Author Topic: Orbits Q&A  (Read 177141 times)

Offline hismaimai8888

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Re: Orbits Q&A
« Reply #340 on: 11/14/2018 03:11 am »
Sometimes satellites loop around a planet to increase velocity.

How does orbiting a planet increase velocity?

Wouldn't its gravity slow it down when departing?
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Offline ChrisWilson68

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Re: Orbits Q&A
« Reply #341 on: 11/14/2018 03:34 am »
First, a clarification: what you are talking about is a gravitational slingshot, which does not involve orbiting a planet.

It increases the speed of the object doing the flyby of the planet by basically zipping past the back of the planet in the planet's orbit around the sun.  The planet is pulled back in its orbit a tiny little bit by the gravity of the spacecraft, and the spacecraft is pulled forward in the orbit of the planet a lot.  It ends out zipping out at a higher speed and going in a different direction.

Offline ChrisWilson68

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Re: Orbits Q&A
« Reply #342 on: 11/14/2018 03:39 am »
Please note that, in the frame of reference of the center of mass of the solar system (basically, the frame of reference of the sun), the energy of the spacecraft will increase, so the energy of the planet must decrease ever so slightly.  That means that the planet must be slowed down a little bit and go into a slightly smaller orbit around the sun.  That's where the energy for the spacecraft to go faster comes from -- it steals a tiny little bit of the energy of the planet in its orbit around the sun.  The planet so much enormously bigger that the little change in orbit of the planet is not noticeable.

It you did a gravitational slingshot with a really, really big object, like the Earth's moon, the planet would lose a noticeable amount of energy and end up in a significantly different orbit.

Offline ChrisWilson68

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Re: Orbits Q&A
« Reply #343 on: 11/14/2018 03:52 am »
Also, just to be pedantic, velocity doesn't increase.  It's the speed that increases.

Speed and velocity are not the same thing.  Speed is a scalar -- that is, it's defined by just a single non-negative real number and a unit, such as 55 mph.  Velocity is a combination of speed and direction.  A combination of a magnitude and a direction is called a vector.

Since a vector is an amount and a direction, it doesn't make sense to compare vectors to see if one is greater than or less than another.  So it's not meaningful to have a vector "increase" or "decrease".  It's the magnitude of the vector, not the vector itself, that can increase or decrease.

So, it's not meaningful to talk about a velocity increasing or decreasing, but it is meaningful to talk about a speed increasing or decreasing.

After a gravitational slingshot, a spacecraft will have a different velocity than it did before the slingshot.  The velocity will have a different direction and a different speed.  The speed will be greater after the slingshot.

A lot of people use the word "velocity" when they really mean "speed" because they don't really know the difference and think of "velocity" as a more scientific-sounding word.  And a lot of people who know better don't think about it and use "velocity" when they really mean "speed".  But there's an important difference, and using the proper word in the proper context will help avoid confusion in some cases and make you sound more like you know what you're talking about to those who know the difference between velocity and speed.
« Last Edit: 11/14/2018 03:54 am by ChrisWilson68 »

Offline Nomadd

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Re: Orbits Q&A
« Reply #344 on: 11/14/2018 03:53 am »
Sometimes satellites loop around a planet to increase velocity.

How does orbiting a planet increase velocity?

Wouldn't its gravity slow it down when departing?
There are a couple of ways to use a planet to get a net speed increase. It will slow you down as you depart, but the amount it speeds you up and slows you down depends on time. So to get a net speed increase, you need to have it slowing you down for less time than it speeds you up.
 One way is by following the planet, like Jupiter in it's orbit and letting it pull you forward and outward, giving you a speed increase, then whipping around the leading edge of the planet to take a more directly outward path relative to the sun. The change in direction means the planet is pulling you back for less time than it pulled you forward, because you're spending less time moving away from it than you spent moving toward it since you're not following it in it's orbit any more.
 Another way is to fire your thrusters as you're receding from your slingshot planet, which will also decrease the time it's pulling you back compared to the time it pulled you in.
« Last Edit: 11/14/2018 03:57 am by Nomadd »
Those who danced were thought to be quite insane by those who couldn't hear the music.

Re: Orbits Q&A
« Reply #345 on: 11/22/2018 08:36 am »
Dear friends,
 I am  studying on space system propuslion. I am interested in orbit.
 I am also trying to simulate an orbit.

can anyone help me to get the matlab files that simulate Low earth Orbit for the spacecraft in which i can use 4 thrusters in the same plane.
how can we compute the distribution matrix in order to correct semi major axis (for example 60meters).
 Thank you all.

