Author Topic: Basic Rocket Science Q & A  (Read 282657 times)

Offline yinzer

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Re: Basic Rocket Science Q & A
« Reply #380 on: 12/28/2009 10:51 PM »
From the report:
Quote
Pre-launch instrumentation staging disconnected prematurely resulting in loss of all booster data measurement.
Sustainer and vernier engine shutdown occurred prematurely during flight.
Range Safety destroyed vehicle at 303 seconds.

So... who knows.  Could have damaged one of the valves that seals off a line that separates during staging, resulting in a leak of fuel or oxidizer or pressurant or something.
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Offline tminus9

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Re: Basic Rocket Science Q & A
« Reply #381 on: 12/29/2009 02:18 PM »
To convert earth-centered inertial cartesian coordinates (e.g., J2000) to geodetic latitude and longitude, I know that the rotation, precession and nutation of the earth since the epoch must be accounted for.

Rotation seems easy enough, but I've yet to find an accessible (i.e., assuming enough physics and math to understand, but not assuming any previous knowledge of astronomy) resource about calculating precession and nutation matrices.

1. Does anyone know of a good (preferably online) description of how to calculate the precession and nutation matrices, given an arbitrary epoch? Included examples would be even better to verify my results. I know the calculation gets pretty hairy, so even an approximation is fine for my purposes.

2. If I neglect precession and nutation, what is the ballpark error in the resulting geodetic calculations? (I assume it depends on the epoch.)

Offline Jorge

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Re: Basic Rocket Science Q & A
« Reply #382 on: 12/29/2009 06:54 PM »
To convert earth-centered inertial cartesian coordinates (e.g., J2000) to geodetic latitude and longitude, I know that the rotation, precession and nutation of the earth since the epoch must be accounted for.

Rotation seems easy enough, but I've yet to find an accessible (i.e., assuming enough physics and math to understand, but not assuming any previous knowledge of astronomy) resource about calculating precession and nutation matrices.

1. Does anyone know of a good (preferably online) description of how to calculate the precession and nutation matrices, given an arbitrary epoch? Included examples would be even better to verify my results. I know the calculation gets pretty hairy, so even an approximation is fine for my purposes.

Vallado's book Fundamentals of Astrodynamics and Applications contains a good discussion of this, and he has graciously put his sample code online, in several languages:

http://www.smad.com/vallado/

The code can be understood without reference to the book, but I highly recommend the book anyway.

Quote
2. If I neglect precession and nutation, what is the ballpark error in the resulting geodetic calculations? (I assume it depends on the epoch.)

Yes, it does depend on the epoch. Generally speaking, the results are good enough for backyard astronomy (that is, you can look up and find the object in question even though the computed position is off) but you wouldn't want to use it for professional applications.
JRF

Offline butters

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Re: Basic Rocket Science Q & A
« Reply #383 on: 01/05/2010 01:38 PM »
Basic orbital mechanics question:


Is it possible to put a satellite in a geostationary orbit over a pole, or in a sort of halo orbit that stays within, let's say, 25 degrees latitude of a (rotational) pole?

Offline Jim

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Re: Basic Rocket Science Q & A
« Reply #384 on: 01/05/2010 01:42 PM »
Basic orbital mechanics question:


Is it possible to put a satellite in a geostationary orbit over a pole, or in a sort of halo orbit that stays within, let's say, 25 degrees latitude of a (rotational) pole?

No

Offline kevin-rf

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Re: Basic Rocket Science Q & A
« Reply #385 on: 01/05/2010 02:15 PM »
But there are other useful orbits that make communications at the poles easier. They require a min of two birds for 24-7 coverage.

google:

molniya orbits ( http://en.wikipedia.org/wiki/Molniya_orbit )

and

tundra orbits ( http://en.wikipedia.org/wiki/Tundra_orbit )
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Offline MP99

Re: Basic Rocket Science Q & A
« Reply #386 on: 01/05/2010 10:52 PM »
Quick question about C3.

