Author Topic: Orbital Mechanics Q&A  (Read 13413 times)

Offline DLR

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Orbital Mechanics Q&A
« on: 03/12/2016 09:56 am »
I have an orbital mechanics question.


Let's say I'm approaching Jupiter's moon Callisto on a Hohmann from Earth and I want to insert myself into a 200km low orbit around Callisto with a single burn.

Can simply use the relative velocity of the spacecraft with respect to Callisto, Callisto's escape velocity at the desired orbit (2.35 km/s) and the desired orbit's orbital velocity (1.66 km/s) to compute my arrival delta v, "ignoring" Jupiter, or do I need to take Jupiter's escape velocity at the alitude of Callisto's orbit into account?

As I approach Jupiter/Callisto, my spacecraft is moving tangentially to Jupiter/Callisto with a velocity of 7.4 km/s. Jupiter's orbital velocity around the Sun is 13 km/s, Callisto's velocity around Jupiter 8.2 km/s. Naturally, I assume, since I am moving slower than Jupiter at that point, I would want to approach Callisto as it is moving towards me in its orbit around Jupiter, so my velocity relative to Callisto would be

13 km/s -8.2 km/s - 7.4 km/s = -2.6 km/s, which means that I am moving 2.6 km/s faster than Callisto at that point.

The orbital capture burn would then require a delta v of SQRT(v_relative^2 + v_esc^2) - v_orbit = SQRT(2.6^2+2.35^2) - 1.66 = 1.85 km/s.

Is that correct?

« Last Edit: 03/12/2016 10:02 am by DLR »

Offline Jim

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Re: Orbital Mechanics Q&A
« Reply #1 on: 03/12/2016 10:01 am »
Jupiter has to be accounted for.

Offline DLR

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Re: Orbital Mechanics Q&A
« Reply #2 on: 03/12/2016 12:02 pm »
How about I capture into a highly elliptical orbit around Jupiter, with the apoapsis at the radius of Jupiter's sphere of influence and the periapsis at Callisto?

I compute capture into this orbit first, and then capture from this orbit into a circular Callisto orbit, and the total delta v is just the sum of the two manoeuvres, even if they are carried out simultaneously, in one burn?


SQRT(v_esc_coj^2+v_rel_j^2) - v_jco + SQRT(v_rel_c^2+v_esc_lco^2) - v_lco = total_delta_v

v_esc_coj: Jupiter's escape velocity at Callisto's altitude
v_rel_j: Spacecraft velocity relative to Jupiter
v_jco: orbital velocity at periapsis of highly elliptical Jupiter capture orbit
v_esc_lco: escape velocity in the low orbit around Callisto
v_rel_c: spacecraft relative velocity to Callisto (after capture into highly elliptical Jupiter orbit)
v_lco: Callisto circular low orbit velocity
« Last Edit: 03/12/2016 12:41 pm by DLR »

Offline DLR

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Re: Orbital Mechanics Q&A
« Reply #3 on: 03/14/2016 08:53 pm »
This spreadsheet gives delta v and trip time for flights from EML2 to Callisto with direct capture into LCO. In the yellow field, you can change the apoapsis of the interplanetary transfer orbit to adjust the trip time.

Orbits are assumed to be circular and coplanar.

The mission profile is as follows:

1) deorbit from EML2
2) TJI burn at 200km altitude
3) direct capture into LCO

I'm not sure whether what I did here is entirely correct, so it would be great if someone knowledgable could look over it and point out/correct any mistakes.
« Last Edit: 03/14/2016 08:57 pm by DLR »

Offline EddieP

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Re: Orbital Mechanics Q&A
« Reply #4 on: 03/18/2016 11:41 pm »
I have the orbital elements of a satellite orbit around earth. It is an inclined elliptical orbit. I have plotted the orbit in 3D and have all the position vectors of the orbit.

