Orbital mechanics and sideways forces

Factual Questions
1

There have been a couple of threads recently that seem to have a lot of confusion caused by misunderstandings about orbital mechanics when an object is pushed off to the side - first the astronaut throwing the tool bag around, and second, the asteroid being blown up/deflected.

I know enough about orbital mechanics to know that it isn’t sufficient to point to Newton’s laws about inertia to describe the new path of the object after having been pushed. I know that the object’s velocity is being changed continuously by the force of gravity so inertia is only part of the equation.

What I’m not clear on is exactly what the effect is (and I’m obviously not alone). Is there a rule of thumb or simplified equation that a layman could use to estimate the change in orbit caused by a change in velocity? Or is this the kind of thing where we really have to bite the bullet and take the course in college?

2

Short form: The object is in a new, slightly different, orbit, and the new orbit intersects with the old one at at least one point, and probably comes close to intersecting at one other point. The orbit, being similar, will also have close to the same orbital period as the original, so an object on the new orbit and one still on the original one (say, a spacecraft and a discarded tool) will come close again (and possibly collide) a half-orbit later.

A more detailed answer would require more details of the orbit involved and the perturbation.

3

All periodic orbits are elliptical. (A circular orbit is a degenerate case of an ellipse, with both foci coincident at the center.) When you add change the momentum of an object in orbit (via some kind of accelerating impulse, like thrust or impact) you change the eccentricity of the ellipse; depending on at what point in the orbit and direction, the eccentricity either increases or decreases. The sum of the potential and kinetic energy of the object in orbit is constant throughout the orbit, though the potential energy gets less with reducing distance from the barycenter (common axis of rotation, located at the foci of the ellipse) of the orbit, and the kinetic energy changes with the velocity, which increases as you approach the barycenter.

The point of all this is when you provide a single impulse to an object, it goes into a new orbit that is slightly more or less elliptical, and has more or less energy corresponding to the energy provided or removed with the impulse, and then remains in this orbit perpetually (not accounting for precession or perturbation). It won’t diverge from this new orbit until another impulse is provided, and will otherwise obey Kepler’s laws of orbital motion.

As mentioned, the above doesn’t account for other influences which will act on the orbit over time. Specifically, an orbit that is inclined (out of plane) with the normal rotation of the system will be subject to precession, i.e. the axes of the ellipse will slowly rotate. The object is also subject to gravitational perturbation from other masses in orbit. These influences will (slowly) alter the orbit of the object. It should also be noted that Kepler’s laws are actually an approximation, as the objects in a two-body system are actually in orbit of a common barycenter. (For an n-body system, the barycenter moves in a complex and generally chaotic fashion, which is why such problems are not generally capable of a closed form solution.)

Prussing’s Orbital Mechanics is my preferred text for basic orbital mechanics. The basic understanding of Keplerian mechanics doesn’t require calculus, and the Lagrangian approach doesn’t require much.

Stranger

4

You make it sound like a non-college educated layman can figure out what the initial orbit was like, and it only gets complicated when that orbit gets changed.

I don’t think so.

5

Actually you can, at least in a qualitative sense. Kepler’s laws are a good approximation for most orbital mechanics problems and are accessible geometrically. Most first order estimates of required direct impulse to change orbits or achieve escape or injection speed start by drawing an ellipse (usually on the back of a budget report or envelope) and guestimating what the optimum low energy maneuver would be. There is nothing in Kepler’s laws that is beyond a high school student that has passed basic algebra and planar geometry, and in fact orbital motion is usually addressed (if briefly) in high school physics classes.

Now, doing actual trajectory planning with non-instantaneous impulse (i.e. you have an engine burn of a significant duration, which is true in most real world changes), gravitational swing-by maneuvers, perturbation due to multiple sources, parametric studies to determine optimum orbit, and statistical orbit determination are much more complicated, and require both some intensive calculus, differential equations, multivariate analysis, plus a solid understanding of numerical methods (or, at least a solid understanding of numerical solvers like Matlab or SciPy). But the governing equations and principles of orbital mechanics, while not intuitive, are fairly simple in concept.

Stranger

6

Essentially really it’s just an intial value problem, i.e. to find and solve the equation of motion of the object given it’s intial postion and velocity. Assuming the object’smass is negligible compared to the Earth’s makes it slightly simpler (but not that much, just makes it easier to express in terms of inertial coordinates)…

7

Nitpick: You don’t necessarily change the eccentricity, and you can change any of the other orbital elements as well. For instance, if the impulse is out of the plane of the orbit, you can turn one circular orbit into another circular orbit with a different inclination.

8

I’m not surprised that there isn’t some kind of rule of thumb, I was just hopeful that there might be. I know that plenty of equations in physics have simplified forms that are easier to use, but only true under certain parameters - like how the standard high school equation for distance traveled under constant acceleration eliminates the need for calculus thanks to the assumption of constant acceleration. That’s great for me; I can pull out my calculus book and do it the hard way if I have to, but I can do the simplified equation in my head.