I understand that this relates the energy of a trajectory to escape velocity (C3=0).

If a mission is described as having a positive C3 (say 1.5km2/s2), is there an easy way to equate that to a delta-V, over and above the delta-V necessary to achieve escape velocity?

Thanks, Martin

Offline ugordan

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Re: Basic Rocket Science Q & A
« Reply #387 on: 01/05/2010 11:05 PM »
If a mission is described as having a positive C3 (say 1.5km2/s2), is there an easy way to equate that to a delta-V, over and above the delta-V necessary to achieve escape velocity?

Try this wiki page, in particular the Voyager 1 example. Since you can deduce V-inf from C3 for a hyperbolic orbit, you should be able to calculate V at periapsis and compare that to circular orbital velocity at that point (and also to escape velocity at that point, it being SQRT(2) greater than orbital)

Offline MP99

Re: Basic Rocket Science Q & A
« Reply #388 on: 01/05/2010 11:54 PM »
Many thanks.

So a C3 of 1.5km2/s2, equates to V-inf of sqrt (2 * 1.5) = 1.73 km/s.

Would it be an over-simplification to say that an EDS would need to burn to escape velocity + 1.73 km/s to achieve that C3?

cheers, Martin

Offline ugordan

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Re: Basic Rocket Science Q & A
« Reply #389 on: 01/06/2010 12:17 AM »
So a C3 of 1.5km2/s2, equates to V-inf of sqrt (2 * 1.5) = 1.73 km/s.

No, V-inf equals sqrt(2 * epsilon). Since C3 already is 2*epsilon (equation further up that page), V-inf becomes sqrt(C3), in your case that's V-inf of 1.22 km/s

Quote
Would it be an over-simplification to say that an EDS would need to burn to escape velocity + 1.73 km/s to achieve that C3?

Yes, and a gross inaccuracy. An EDS would need a smaller burn than that because of the Oberth effect. In effect having a small boost above V-esc you get out of the gravity well a lot quicker and that translates into a larger velocity at infinity.

That's why you need the equation in that Voyager 1 example to find the actual velocity at periapsis of that hyperbola. If you do the calculation, you'll find that an EDS needs substantially less than 1.22 km/s to reach that specific velocity at infinity. The deeper into the gravity well you are when burning to escape, the more efficient this effect is. It's one of the reasons parking orbits are just low enough not to decay over a period until the 2nd burn.

Offline toddbronco2

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Re: Basic Rocket Science Q & A
« Reply #390 on: 01/06/2010 04:10 AM »
Basic orbital mechanics question:


Is it possible to put a satellite in a geostationary orbit over a pole, or in a sort of halo orbit that stays within, let's say, 25 degrees latitude of a (rotational) pole?

Jim is right that it's not possible to be geostationary above a rotational pole, but there are orbits that are loosely associated with the L3 family of Halo orbits that are similar to your description, in that they dwell for long periods of time above the north-pole.  Notice the nearly vertical orange colored orbits that pass closest to 'Earth' (the black dot).  Source: D. Grebow (https://engineering.purdue.edu/people/kathleen.howell.1/Publications/publications.html)

Offline MP99

Re: Basic Rocket Science Q & A
« Reply #391 on: 01/07/2010 04:41 PM »
Constellation assumes that TLI will take advantage of the lowest delta-V opportunities. I've seen contradictory comments that these opportunities occur either once or twice per month. Which is correct?

How does the delta-V vary between these opportunities - say if you wanted to just pick the best opportunity on any particular day?

Many thanks, Martin

Offline Namechange User

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Re: Basic Rocket Science Q & A
« Reply #392 on: 01/07/2010 04:47 PM »
Constellation assumes that TLI will take advantage of the lowest delta-V opportunities. I've seen contradictory comments that these opportunities occur either once or twice per month. Which is correct?


Who cares.  Either is more frequent than actual launch rates will be. 
Enjoying viewing the forum a little better now by filtering certain users.