I am now trying to plot a ground track of the orbit on MATLAB and I am unsure where to start. I am looking for guidance on the steps required to achieve this

Offline Remes

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Re: Orbital Mechanics Q&A
« Reply #5 on: 03/19/2016 09:20 pm »
and have all the position vectors of the orbit.
So you have a vector from earth center to object?

One possibility is to shorten the vector length to earth radius. Then you would have the point on earth surface.

Or you calculate the angle between your vector and the equatorial plane and the angle between your vector and the plane which goes through greenwich (0 degree). That should be latitude and longitude.
(edit: i.e.: projection of the vector to the equatorial plane, then calculating the angle between your vector and the projection. To get the other value: calculate the angle between the projection and the vector going through 0 degree [in the equatorial plane]).


Putting the later coordinates in a map is a little bit tricky, as maps are distorted (sphere->rectangle). Never did that.
« Last Edit: 03/19/2016 09:57 pm by Remes »

Offline Jorge

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Re: Orbital Mechanics Q&A
« Reply #6 on: 03/20/2016 06:09 am »
I have the orbital elements of a satellite orbit around earth. It is an inclined elliptical orbit. I have plotted the orbit in 3D and have all the position vectors of the orbit.

I am now trying to plot a ground track of the orbit on MATLAB and I am unsure where to start. I am looking for guidance on the steps required to achieve this

What coordinate system are the orbital elements/position vectors expressed in? Most likely they are in an Earth Centered Inertial (ECI) system, such as J2000. You need to transform the vectors to J2000 if they are not already. Then transform the vectors to an Earth Centered Earth Fixed (ECEF) system. Then convert the position vectors from Cartesian to latitude/longitude/altitude. Then plot latitude/longitude using Matlab's mapping toolbox.

Fundamentals of Astrodynamics and Applications (Vallado) describes the algorithms in detail. The algorithms are also available for download in Matlab and several other languages here:

http://www.celestrak.com/software/vallado-sw.asp

The two key functions are eci2ecef.m (transform ECI to ECEF) and ijk2ll.m (transform cartesian ECEF to latitude/longitude). You can plug-n-chug your position vectors through these routines, but I do highly recommend the book if you want to understand what's going on.
JRF

Offline whitelancer64

Re: Orbital Mechanics Q&A
« Reply #7 on: 04/06/2016 07:56 pm »
If someone built a solid sphere of Iron 300 meters in diameter at some convenient location on the surface of the Earth and built a vacuum chamber around it, would it then be possible to put something small (say, a 0.1 mm diameter bb) into orbit around it? In other words, would this large Iron sphere have a Hill Sphere?

If not, how large would such an Iron sphere have to be to have a Hill Sphere? Would Tungsten be a better (though probably not cheaper) material, thanks to its density, to build something like this with?

I know that due to the 3-body problem this orbit will not be stable. For the purposes of my question, once around would be fine to be an orbit. I know it's wildly impractical idea and probably not possible to build from an engineering standpoint. Think of this as a grand science experiment in gravity and orbital mechanics, and it is paid for and agreed upon by all nations that this is a good idea.
"One bit of advice: it is important to view knowledge as sort of a semantic tree -- make sure you understand the fundamental principles, ie the trunk and big branches, before you get into the leaves/details or there is nothing for them to hang on to." - Elon Musk
"There are lies, damned lies, and launch schedules." - Larry J

Offline the_other_Doug

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Re: Orbital Mechanics Q&A
« Reply #8 on: 04/06/2016 08:21 pm »
If someone built a solid sphere of Iron 300 meters in diameter at some convenient location on the surface of the Earth and built a vacuum chamber around it, would it then be possible to put something small (say, a 0.1 mm diameter bb) into orbit around it? In other words, would this large Iron sphere have a Hill Sphere?

If not, how large would such an Iron sphere have to be to have a Hill Sphere? Would Tungsten be a better (though probably not cheaper) material, thanks to its density, to build something like this with?

