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## Transfer orbits for dummies! A hillbilly tutorial.

It sounds like the first rocket is taking advantage of the gravity well of a planet by holding on to its fuel's mass until most of its potential energy has become kinetic energy, then blowing the fuel away contrary to the spaceship's velocity, which transfers some (how much depends on the exhaust speed) of the fuel mass's kinetic energy into the spaceship. The ship then rises out of the gravity well still having all of the energy from the fall, the energy from the thrust, plus the gravitational potential energy stolen from the fuel.

For the second spaceship, there's also an advantage to losing the weight of the propellant as fast as possible. The least efficient (zero efficiency) launching rocket is one that can do no more than hover by making thrust equal only to its own weight. When launching from a planet with an atmosphere, though, there may be a tradeoff between thrust and aerodynamic pressure, so that you'd want to get above the soupy part of the atmosphere before flooring the pedal.

I'm really not much of an expert in astrodynamics. I really am a hillbilly. I live in the Alleghenies near Hillsboro, West Virginia, where I raise goats and grow potatoes and apple and walnut trees. Now and then I like to figure things out, like I did with transfer orbits a few years ago.

Jerry Abbott
 Depends on what you want I guess. If you want to gain speed using planetary a flyby, it's more advantageous to fire at pericentre. This is because speed is largest there. The speed that you add by burning fuel will always be the same (it depends on the fuel, engine, etc). So, because kinetic energy goes with the square of the velocity, adding that same velocity to a large speed will give you a larger increase in kinetic energy (and because potential energy stays constant during the burn, the total energy increase is the same as the kinetic energy increase for both cases.)
 Recognitions: Science Advisor For the case of the modest rocket going straight up and straight down in a constant-g field... If we ignore the effect of an atmosphere, we ought to be able to graph apogee vs. thrust and get some sort of curve, with fixed parameters being structure mass, propellant mass, and total impulse, and the constraint that thrust times burn time equals the total impulse. As Jenab noted, "The least efficient (zero efficiency) launching rocket is one that can do no more than hover by making thrust equal only to its own weight." To my intuition, that means the curve will not have its peak over toward the left side of the graph where the thrust is low. In fact, my intuition suggests the curve is monotonically increasing with increasing thrust. (In the real world, structure mass would have to increase when thrust gets bigger to keep the rocket from collapsing or blowing up, so that its non-constancy would have to be taken into account, so the real-world version of the graph would actually have some maximum apogee for some finite value of thrust, beyond which increasing thrust lowers the apogee; "the point of diminishing returns" you might call it.) Remcook, when you say, "because potential energy stays constant during the burn," are you making the approximation that the burn time is short compared to the time it takes for the gravitational potential to change appreciably when the probe is near its minimum point in the potential field? And another question for Jenab: can a West Virginian still grab onto a slow-moving coal hopper and go places in your state? Or are those days long gone? As it happens, I recently read an article on M&K Junction in West Virginia, across the Cheat River from Rowlesburg.

 Remcook, when you say, "because potential energy stays constant during the burn," are you making the approximation that the burn time is short compared to the time it takes for the gravitational potential to change appreciably when the probe is near its minimum point in the potential field?
Yes

Both in apogee and in perigee the velocity is perpendicular to the gravity acceleration anyway (no gravity loss). But the satellite moves quicker at perigee of course. But if the burn time is small...

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 Quote by Janitor And another question for Jenab: can a West Virginian still grab onto a slow-moving coal hopper and go places in your state? Or are those days long gone? As it happens, I recently read an article on M&K Junction in West Virginia, across the Cheat River from Rowlesburg.
I never tried jumping on a coal hopper. I'm in Pocahontas County, about 4 miles NW of Hillsboro, or 2 miles NW of Mill Point, or 8 miles SW of Marlinton, or 33 miles north of Lewisburg. The nearest river is the Greenbriar, which runs N-S a little to the east of Hwy 219. I moved up here from Alabama in 1998.