Offline jongoff

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Re: Basic Rocket Science Q & A
« Reply #393 on: 01/07/2010 05:08 PM »
Constellation assumes that TLI will take advantage of the lowest delta-V opportunities. I've seen contradictory comments that these opportunities occur either once or twice per month. Which is correct?


Who cares.  Either is more frequent than actual launch rates will be. 

MP99, it varies depending on if you're launching from the surface or from LEO.  If you're launching from *the surface*, you get daily opportunities, because you can pick your launch time such that when your assembly is ready to depart LEO, your stack's orbital plane intersects the point where the moon will be when your stack arrives in its vicinity.  Once you're in LEO though, you've got a plane, and planes precess at a rate that gives you launch windows on the order of 2-3 times per month.  Depends a lot on inclination and altitude.  So, if you're doing an EOR mission (say one with a depot for instance), you only get 2-3 good departure dates per month. 

But OV-106 is right.  Having launch window frequency being a real issue would be a nice problem to have.  If it ever becomes a problem (say with a depot-based architecture), you always have the option of putting up 1-2 more depots in additional orbital planes.  Then you can send missions every 2-3 days....but we're a *long* way from that being important.

~Jon

[edit: corrected a dumb error in line two...sigh]
« Last Edit: 01/08/2010 04:33 PM by jongoff »

Offline MP99

Re: Basic Rocket Science Q & A
« Reply #394 on: 01/07/2010 07:30 PM »
Constellation assumes that TLI will take advantage of the lowest delta-V opportunities. I've seen contradictory comments that these opportunities occur either once or twice per month. Which is correct?


Who cares.  Either is more frequent than actual launch rates will be. 

MP99, it varies depending on if you're launching from the surface or from LEO.  If you're launching from LEO, you get daily opportunities, because you can pick your launch time such that when your assembly is ready to depart LEO, your stack's orbital plane intersects the point where the moon will be when your stack arrives in its vicinity.  Once you're in LEO though, you've got a plane, and planes precess at a rate that gives you launch windows on the order of 2-3 times per month.  Depends a lot on inclination and altitude.  So, if you're doing an EOR mission (say one with a depot for instance), you only get 2-3 good departure dates per month. 

But OV-106 is right.  Having launch window frequency being a real issue would be a nice problem to have.  If it ever becomes a problem (say with a depot-based architecture), you always have the option of putting up 1-2 more depots in additional orbital planes.  Then you can send missions every 2-3 days....but we're a *long* way from that being important.

~Jon

I'm thinking of CxP modified for dual launch.

If the lander launches to take advantage of a minimum delta-V window (max payload), could the crew launch a few days later, even though it's not optimal?

How much extra delta-V would be required? How does it vary as the month progresses?

cheers, Martin

Offline jongoff

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Re: Basic Rocket Science Q & A
« Reply #395 on: 01/08/2010 04:39 AM »
Constellation assumes that TLI will take advantage of the lowest delta-V opportunities. I've seen contradictory comments that these opportunities occur either once or twice per month. Which is correct?


Who cares.  Either is more frequent than actual launch rates will be. 

MP99, it varies depending on if you're launching from the surface or from LEO.  If you're launching from LEO, you get daily opportunities, because you can pick your launch time such that when your assembly is ready to depart LEO, your stack's orbital plane intersects the point where the moon will be when your stack arrives in its vicinity.  Once you're in LEO though, you've got a plane, and planes precess at a rate that gives you launch windows on the order of 2-3 times per month.  Depends a lot on inclination and altitude.  So, if you're doing an EOR mission (say one with a depot for instance), you only get 2-3 good departure dates per month. 

But OV-106 is right.  Having launch window frequency being a real issue would be a nice problem to have.  If it ever becomes a problem (say with a depot-based architecture), you always have the option of putting up 1-2 more depots in additional orbital planes.  Then you can send missions every 2-3 days....but we're a *long* way from that being important.

~Jon

I'm thinking of CxP modified for dual launch.

If the lander launches to take advantage of a minimum delta-V window (max payload), could the crew launch a few days later, even though it's not optimal?