It’d be awesome if there was some correlation like a 1% change to velocity in direction x increased the minor and major axes of the ellipse by 2%. I don’t necessarily expect that to be the case, but it would be nice. It sounds like maybe I already know all of the shortcuts I can without a fuller understanding of the underlying math. (I once wrote a sci-fi story and received a compliment from an engineer who said [paraphrased] “Congratulations! While your story glosses over orbital mechanics entirely, you didn’t say anything wrong either.”)

I suppose at some point I’ll have to just sit down and learn it all in depth. Thanks, SOAT, for the recommendation on the book.

9

I stand corrected. To clarify, any impulse change in the plane of the existing orbit will result in a change of eccentricity; to return to the same eccentricity–say, to go from a circular orbit to a wider circular orbit–will require two or more impulses. Apply enough impulse, and you’ll end up in a parabolic or hyperbolic (escape) trajectory, never to return. If you want to go from one sphere of influence to another–say, from Earth to Lunar orbit–you generally used what is called a “patched conic” method, where you figure out the initial and final orbits and then match a curve that is tangent to both at some point and back out the impulses required. Going from Earth to Mars would require two patched conics (Earth to solar orbit, solar to Mars), and is usually optimized to find a minimum energy trajectory for the impulses.

Nope, it doesn’t work that way. The impact on the orbit depends on specifically where in the orbit you apply the impulse, and the specific vector of the momentum change. Here’s the thing, though; you can start at any point in the orbit of a known object with an initial position and velocity and determine the orbit a priori; you don’t need to know anything about the last position. So if you plot the velocity of an object in one orbit at a given time, add an instantaneous momentum change of a known quantity and vector, you can calculate the parameters of the new orbit.

The only general thing we can say about orbital mechanics of elliptical orbits is parahrased from Larry Niven’s Integral Trees: in terms of impulse vector on motion and speed, retrograde (against the direction of motion) takes you down, down takes you prograde, prograde takes you up, and up takes you retrograde. Or, in other words, “You speed up to slow down, and you slow down to speed up.”

Robert Heinlein and his wife, Virginia (who held degrees in aerospace engineering and mathematics) once purportedly spent three days calculating out an orbital transfer maneuver on rolls of butcher paper for two lines of text. Given how easy it is to gloss over the details of such maneuvers in fiction, I’d say it’s generally not worth it unless there it is in some way critical to the plot.

Stranger

10

Here are the qualitative effects of an impulse. If you are orbiting the Earth (facing the direction of motion) and throw a wrench:

  1. Ahead of you. From your perspective the wrench will move up (away from the Earth). It will reach its maximum distance above you after half an orbit and will come back to your hand after a full orbit.

  2. Behind you. From your perspective the wrench will move down (towards the Earth). It will reach its maximum distance below you after half an orbit and will come back to your hand after a full orbit.

  3. To the side. From your perspective the wrench will move to the side you threw it. It will reach its maximum distance after a QUARTER of an orbit, will return to your had after half an orbit and then swing out to the other side reaching maximum distance on the other side after 3/4 of an orbit and return back to your hand again after a full orbit.

As for quantitative results, from the OP you seem most interested in case 3, so lets look at that. Lets make it really easy and say that you are in a circular orbit, you instantaneously throw your wrench, and that you throw it slightly backward so that it has the same speed as you (just in a different direction that is not inclined towards or away from the Earth).

The angle of the wrench’s velocity vector to yours is 2asin(S/2V) where S is the new speed you impart to it and V is the magnitude of its velocity vector before and after you throw it. If you want to know the angle between your two orbits, we make a small angle assumption and then convert from radians to degrees and we get that the angle is 180S/V. The maximum distance the wrench will be from you is rS/V where r is the radius of your orbit (note that the maximum distance does not assume small angles).

That tells you that the maximum straight line distance your wrench will be from you is the percentage of your orbital speed that you throw it times the radius of your orbit and the angle is the percentage of your orbital speed that your throw it times 180.

I will cover forward an back throws in a minute.

11

OK, there are a couple errors in 1 and 2 which I will correct, but part 3 and the math after is correct.

12

No, you can keep the same eccentricity there, too. Given any two orbits that intersect, you can switch from one to the other with a single impulse. Consider two different coplanar elliptical orbits, oriented differently, for instance.

13

The vis-viva equation (described here Vis-viva equation - Wikipedia) is pretty useful for this kind of thing. It’s

(magnitude of orbital velocity) squared=G*(M+m) (2/r-1/a)

where r is the current distance between the orbiting object and the object being orbited, a is the semimajor axis, and M and m are the masses of the orbiting object and the object being orbited. For a circular orbit, a=r, so you automatically get the velocity for a circular orbit out of this equation, and if you change velocity, you can see how a changes (just add the velocities correctly - remember that velocity is a vector. It’s easy to use this equation to figure out a least-energy transfer orbit, for example - if you’re in a circular orbit of radius r1 and you want to get to an orbit of radius r2, first calculate the difference between the velocity for a circular orbit of r1 and the velocity for an orbit with a of (r1+r2)/2 at r=r1. That difference is the amount of velocity that needs to be added to move to the transfer orbit. Reverse the process to calculate the delta-v to get from the transfer orbit to an orbit with a=r2 and r=r2, and you’re done.

