclination change or 10, 15, 20%, etc?

* Edit: "This pass increases the inclination of MESSENGER's orbit to match that of Mercury; Venus is mostly sunlit during the spacecraft's approach. "
http://spacex.9f.com/messenger.htm
So used Venus gravity assist to change inclination and also used the burn at Mercury to change it:
http://messenger.jhuapl.edu/the_mission/orbit.html
« Last Edit: 05/30/2013 06:15 AM by gbaikie »

Offline spacejeff

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Re: Basic Rocket Science Q & A
« Reply #799 on: 06/03/2013 12:14 PM »
Hi all.  I am not an aerospace engineer but I have followed the space program with interest all my life and am trying to understand some of the fundamental principles of rocketry.  I apologize in advance for any stupid mistakes or misunderstandings ;-)

I'm trying to understand the rocket equation and how a delta-V budget relates to the energy it takes to get into orbit.  I'm also confused by the results of a paper called "High Altitude Launch for a Practical SSTO" by Landis and Denis from 2003, probably because I don't understand delta-V.

First off, reading Don Petit's "The Tyranny of the Rocket Equation" it seemed like there was a relationship between orbital velocity and the delta-V needed to get there.  Don talks about needing 8km/s to get into LEO (not including drag, etc), and I understand the orbital velocity of LEO is about 8km/s.  But according to Wikipedia (Delta-v_budget) it takes an additional 4km/s of delta-V to get to GEO.  However, higher orbits have a lower velocity, yes?  According to Wikipedia, velocity at GEO is about 3km/s, though it takes 8km/s + 4km/s delta-V to get there.  Is that right?  So is it just a coincidence that the delta-V needed for LEO happens to be close to the actual velocity of LEO?

Now about the delta-V budget needed for a given orbit: I tried to use the rocket equation to help me understand the following (simple?) example: give a single stage rocket a super-powerful first booster stage - something that in a few seconds would zoom it up to 100m/s.  I imagined this would have a significant impact on payload since rockets seem to take forever to get going, but then pick up speed faster later on.

Using Isp = 310sec and g=9.8m/s^2, I solve for mass ratio:

delta-V = Isp * g * ln( MR )
delta-V = 3038 * ln( MR )
delta-V / 3038 = ln( MR)
e^(delta-V / 3038) = MR

Now I use the delta-V needed for LEO (approximately including drag and gravity) = 9km/s and I get a mass ratio of 19.34.  Then, to see the effect of the super-booster, I add the 100m/s from the super-booster to delta-V and solve again.  This gives a mass ratio of 19.99 (3.3% more than without the super-booster).  Is it ok to do this?  To just give the rocket an initial upward velocity and add that to the total delta-V budget?

Can I conclude that the super-booster is worth a 3.3% change in mass ratio?  This would translate to a 3.3% change in mass to orbit, yes?

Once I understand this example I'll ask the question about the High Altitude Launch paper :-)

Thanks,

Jeff