How much extra delta-V would be required? How does it vary as the month progresses?

cheers, Martin

Well, my orbital dynamics fu is pretty weak, but AIUI, launching at a time that is more than a day or so away from a good launch window makes the delta-V costs prohibitive.  As in several times higher dV requirements.  There *may* be some clever trick, but I'm not smart enough (and don't have good enough tools) to know for sure.

~Jon

Offline MP99

Re: Basic Rocket Science Q & A
« Reply #396 on: 01/08/2010 02:07 PM »
Going back to my earlier question re C3 - let's see if I've got it now.

C3 is numerically twice the kinetic energy per Kg that an escaping object would attain if it was allowed to continue to infinity. Measured in J/Kg, which is also equivalent to m2/s2 (meters per second, squared).



Specific orbital energy is the total orbital energy of the object per Kg of mass, composed of kinetic and potential terms. Kinetic is always positive. Potential is negative, tending to zero at inifinity. (For a circular orbit, kinetic energy is half the potential, but of opposite sign.)

An object on the Earth's surface has potential energy but no kinetic (baring that from the rotational speed for the Earth's latitude).

To lift an object to LEO it needs to gain a small increase in potential energy, and a large amount of kinetic. The net delta-V can be equated to the specific energy gain via 1/2mv2, where m=1, ie 1/2v2. The energy of this orbit is -GM/2a, where a = semi-major axis. AFAICT, this is the delta-V to reach an orbit with the same inclination as the latitude of the launch site, further delta-V will be  required for plane changes.

For injection out of Earth's gravity from LEO, add the energy required to escape (+GM/2a) to the additional energy required to reach the destination (half of the C3 value). Again, assumes that all orbital parameters at start of the injection are ideal for the required orbit. Again, delta-V can be equated to this total specific energy via 1/2v2

For TMI via a hohmann orbit, I presume this would equate to the difference in energy between the orbits of Earth & Mars, within the Sun's gravitational field.



OK, how did I do?

cheers, Martin
« Last Edit: 01/08/2010 10:27 PM by MP99 »

Offline mmeijeri

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Re: Basic Rocket Science Q & A
« Reply #397 on: 01/09/2010 11:51 AM »
Let's see how well (or not) this amateur can do...

To relate delta-v to energy you can't just say 0.5 * delta_v ^2 = delta-E. The efficiency of the burn depends strongly on your current velocity.

If we idealise things and consider only infinitely short, impulsive burns then we obtain instantaneous changes in velocity while positions remain unchanged. In real life situations with finite duration burns of significant thrust we get small changes to position and potentially large changes to velocity. If positions remain unchanged, then potential energy remains unchanged too. Kinetic energy does change of course. The total change in energy due to an impulsive burn therefore consists entirely of a change in kinetic energy. And since kinetic energy is quadratic in velocity, the increase in energy for a given delta-v increases linearly with the starting velocity. The faster you go already, the more benefit you get from your burn. This is the famous Oberth effect. For true orbits (i.e. not suborbital trajectories) this means the most efficient place to increase your energy is at perigee. On suborbital trajectories you can hardly expect to be able to execute a subterranean perigee burn...

Of course there's more to it than just energy. To begin with velocity is a vector. You may also want to change the orientation of your orbital plane and the orientation of your major axis within that plane. It's no use flying in the wrong direction with the right speed!

You typically need more than one burn to get into the desired orbit. In theory you could get into a sea level Earth orbit with a single impulsive burn that raises your perigee from the center to the surface of the Earth. In practice your burn cannot be impulsive and even if it could be you would burn up in the atmosphere and smash into mountains and other obstacles. To get into circular LEO you would need at least two impulsive burns: one burn to raise either apogee or perigee to the desired altitude, and once at the new apogee another circularisation burn to raise the new perigee to the desired altitude. In the real world you need to raise your current apogee by enough to escape the effects of drag and heating and to give your engines enough time to raise your perigee.