Jerry Abbott
 Recognitions: Science Advisor Thanks for confirming, Remcook. Jenab, M&K Junction is three counties north of you in Preston. Ever been to Harper's Ferry or Spruce Knob? I have not, but those are places on my to-do list, to get to one of these years.
 Recognitions: Science Advisor I don't travel much, except locally to shop for groceries, animal feed, to buy building supplies/tools, or to send and receive mail. I came originally to West Virginia to work for Dr. William Pierce as an editor for National Vanguard Books. When he died, the National Alliance came under new management, and I soon discovered that I had irreconcilable differences with the new chief, so I quit. Now I live very pastorally on my ranch/orchard, and circumstances don't give me a lot of time for sightseeing. Once I chased a presumptuous young black bear away from my trash can, off my porch and into the woods. Another time, I shot a bobcat as he was stalking my chicken, Mr. Bronfman; but unfortunately the same chicken met an untimely demise two weeks later under the talons of a hawk. My previous cat, Father Wiggly, had a run-in with the bobcat a week before I had my own show-down with it. ("Bobcat, this ranch isn't big enough for the two of us.") Wiggles managed to get away, but he was scarred up some. But he disappeared later that year and never showed up again. So now I have two new kittens, Spooky and Goblin. Most of my trouble with wildlife, though, comes from deer entering the property to nibble on my young apple trees. I have each tree caged inside a circle of fencing 3 feet wide and six feet high. That works, most of the time, but I lost three of my 30 trees last winter to rabbits nibbling off the bark at the bottom. So now I've added another circumferential layer of chicken wire down there. Jerry Abbott
 Recognitions: Gold Member Staff Emeritus Sounds like a beautiful life. Are you anywhere near Spencer? That's where my father grew up. He used to spend summers on his uncle's farm and had a bunch of stories just like yours that he used to tell me.

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 Quote by selfAdjoint Sounds like a beautiful life. Are you anywhere near Spencer? That's where my father grew up. He used to spend summers on his uncle's farm and had a bunch of stories just like yours that he used to tell me.
It's a good place to live, especially if you like being close to nature. I like to sit on my front porch and read. My house is on a rise, with a road coming up from a valley to the south and halfway circling my house on a spiral upward. The hilly nature of the area is such that my front porch faces an even higher rise, beyond the road, to the NE. Yet higher (and more distant) hilltops are visible to the north and west. And there's woods covering most of the hills.

On a windy day, the leaves make a very nice susurration, which is a kind of leafy "white noise," during which the trees seem to oscillate between dark green (when you see the tops of the leaves) and very light greenish white (when the wind shows you the lighter flip side of the leaves). And on top of that, there's cloud-play alternately shadowing the foreground, then the background, etc.

But the best thing about this area is that it is smack in the middle of a demographic safe-zone. When fossil fuels become, in 20 years or so, too expensive for government subsidies to keep mechanized agriculture going, there's going to be famine along the coast and in urban areas generally, which means that the same people who will rob you now for the money in your wallet will be breaking into homes looking for cans of beans or sacks of rice.

These "safe-zones" I refer to are areas with low population density, especially in regard to demographic groups that have an elevated statistical propensity for causing crime. Pardon the circumlocution. We'll have a food shortage here, but we have a chance at maintaining ourselves with our own efforts, and there will be fewer bandits to contend with.

Spencer is west of me. It's in western West Virginia, over by Ohio. I'm in eastern West Virginia, over by Virginia.