Re: Orbits Q&A
« Reply #346 on: 11/28/2018 07:58 am »
Dear friends,
If anyone has a model to use Bang bang control for orbital maneuver.
MAtlab code
thanks in advance.

Offline libra

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Re: Orbits Q&A
« Reply #347 on: 07/21/2020 11:35 am »
Hello

I have a very simple question...

Launching into GTO takes 2.5 km/s, but how much delta-v does it take to launch into a Molniya orbit ? 

Sounds like a straightforward question but mind you, it is next to impossible to find an answer on the web...  :(

Offline jasperval

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Re: Orbits Q&A
« Reply #348 on: 07/22/2020 06:27 pm »
I have an orbit question: How do you calculate the orbital inclination of a dog-leg launch?

Say for some reason you needed to reach the ISS from Vandy. Because inclination = arccos(cos(launch latitude)*sin(launch azimuth)), if we want the inclination to be 51.6, and Vandy's latitude is 34.742, then the required launch azimuth would be 49.1 degrees or 130.9 degrees. That would take it either just over Salt Lake City or just over San Diego. Neither would be good, obviously. And certainly it's outside of Vandy's approved launch azimuths of 158-202 degrees.

Bu let's say we got approval to do a "coast hugging" maneuver to do a dogleg. Obviously that's a lot more dynamic, and I bet there's an insane delta-V penalty. But it looks like a coast hugging trajectory is a launch azimuth of about 140 (ignoring going over the islands south of Vandy), resulting in about a 61 degree inclination on direct assent. If we instead traveled that, and then turned more to the east after clearing Baja,what the inclination would end up being? Say you traveled a launch azimuth of 140 at launch, then after traveling 3500 km, you turned to 110 degrees until reaching orbital speed. What would your resultant orbital inclination be? And what would the delta-v penalty be for the maneuver (say, if it was a Falcon 9 doing it).

Offline Hog

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Re: Orbits Q&A
« Reply #349 on: 02/15/2023 08:07 pm »
If a vehicle is loitering in high lunar orbit, is that vehicle also in an Earth orbit?
Paul

Offline 1

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Re: Orbits Q&A
« Reply #350 on: 02/15/2023 10:07 pm »
Yes, by virtue of the fact that the moon itself can be considered to be orbiting the Earth.

Practically, of course, your choice of reference frame will likely be driven by whatever's most convenient. Someone on the lunar surface attempting to communicate with your vehicle would probably do well to consider the vehicle's trajectory from a lunar-centric coordinate system.

But someone on Earth attempting to do the same might find it easier to use a geocentric system and, if needed, simply introduce a small epicycle-like perturbation term due to the moons influence. Likewise, a hypothetical distant observer might ignore both Earth and moon, and simply consider your vehicle to be described well enough by it's greater solar orbit.

So really, you're free to use whatever is good enough for your application as long as you take care to confirm that it is, indeed, sufficiently good. Just keep it in the back of your mind that you might need to change your approach if calculations in your chosen system deviate too far from required fidelity (or reality).

Offline Jim

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Re: Orbits Q&A
« Reply #351 on: 02/15/2023 10:11 pm »
I have an orbit question: How do you calculate the orbital inclination of a dog-leg launch?

Say for some reason you needed to reach the ISS from Vandy. Because inclination = arccos(cos(launch latitude)*sin(launch azimuth)), if we want the inclination to be 51.6, and Vandy's latitude is 34.742, then the required launch azimuth would be 49.1 degrees or 130.9 degrees. That would take it either just over Salt Lake City or just over San Diego. Neither would be good, obviously. And certainly it's outside of Vandy's approved launch azimuths of 158-202 degrees.

Bu let's say we got approval to do a "coast hugging" maneuver to do a dogleg. Obviously that's a lot more dynamic, and I bet there's an insane delta-V penalty. But it looks like a coast hugging trajectory is a launch azimuth of about 140 (ignoring going over the islands south of Vandy), resulting in about a 61 degree inclination on direct assent. If we instead traveled that, and then turned more to the east after clearing Baja,what the inclination would end up being? Say you traveled a launch azimuth of 140 at launch, then after traveling 3500 km, you turned to 110 degrees until reaching orbital speed. What would your resultant orbital inclination be? And what would the delta-v penalty be for the maneuver (say, if it was a Falcon 9 doing it).

It is not a simple equation or easy to get an accurate answer.  The easy approximation is to use the minimum launch azimuth allowed from Vandenberg (158 degrees) to find the lowest launch inclination (72 degrees) and do a plane change once on orbit.