All in all you'd need more delta-v than you'd expect from just comparing energies.
« Last Edit: 01/09/2010 11:53 AM by mmeijeri »
We will be vic-toooooo-ri-ous!!!

Offline MP99

Re: Basic Rocket Science Q & A
« Reply #398 on: 01/09/2010 05:55 PM »
Let's see how well (or not) this amateur can do...

To relate delta-v to energy you can't just say 0.5 * delta_v ^2 = delta-E. The efficiency of the burn depends strongly on your current velocity.

Yeah, I'd found that by working through examples on the wiki page that ugordan pointed to.

It's applicable to the C3 part of the calculation, but not the escape element.


Quote
If we idealise things and consider only infinitely short, impulsive burns then we obtain instantaneous changes in velocity while positions remain unchanged. In real life situations with finite duration burns of significant thrust we get small changes to position and potentially large changes to velocity. If positions remain unchanged, then potential energy remains unchanged too. Kinetic energy does change of course. The total change in energy due to an impulsive burn therefore consists entirely of a change in kinetic energy. And since kinetic energy is quadratic in velocity, the increase in energy for a given delta-v increases linearly with the starting velocity. The faster you go already, the more benefit you get from your burn. This is the famous Oberth effect. For true orbits (i.e. not suborbital trajectories) this means the most efficient place to increase your energy is at perigee. On suborbital trajectories you can hardly expect to be able to execute a subterranean perigee burn...

Of course there's more to it than just energy. To begin with velocity is a vector. You may also want to change the orientation of your orbital plane and the orientation of your major axis within that plane. It's no use flying in the wrong direction with the right speed!

You typically need more than one burn to get into the desired orbit. In theory you could get into a sea level Earth orbit with a single impulsive burn that raises your perigee from the center to the surface of the Earth. In practice your burn cannot be impulsive and even if it could be you would burn up in the atmosphere and smash into mountains and other obstacles. To get into circular LEO you would need at least two impulsive burns: one burn to raise either apogee or perigee to the desired altitude, and once at the new apogee another circularisation burn to raise the new perigee to the desired altitude. In the real world you need to raise your current apogee by enough to escape the effects of drag and heating and to give your engines enough time to raise your perigee.

All in all you'd need more delta-v than you'd expect from just comparing energies.

Taking the 100km circ LEO example where SOE = −30.8 MJ/kg and v = 7.85km/s.

To escape, another total ∆v of 7.85km/s would be required by 1/2(∆v)2.

Instead, Oberth gives ∆SEO = (v * ∆v) + 1/2(∆v)2 and ∆v of 3.25 km/s is required for escape, as you'd expect.



Starting from the same LEO, a ∆v of 4.25 km/s gives me a ∆SEO of 42.4 MJ/Kg. Presumably, I have this right that this results in a C3 = (42.4 - 30.8) * 2 = 23.2 km2/s2?

cheers, Martin

Offline yinzer

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Re: Basic Rocket Science Q & A
« Reply #399 on: 01/09/2010 06:30 PM »
It all boils down to the equation for specific orbital energy:
Quote
C3/2 = E = (v^2/2) - /r

Trying to relate delta-energy and delta-velocity is going to require calculus, so is probably best avoided.  Just figure out what terms of the specific orbital energy equation you know and solve for the remaining one.

For escape, C3 = 0 so you can solve the above equation with appropriate values of and r.  100 km from Earth gives 0 = v^2/2 - (3.99e6 / 6.46e3), or v = 11.1 km/sec.  If your orbit has you going 7.85 km/sec at 100 km, you indeed need a delta-V of 3.25 km/sec.

For your other example, after adding 4.25 km/sec from your LEO, you are at 100km altitude and travelling 12.1 km/sec.  v^2/2 - /r, or (12.1^2)/2 - 3.99e6/6.46e3, or 11.47 km^2/sec^2.  This is energy, so double it for C3, and you end up with roughly the same number (22.9 vs. 23.2, but that might be rounding errors).
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