Jerry Abbott
 Recognitions: Science Advisor Our federal government in the form of the I.R.S. and welfare agencies has a bad track record of rewarding people for having lots of kids. My opinion is that coming shortages of fuel plus coming shortages (out West anyway) of water are aggravated by population growth. I doubt whether more than a tiny handful of congresspersons would agree with me on that, though.
 Recognitions: Science Advisor It looks like I allowed roundoff error to accumulate as I lazily truncated decimals to save button pushes on the calculator. I just now programmed the transfer orbit procedure and got a transit time of 224.85 days for the spaceship, as compared with 225.1 days for the Earth. So my elliptical transfer orbit was better than I thought. Transfer orbit from Vesta (JD 2453040.3) to Earth (JD 2453265.4). Aphelion at departure. a = 1.319533 AU e = 0.6519092 i = 0.2289958 degrees L = 353.5454 degrees w = 112.0761 degrees True anomaly of arrival: 247.9239 degrees. Delta-vee at departure (HEC) dV1x = -8.772 km/sec dV1y = -1.514 km/sec dV1z = +2.619 km/sec Delta-vee at arrival (HEC) dV2x = +20.487 km/sec dV2y = +1.515 km/sec dV2z = -0.103 km/sec A coordinate rotation will get the velocity vector into celestial coordinates, which will permit the spaceship pilot to orient his thrust by observing the starfield into which he must accelerate, in order to enter the transfer orbit. Jerry Abbott
 Recognitions: Science Advisor If the spaceship pilot has a star atlas referred to ecliptic coordinates, he won't need to do this step. But since most star atlases use celestial coordinates, I thought it best to include the rotation from ecliptic to celestial. dVx' = dVx dVy' = dVy cos q - dVz sin q dVz' = dVy sin q + dVz cos q Where [q] is Earth's obliquity at the moment of thrust. The J2000 value of the obliquity is q = 23.439281 degrees = 0.40909263 radians The magnitude of the delta-vee is independent of the rotation, of course, dV' = dV = { dVx^2 + dVy^2 + dVz^2 }^0.5 The right ascension of the delta-vee (hence also the thrust) is dVRA = arctan2( dVy' , dVx' ) The declination of the delta-vee (hence also the thrust) is dVDEC = arcsin( dVz' / dV' ) Putting in the numbers... Departure. dV1x = -8.772 km/sec dV1y = -1.514 km/sec dV1z = +2.619 km/sec dV1 = 9.279 km/sec dVRA1 = 13h 1m 57.4s dVDEC1 = +11d 11m 8s The departure thrust will be roughly toward Vindemiatrix, Virgo. Arrival. dV2x = +20.487 km/sec dV2y = +1.515 km/sec dV2z = -0.103 km/sec dV2 = 20.544 km/sec dVRA2 = 0h 4m 11.0s dVDEC2 = 0d 1m 29s The arrival thrust will be toward a point in Pisces. Jerry Abbott
 Jenab: if I may ask: why did you post this stuff? I mean..how did you come up with this? Why would you want to calculate this? Ehm..can't find the right question to ask. Hope I am not rude. you're doing a great job just wondering...

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 Quote by Jenab *SNIP dV1 = 9.279 km/sec dVRA1 = 13h 1m 57.4s dVDEC1 = +11d 11m 8s The departure thrust will be roughly toward [arrgh! I don't have a star atlas!] The arrival thrust will be roughly toward [arrgh! I don't have a star atlas!]
I can't vouch for (or against) any of these, but some are free, and all can work on your PC (so they claim):
star atlases

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 Quote by remcook Jenab: if I may ask: why did you post this stuff? I mean..how did you come up with this? Why would you want to calculate this? Ehm..can't find the right question to ask. Hope I am not rude. you're doing a great job just wondering...
There's really two questions here. First, what provoked a goatherd into learning celestial mechanics? Second, what prompted me to post a demonstration on calculating transfer orbits here?

It would be poetic for me to point out that herdsmen have been watching the sky for thousands of years and must somehow acquire a natural curiosity about how things work up there. But really what got me started on celestial mechanics were two books by Robert A. Heinlein: The Moon is a Harsh Mistress and The Rolling Stones. In both books, celestial mechanics plays a role in the story.

I decided that I'd see how tough it was to learn how to fly a spaceship by the seat of my pants, even if our ridiculous Government would never let me get close to one. It turns out that basic astronavigation isn't as hard as it's usually cracked up to be. The concepts are fairly simple. It's the presentation that's usually wrong: too bent on throwing the history and derivation behind every physical law discovered since Ptolemy into unprepared faces.

The way to teach celestial mechanics is the way I've done it here. Put all the important equations out in the open. Minimize on the derivations - those can come later as advanced, further reading topics. Keep the notation simple, at the high school algebra level as much as possible. Follow the essential equations with a fully worked-out example. Dish the work out in bite-sized chunks.

Voila, now anybody can follow the procedure. Or almost.