I know that due to the 3-body problem this orbit will not be stable. For the purposes of my question, once around would be fine to be an orbit. I know it's wildly impractical idea and probably not possible to build from an engineering standpoint. Think of this as a grand science experiment in gravity and orbital mechanics, and it is paid for and agreed upon by all nations that this is a good idea.

Your iron mass would have such little mass that its own gravitational field would be barely detectable.  The orbital speed about such a mass, in such a minuscule gravitational field, would be extremely slow, probably (a gut-level guess, not having run any numbers) on the order of millimeters per second, at best.

Because of the vast difference in strength between Earth's mass and your iron sphere's mass, the Earth's gravity would overwhelm the tiny effect of gravitation by the iron sphere in a very short time, probably milliseconds.

So, to be honest, I don't think you could even get a fraction of a millimeter in such an orbit before your orbiting body acted as if the only gravitational field affecting its motion was that of Earth's.

If someone can please run the math to prove it, I'd be grateful, but my gut-level apprehension of the scale between the two gravitational fields tells me you would not be able to achieve your stated goal.
-Doug  (With my shield, not yet upon it)

Offline whitelancer64

Re: Orbital Mechanics Q&A
« Reply #9 on: 04/06/2016 08:31 pm »
I know some very small (~100 m) asteroids have moons, though they usually aren't anywhere near the perturbing effects of Earth's gravity.
"One bit of advice: it is important to view knowledge as sort of a semantic tree -- make sure you understand the fundamental principles, ie the trunk and big branches, before you get into the leaves/details or there is nothing for them to hang on to." - Elon Musk
"There are lies, damned lies, and launch schedules." - Larry J

Offline ChrisWilson68

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Re: Orbital Mechanics Q&A
« Reply #10 on: 04/06/2016 08:51 pm »
If someone built a solid sphere of Iron 300 meters in diameter at some convenient location on the surface of the Earth and built a vacuum chamber around it, would it then be possible to put something small (say, a 0.1 mm diameter bb) into orbit around it? In other words, would this large Iron sphere have a Hill Sphere?

If not, how large would such an Iron sphere have to be to have a Hill Sphere? Would Tungsten be a better (though probably not cheaper) material, thanks to its density, to build something like this with?

I know that due to the 3-body problem this orbit will not be stable. For the purposes of my question, once around would be fine to be an orbit. I know it's wildly impractical idea and probably not possible to build from an engineering standpoint. Think of this as a grand science experiment in gravity and orbital mechanics, and it is paid for and agreed upon by all nations that this is a good idea.

Lets say you put your sphere 1 foot above the Earth.  For an object to make it once around your sphere, in the part of the orbit between the sphere and the Earth you would have to have a more powerful gravitational pull from your sphere than the gravitational pull of the Earth.  If your sphere had the same density as the Earth, that could only happen if your sphere was larger than the Earth.

Of course, if you make it super-dense, you can get away with a much smaller mass -- a tiny black hole will be able to orbit something in a tiny orbit and small mass.  But if you're talking reasonable density, it's on the order of the density of the Earth, and that means a sphere larger than the Earth to get something to orbit it while it's very close to the Earth's surface.

Things can orbit the Moon even though the Moon is smaller than the Earth only because the Moon is far enough away from the Earth that Earth's gravity drops off a lot.  On the surface of the Earth, you need something roughly Earth-sized.

Of course, having something Earth-sized right next to the surface of the Earth is not only impractical, but it would immediately rip the whole planet to shreds.
« Last Edit: 04/06/2016 08:52 pm by ChrisWilson68 »

Offline Bob Shaw

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Re: Orbital Mechanics Q&A
« Reply #11 on: 04/06/2016 09:01 pm »
The late, great Iain M Banks described exactly this scenario in *The Hydrogen Sonata'. It didn't end well.

Offline ChrisWilson68

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Re: Orbital Mechanics Q&A
« Reply #12 on: 04/06/2016 09:22 pm »
Note that if you have your iron sphere orbiting the Earth instead of on the surface of the Earth, then it becomes more practical to have something orbit it.  At that point, you only need to worry about the tidal force, which is the difference between Earth's gravity on one side of your sphere and Earth's gravity on the other.  If your sphere's gravity is greater than that difference, you're in business.