Offline laszlo

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Re: Orbits Q&A
« Reply #352 on: 02/16/2023 11:47 am »
I have an orbit question: How do you calculate the orbital inclination of a dog-leg launch?

Say for some reason you needed to reach the ISS from Vandy. Because inclination = arccos(cos(launch latitude)*sin(launch azimuth)), if we want the inclination to be 51.6, and Vandy's latitude is 34.742, then the required launch azimuth would be 49.1 degrees or 130.9 degrees. That would take it either just over Salt Lake City or just over San Diego. Neither would be good, obviously. And certainly it's outside of Vandy's approved launch azimuths of 158-202 degrees.

Bu let's say we got approval to do a "coast hugging" maneuver to do a dogleg. Obviously that's a lot more dynamic, and I bet there's an insane delta-V penalty. But it looks like a coast hugging trajectory is a launch azimuth of about 140 (ignoring going over the islands south of Vandy), resulting in about a 61 degree inclination on direct assent. If we instead traveled that, and then turned more to the east after clearing Baja,what the inclination would end up being? Say you traveled a launch azimuth of 140 at launch, then after traveling 3500 km, you turned to 110 degrees until reaching orbital speed. What would your resultant orbital inclination be? And what would the delta-v penalty be for the maneuver (say, if it was a Falcon 9 doing it).

It is not a simple equation or easy to get an accurate answer.  The easy approximation is to use the minimum launch azimuth allowed from Vandenberg (158 degrees) to find the lowest launch inclination (72 degrees) and do a plane change once on orbit.

A book whose title I unfortunately do not recall (since I read it in the 70's) indicated that such problems back then were solved numerically, not analytically. So instead of deriving a formula you'd write a (FORTRAN) program that numerically integrated the problem and then try it with a bunch of different initial conditions until you got the solution you wanted. The example given was the powered descent for Apollo which was set up as an ascent into lunar orbit and integrated backwards to get the desired descent trajectory.

Whether this actually happened or not, the approach should still be good and give a pretty accurate answer. Today's computers would make it a lot easier to set up the problem and faster to run it. Sounds like a job for MATLAB.

Offline Kyra's kosmos

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Re: Orbits Q&A
« Reply #353 on: 05/09/2023 01:40 am »
Not really an Orbital Mechanics question, more an Attitude & Pointing Question.

I have a set of NASA supplied unit vectors I am converting to Right Ascension and Declination. I am using these formulae ρ2=x2+y2+z2,tanθ=yx, and φ=arccos(z√x2+y2+z2). To check the equations I started with a couple I could check with.

11   -1.2   -0.8   -0.17832207   0.94125671   -0.28677002   SIRIUS (ALP CMA)
12   -0.4   -0.1   -0.06034819   0.60343258   -0.79512717   CANOPUS (ALP CAR)

I started with a couple of stars we know the RA and DEC for. ρ=1 So the data is good, but θ and φ are not making any sense.


Offline Proponent

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Re: Orbits Q&A
« Reply #354 on: 05/09/2023 09:19 pm »
I'm not quite sure how your xyz-coordinate frame is oriented.  Assuming that φ is RA and θ declination, and that the +x-direction corresponds to φ = θ = 0o, the +y-direction to φ = 90o and θ = 0o, and the +z-direction to θ = 90o, then I would expect

     tan φ = y/x

and

     tan θ = z/(x2 + y2)1/2 .

Offline Jorge

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Re: Orbits Q&A
« Reply #355 on: 05/09/2023 11:03 pm »
I'm not quite sure how your xyz-coordinate frame is oriented.

In the space shuttle program, the standard inertial frame was Mean of 1950. +X along mean vernal equinox (intersection of equatorial and ecliptic planes), +Z along mean rotational axis (north pole), +Y completes a right-handed system (and therefore both X and Y are in the mean equatorial plane).

Really should use the atan2 function to get an unambiguous, quadrant-correct solution for RA:

RA = atan2(y,x)

And since it's already established that these are unit vectors, the solution for DEC reduces to:

DEC = asin(z)

Offline Kyra's kosmos

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Re: Orbits Q&A
« Reply #356 on: 05/10/2023 07:13 pm »
I'm not quite sure how your xyz-coordinate frame is oriented.

In the space shuttle program, the standard inertial frame was Mean of 1950. +X along mean vernal equinox (intersection of equatorial and ecliptic planes), +Z along mean rotational axis (north pole), +Y completes a right-handed system (and therefore both X and Y are in the mean equatorial plane).