14

OK, 1 and 2 should be this (assuming an initially circular orbit):

  1. Ahead of you. Initially it will move ahead of you and up (away from the Earth). As you move around the Earth, though, it will keep moving further away from the Earth but you will overtake it and move ahead of it. After about one half revolution it will reach its furthest distance above you and start coming back down, but you will keep pulling ahead of it. In the last bit of the orbit it will start to catch up with you, but will never make it. It will eventually reach the exact point you threw it away a little while after you have already passed through that point yourself.

  2. Behind you. Initially it will move behind you and down (towards the Earth). As you move around the Earth, though, it will keep moving closer to the Earth but you will fall back and behind it. After about one half revolution it will reach its furthest distance below you and start coming back up, but you will keep falling behind it. In the last bit of the orbit you will start to catch up with it, but you will never make it. It will eventually reach the exact point you threw it away a little while before you pass through that point yourself.

15

[quote=“Stranger_On_A_Train, post:9, topic:550589”]

I was just going to post this. IIRC, there was even a litany that had been memorized but I don’t have my copy handy.

16

[quote=“standingwave, post:15, topic:550589”]

East takes you Out,
Out takes you West,
West takes you In,
In takes you East,
Port and Starboard bring you back"

See here for discussion http://www.larryniven.net/physics/img34.shtml

17

I solved a closed-form solution to this. It holds for velocity changes in or opposite the direction of motion when you start with a circular orbit. It gets a bit more complex when you start with an elliptical orbit and depends on where in the orbit you make the change. Anyway, it looks like this:

%r = (%V)^2/(2 - (%V)^2)

where %r is the percent change (divided by 100) in the distance to the Earth at the opposite end of the orbit and %V is the percent change (divided by 100) in velocity. So, if I throw the wrench forward at 1% of my velocity (%V = 1.01), the distance from the Earth at the opposite side of its orbit (the most extreme change) will be 4.1%.

For small changes (which is almost always going to be the case, the percentage change in radius is 4 times the percentage change in velocity. I think this is what you were looking for.

18

Oh, at the four points at which they intersect, sure. The reality, though, is that the lowest energy impulse is going to be done at or near periapsis.

Stranger

19

Nitpick: I’m pretty sure that two ellipses with one shared focus will intersect in two points at most.

20

The mathematics of such phrases as “instantaneous impulse”, “in the direction of orbital motion”, and “perpendicular to the plane of the ellipse” should be mentioned.

Most objects separated from spacecraft were put in a convenient spot with respect to an astronaut, not thrown. But, they were not put perfectly at rest with respect to the astronaut, or the vehicle. When noticed, most were within a few meters of the astronaut, and not drifting at a visible rate. They were still lost irretrievably, because you can’t just go fetch it.

The thing is, the vector was very close to random in magnitude and direction. While not specifically impossible to solve, it’s not something you can do on a mission, because to do so you have to change the orbit of the spacecraft, and then calculate a new orbit to return to your mission rendezvous. Then you have to risk an astronaut’s life, and waste atmosphere, to go EVA to fetch it. Not really within reasonable parameters for most current mission capabilities.

On the other end of the range of problems, nuking an object as it moves toward Earth, the object will not break up into even sized pieces, but will rather break along its existing structural weak points, into objects of highly variable size. It is a necessary consequence of the same mathematics that the largest fragments will change vectors the least, barring some extraordinary coincidence of blast parameters, and object geometry. While the gut level expectation is for it to be a cone of objects blasting to one side or another, the blast characteristics in space will be very dependent on shape, surface characteristics, and tensile and compression strength of the object, which are pretty well unknowable.

Example: The object has a layer of accreted ice, and rock material several meters in depth. The bulk of it’s mass, however is iron, and relatively solid. The nearby nuke blast will not move the central mass nearly as much as it will the portions of the accreted material close inside the tangent cone from the blast center to the radial edges of the object. Even accreted material in the center surface (with respect to the blast center) would be moved from their current trajectory less than the edges. Computing how much blast, and just where to place it is an enormous engineering problem, and necessarily ends up being best guesses, and approximations.

Each fragment has a unique orbit after the first blast. Second attempts must involve immediate recalculation for each significant sized piece, and that requires observations of the new orbits for each as well. Then new calculations must be made for blast size, and placement for each piece that still represents a danger on the current orbital pass, and none of that tells you what happens when the detritus comes around again.

By the way, you have to have your second salvo in orbit, and ready, unless you start on the object several months before impact at a minimum. In the case of Comet Shoemaker-Levy 9, estimates made in six days or less before actual observed impacts were within a range of 1 - 7 minutes for impact with Jupiter. These objects had been under frequent observation by multiple astronomical instruments for more than two years. Three of the objects were later characterized as “missing” since there was no data on impacts for them, or observations showing where they were, if they missed Jupiter.

The math isn’t beyond us theoretically, but pragmatically, it’s too big a problem to solve in the time we would have to solve it by several orders of magnitude.

Tris