 Quote by Nereid I can't vouch for (or against) any of these, but some are free, and all can work on your PC (so they claim): star atlases
Thanks! I downloaded one of the free ones. And I bought a 20th edition Norton's Star Atlas from Amazon.com.

Jerry Abbott
 Recognitions: Science Advisor In the example used for calculating a transfer orbit, we had a spaceship departing from Vesta at a certain time. We used Vesta's orbital elements and the departure time to calculate the preburn state vector for Vesta (hence also for the rocket) immediately before the rocket fires its engines to enter the transfer orbit. Vesta is in an elliptical orbit, and the method shown for obtaining the preburn state vector was the method appropriate for elliptical orbits. But suppose the rocket had been in a hyperbolic orbit, relative to the sun, instead? The calculation must proceed somewhat differently, in that case. There is, of course, no period associated with a hyperbolic orbit. But we can determine an equivalent to the mean motion: m = ( GMsun / a^3 )^0.5 Where GMsun = 1.32712440018E+20 m^3 sec^-2 and remember to enter the semimajor axis of the hyperbolic orbit in meters. One astronomical unit equals 1.49597870691E+11 meters. Likewise, the mean anomaly has no special geometric meaning for a hyperbolic orbit, but it nonetheless remains mathematically convenient as an intermediate quantity. The mean anomaly is zero at perihelion, negative prior to perihelion, and positive after perihelion. M = m (t - T) where [t] is the moment of interest (e.g. the time of departure) and [T] is the time of perihelion passage. This difference of time is entered in seconds, and M will result in radians. Writing the equation fully: M = {GMsun / a^3)^0.5 (t - T) If you'd rather input astronomical units for [a] and days for (t-T), then M = 0.01720209895 (t-T) a^-1.5 AU^1.5 day^-1 and, again, M will be in radians. It is important to remember that we do not correct the mean anomaly of hyperbolic orbits to the interval [0,2 pi). If it comes out negative, leave it that way. Kepler's equation for hyperbolic orbits is M = e sinh u - u Where (u) is the hyperbolic eccentric anomaly, which, along with (M), must be in radians. As it was in the elliptical case, the equation is transcendental in the variable that we are trying to find, and we must use a differential calculus method for solving it. Danby's Method for finding the eccentric anomalies of hyperbolic orbits. u(0) = 0 Repeat over index j f0 = e sinh u(j) - u(j) - M f1 = e cosh u(j) - 1 f2 = e sinh u(j) f3 = e cosh u(j) d1 = -f0 / f1 d2 = -f0 / [ f1 + (d1)(f2)/2 ] d3 = -f0 / [ f1 + (d1)(f2)/2 + d2^2 (f3)/6 ] u(j+1) = u(j) + d3 Until |u(j+1)-u(j)| < 1E-12 The converged value for (u) from this loop is the eccentric anomaly for the hyperbolic orbit. We don't correct (u) to the interval [0,2 pi) either; if it comes out negative, we leave it that way. Finding the canonical position vector. The true anomaly is found from Q' = arccos { (e - cosh u) / (e cosh u - 1) } if u>0 then Q = Q' if u=0 then Q = 0 if u<0 then Q = 2 pi - Q' The heliocentric distance is r = a (e cosh u - 1) The canonical position vector is x''' = r cos Q y''' = r sin Q z''' = 0 The canonical velocity vector is Vx''' = -(a/r) { GMsun / a }^0.5 sinh u Vy''' = +(a/r) { GMsun / a }^0.5 (e^2 - 1)^0.5 cosh u Vz''' = 0 The triple-primed position and velocity, although relative to the sun, are not yet presented in the heliocentric ecliptic coordinate system. They are each rotated (negatively) by the orbital elements w (about the z''' axis), i (about the x'' axis), and L (about the z' axis) in order to appear in the (unprimed) HEC system. A check on the magnitude of the velocity (i.e., the sun-relative speed) is available: Vx^2 + Vy^2 + Vz^2 = GMsun { 2 / (x^2 + y^2 + z^2)^0.5 + 1/a } The state vector of an object in a hyperbolic orbit of elements [ a , e , i , L , w , T ] at the moment of interest [t] is [ x , y , z , Vx , Vy , Vz ] Jerry Abbott