Offline whitelancer64

Re: Orbital Mechanics Q&A
« Reply #13 on: 04/06/2016 09:50 pm »
Lets say you put your sphere 1 foot above the Earth.  For an object to make it once around your sphere, in the part of the orbit between the sphere and the Earth you would have to have a more powerful gravitational pull from your sphere than the gravitational pull of the Earth.  If your sphere had the same density as the Earth, that could only happen if your sphere was larger than the Earth.

Of course, if you make it super-dense, you can get away with a much smaller mass -- a tiny black hole will be able to orbit something in a tiny orbit and small mass.  But if you're talking reasonable density, it's on the order of the density of the Earth, and that means a sphere larger than the Earth to get something to orbit it while it's very close to the Earth's surface.

Things can orbit the Moon even though the Moon is smaller than the Earth only because the Moon is far enough away from the Earth that Earth's gravity drops off a lot.  On the surface of the Earth, you need something roughly Earth-sized.

Of course, having something Earth-sized right next to the surface of the Earth is not only impractical, but it would immediately rip the whole planet to shreds.

This is quickly turning into an xkcd-like What If scenario. Maybe I should send this question to them!

*edit* and I have done so.
« Last Edit: 04/06/2016 09:59 pm by whitelancer64 »
"One bit of advice: it is important to view knowledge as sort of a semantic tree -- make sure you understand the fundamental principles, ie the trunk and big branches, before you get into the leaves/details or there is nothing for them to hang on to." - Elon Musk
"There are lies, damned lies, and launch schedules." - Larry J

Offline gargoyle99

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Re: Orbital Mechanics Q&A
« Reply #14 on: 04/06/2016 10:08 pm »
Note that if you have your iron sphere orbiting the Earth instead of on the surface of the Earth, then it becomes more practical to have something orbit it.  At that point, you only need to worry about the tidal force, which is the difference between Earth's gravity on one side of your sphere and Earth's gravity on the other.  If your sphere's gravity is greater than that difference, you're in business.

The relevant equations are in the Wikipedia article for Hill Sphere if you want to calculate it exactly:

https://en.wikipedia.org/wiki/Hill_sphere

To summarize, as other commenters have noted, sadly, it won't work for any reasonable sized iron sphere anywhere below LEO.


Offline whitelancer64

Re: Orbital Mechanics Q&A
« Reply #15 on: 04/06/2016 10:20 pm »
The relevant equations are in the Wikipedia article for Hill Sphere if you want to calculate it exactly:

https://en.wikipedia.org/wiki/Hill_sphere

To summarize, as other commenters have noted, sadly, it won't work for any reasonable sized iron sphere anywhere below LEO.
Yeah, I earlier found a Hill Sphere calculator

http://orbitsimulator.com/formulas/hillsphere.html

and I think a 300 meter wide Iron sphere on the surface of the Earth has a Hill Sphere of about 0.0009 mm. Unless I did the math wrong, which I probably did.

I'm also pretty sure that's about as small as you can make an Iron sphere that has a Hill Sphere larger than its own diameter on the surface of the Earth. Unless, again, I did the math wrong. Which I, again, probably did. Which is, of course, why I'm asking on NSF. I'm certain someone, somewhere on here can math better than I can.
"One bit of advice: it is important to view knowledge as sort of a semantic tree -- make sure you understand the fundamental principles, ie the trunk and big branches, before you get into the leaves/details or there is nothing for them to hang on to." - Elon Musk
"There are lies, damned lies, and launch schedules." - Larry J

Offline DanielW

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Re: Orbital Mechanics Q&A
« Reply #16 on: 04/06/2016 10:41 pm »
You can't do this as proposed. Hill sphere's usually relate to an object that is itself in orbit. because of this you get the benefit of it being in equilibrium with the heavier object by virtue of it's inertia. In the case of a relatively stationary object on the surface of the earth you need to have a surface gravity on your object greater than g.