Really should use the atan2 function to get an unambiguous, quadrant-correct solution for RA:

RA = atan2(y,x)

And since it's already established that these are unit vectors, the solution for DEC reduces to:

DEC = asin(z)

Jorge, Thank you for your brilliant explanation and reduction of the situation here. I knew they would have spacified if they were Starball, and that why some anwers seemed right and others were in wonky had something to do with ambiguity. ATAN2 saves the day!


Offline Jorge

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Re: Orbits Q&A
« Reply #357 on: 05/11/2023 05:35 am »
I'm not quite sure how your xyz-coordinate frame is oriented.

In the space shuttle program, the standard inertial frame was Mean of 1950. +X along mean vernal equinox (intersection of equatorial and ecliptic planes), +Z along mean rotational axis (north pole), +Y completes a right-handed system (and therefore both X and Y are in the mean equatorial plane).

Really should use the atan2 function to get an unambiguous, quadrant-correct solution for RA:

RA = atan2(y,x)

And since it's already established that these are unit vectors, the solution for DEC reduces to:

DEC = asin(z)

Jorge, Thank you for your brilliant explanation and reduction of the situation here. I knew they would have spacified if they were Starball, and that why some anwers seemed right and others were in wonky had something to do with ambiguity. ATAN2 saves the day!



You're welcome! This is my old stomping ground so feel free to ask more.

You'll probably notice small discrepancies between the RA/DECs in the shuttle star catalog when comparing to most modern star catalogs since they're based on the B1950 epoch instead of J2000. The Earth's axes precessed a bit during those 50 years.

The solution to your problem was actually buried within another document from the same set you got the star catalog from, if you're curious.
JRF

Offline AS_501

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Re: Orbits Q&A
« Reply #358 on: 11/03/2023 01:09 am »
My grasp of orbital mechanics is limited, so I'm trying to understand the purpose of a perigee raise maneuver, in particular when heading to the Moon.  When an upper stage or spacecraft initially climbs into a high elliptical orbit, is the perigee of this orbit typically inside the Earth's atmosphere?  Thus, you fire the upper stage or spacecraft engine(s) at apogee to raise the perigee above the atmosphere?  I assume that launching into a high elliptical orbit where perigee is already above the atmosphere is not as fuel-efficient.  I hope I've properly stated the question here.
Thx
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Offline tbellman

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Re: Orbits Q&A
« Reply #359 on: 11/03/2023 09:09 am »
My grasp of orbital mechanics is limited, so I'm trying to understand the purpose of a perigee raise maneuver, in particular when heading to the Moon.  When an upper stage or spacecraft initially climbs into a high elliptical orbit, is the perigee of this orbit typically inside the Earth's atmosphere?  Thus, you fire the upper stage or spacecraft engine(s) at apogee to raise the perigee above the atmosphere?  I assume that launching into a high elliptical orbit where perigee is already above the atmosphere is not as fuel-efficient.  I hope I've properly stated the question here.
Thx

I don't think there is any "typical" orbits that most missions follow to go to the Moon.  For example, Chandrayaan 3 inserted into an elliptical orbit where perigee was well outside of the Earth's atmosphere, and did not have a perigee raise maneuver; see the attached animation, taken from Wikipedia.

Apollo inserted into a near-circular orbit with an altitude of about 185 km (100 nautical miles), and from there did a direct trans-lunar injection.  185 km is usually considered to be outside of Earth's atmosphere, and a spacecraft in such an orbit will not fall down immediately.  There is however enough air that it would drop within just a few revolutions around Earth, so they couldn't stay there very long.

SLS with Orion is usually described as having a perigee raise maneuver.  The SLS core inserts ICPS/EUS+Orion into an elliptical orbit where perigee is well inside the Earth's atmosphere.  Then about a minute after that (time enough for Orion to deploy its solar panels), ICPS/EUS performs a shortish burn to raise perigee, so it is outside of the atmosphere.  During Artemis 1, that was followed half an hour later by a TLI burn; at perigee, I assume, but I don't have time to verify that right now.  From what I can remember, Artemis 2 is planning to spend a bit more time in that orbit, to have time checking out that Orion works as intended before performing the TLI burn.

SLS is a bit of an oddity, though, as its core stage is a sustainer that takes it all the way into (an) orbit, and its upper stage is intended to mostly perform the departure from Earth.  I don't think I have seen any other rockets/missions that do it like that and inserts into orbits with a high apogee, while perigee is well inside the atmosphere.  (I fully expect to be corrected about that, though. :) )

Compare also to the launch of the James Webb Space Telescope.  It didn't have a staging orbit at all; Ariane 5 inserted it directly into its trajectory towards L2.  (But there was no other choice, as the ESC upper stage used can't restart its engine.)

Tags: inclination 
 

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