Offline Alf Fass

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Re: Orbital Mechanics Q&A
« Reply #17 on: 04/18/2016 07:01 am »
If someone built a solid sphere of Iron 300 meters in diameter at some convenient location on the surface of the Earth and built a vacuum chamber around it, would it then be possible to put something small (say, a 0.1 mm diameter bb) into orbit around it? In other words, would this large Iron sphere have a Hill Sphere?

If not, how large would such an Iron sphere have to be to have a Hill Sphere? Would Tungsten be a better (though probably not cheaper) material, thanks to its density, to build something like this with?

I know that due to the 3-body problem this orbit will not be stable. For the purposes of my question, once around would be fine to be an orbit. I know it's wildly impractical idea and probably not possible to build from an engineering standpoint. Think of this as a grand science experiment in gravity and orbital mechanics, and it is paid for and agreed upon by all nations that this is a good idea.

I'll use a sphere with a surface gravity equal to Earth's as a starting point, and to make things a little easier give it a diameter one 10 thousandth of Earth ie. 1.2 km.

Because of the inverse square law, to have the same surface gravity as the Earth the object needs 1/10,000^2 of Earth's mass that is a hundred millionth of the Earth's mass.

Earth's mass is ~6 x 10^24 kg, so the objects mass needs to be 6 x 10^24 / 100,000,000 kg which is 6 x 10^16 kg.

The spherical objects volume is 4/3 pi r^3 which is 900,000,000 m^3.

The objects density would be 6 x 10^16 kg / 900,000,000 m^3 = 66,666,666 kg/m^3 or ~67,000 tonnes a cubic meter.

 Metals like gold, tungsten and platinum have densities around 20 tonnes / m^3, so a 1.2 km diameter sphere would need a density over 3,000 times that of the densest metals to be able to compete with Earth's gravity at Earth's surface.
« Last Edit: 04/18/2016 09:18 pm by Alf Fass »
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Offline dcporter

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Re: Orbital Mechanics Q&A
« Reply #18 on: 08/02/2016 08:13 pm »
What happens if you hit a golf ball off of the space station directly towards Earth? It'll be in the "wrong orbit" for its velocity, but what does that mean? What happens next?

I have a good picture of what happens when you apply a v-bar force to an object in orbit: one of the orbit's foci moves and you immediately enter a new stable orbit — or you escape or collide, but at any rate you enter a new conical cross-section, and you stay there until another force acts on you.

I think I got z-bar forces too — you end up in a non-coplanar version of the orbit you were just in. You basically turn right a little bit, and then keep going on your way. (Not much different than pushing an ice skater sideways, just, on the surface of a sphere.)

But I can't for the life of me picture what happens if you apply an r-bar force. The famous space station golfer shot the ball retrograde so it would enter a lower orbit and degrade, but what would the golf ball have done if he shot it directly down the r-bar towards Earth?

(Let's be physicists and assume no atmosphere.) (Also thanks whoever moved this to the right place.)
« Last Edit: 08/02/2016 08:27 pm by dcporter »

Offline jabe

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Re: Orbital Mechanics Q&A
« Reply #19 on: 08/02/2016 08:29 pm »
I'll take a crack at it...
assume you are in leo in a circular orbit where your v-bar is perpendicular to the r-bar.  your velocity is about 8 km/s.
To find your final velocity if deltav is along the r-bar you simply use Pythagorean theorem to our new slightly higher velocity velocity.  neat thing is same semi major axis regardless if along r- positive ot r - negative.
So now you have a higher energy so your semi major axis is larger. but no longer in a circle but in a small non zero eccentric orbit.  I'll add screen shots in a bit to show different orbit changes.
jb
edit:added pic..green is original circular orbit.  red is new orbit.  i believe focal point is 90's from where rbar vector. other pic is delta v in opposite direction.. of 1st burn
« Last Edit: 08/02/2016 08:45 pm by jabe »

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