Transfer orbits for dummies A hillbilly tutorial.

In summary: L = 257.9806...degrees w = 138.812...degrees T = JD 2453009.3 In summary, Jerry Abbott provides orbital elements for Earth and Vesta, and notes that ellipses and hyperbolas are the only orbits that will be considered by his method. He provides the coordinates in heliocentric ecliptic coordinates, which assumes that the XY plane is the current plane of Earth's orbit. He also notes that all transfer orbits will have an apside at one, but not at both, endpoints of the intended trajectory.
  • #36
A transfer orbit example from "The Rolling Stones" by Robert A. Heinlein

I regard Robert A. Heinlein as one of the greatest science-fiction writers ever. His science was as meticulous as his stories were fun to read. Someone who didn't have a scientific background might think he was only "hand-waving" at number-crunching that he had not actually done. On the contrary; he did it. He just didn't flaunt it.

A good example of his low-profile diligence is the celestial mechanics that forms part of the story of his novel The Rolling Stones. In order to get the full picture of the world behind this story, one would also need to read two other books, namely The Moon is a Harsh Mistress and The Cat Who Walked Through Walls.

Here's some historical background. Hazel Meade Stone was born on 25 Dec 2063, and, as is chronicled in The Moon is a Harsh Mistress she fought the evil Terran military goons with great distinction during the Earth-Moon War of 2076. She later married Slim Lemke. Their children were Roger (b. 22 Sep 2078, an early first child) and Ingrid (born later).

Roger Stone (family name taken from his mother?) married Edith (maiden name unknown) while he was mayor of Luna City (2122-2130). Their children were Meade (b. 2130), the twins Castor and Pullox (b. 2133), and Lowell (b. 2144).

As The Rolling Stones opens, it is early to mid-2148. Castor and Pullox are both 15, however very well-educated by our standards today, especially in mathematics. The twins plan to buy a spaceship and fly off to the asteroid belt, there to make a fortune mining high-grade metal ore. But their father gets wind of the plan and scotches it. The idea of an extended family outing in a larger, more expensive space yacht takes root, however, and before long Roger, assisted by Hazel (who knows how to arm-twist spaceship merchants), has bought a spaceship and is calling himself 'Captain.'

After some harranguing, the ship is named "The Rolling Stone," and Mars is selected as the first destination because the launch window for the minimum energy trajectory from Earth to Mars will soon be open.

In fact, that's quite correct. It opened (in the real world "will open") for departure from Earth around September 2148.

There is a valid transfer orbit, an ellipse with perihelion at departure (6 September 2148), from Earth to Mars, with a transit time of 259 days, with arrival occurring on 23 May 2149. The heliocentric longitude of Earth at departure will be 341.69 degrees; that of Mars at departure will be 23.76 degrees; and that of Mars at arrival will be 152.08 degrees.

Orbital elements of Earth.
a 1.00000011
e 0.01671022
i 0 (zero)
L 0 (zero)
w 102.94719 deg
T JD 2453009.3

Orbital elements of Mars.
a 1.523688
e 0.093405
i 1.8497 deg
L 47.5574 deg
w 286.5016 deg
T JD 2452873.0

Orbital elements of the transfer orbit.
a 1.3411728 AU
e 0.248072
i 10.608 deg
L 341.687 deg
w 0 (zero) deg

The magnitude of the departure delta-vee is 6.8277 km/sec.*
The magnitude of the arrival delta-vee is 4.2570 km/sec.

*Does not take into account the difference between preburn orbital speed and the local escape speed relative to Earth. The "departure" is really the thrust applied near Earth after having dropped toward perigee from the moon, so probably another half kilometer per second (or thereabout) may be needed.

Jerry Abbott
 
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  • #37
Ah, the good old days of science fiction, when any scientist worth his salt could design an interplanetary spacecraft , and build it in his basement or back yard.
 
  • #38
Did Heinlein slip up on the Mars-to-Hallelujah Node transfer orbit?

The main task of a fiction writer is to produce a good story. Science fiction writers are no exception. Heinlein probably liked to get his details straight, and his figuring for the Earth-to-Mars transfer orbit was impeccable.

However, he afterward writes about the Stone family taking their space yacht out to a cluster of asteroids known as "the Hallelujah Node," where a rich strike of "uranium and core metal" (RS, p. 158) had recently been discovered. They decide to go out there, mostly because they don't like Mars (it's one giant, over-regulated tourist trap) and partly because they've never seen asteroids up close. Castor and Pullox figure on selling supplies, girlie pin-ups, chocolate, and cute little furry pets (Martian flatcats) to the miners.

As mentioned in my previous post on this story, the spaceship Rolling Stone reaches Mars on 23 May 2149. They spend one day getting disposed in various ways, with Hazel, Castor, Pullox, Meade, and Lowell taking the shuttle from the Phobos port station down to Mars surface. Moving very quickly, they examine several different hotels for possible accomodations, rejecting them all because of exorbitant prices, until, late in the day, they rent a small apartment at "Casa Manana Apartments" from a Mr. d'Avril (RS, pp. 148-150).

By this time, it is probably 24 May 2149. It is mentioned (presumably the narrator is revealing Hazel's thoughts) that the time for making an economical transit from Mars to Venus will begin 96 "Earth standard" days hence.

(UPDATE: It looks like that Mars-Venus transit was an instance of artistic license by Heinlein. The only transfer orbit that I found from Mars to Venus, with a departure on JD 2506204.2 is a retrograde elliptical orbit with aphelion at Mars and a transit time of 137.2 days. The delta-vee magnitude at departure would be 34.3 km/sec, and the one at arrival would be 68.3 km/sec. That certainly not a minimum energy trajectory.)

Somewhat later in the book, the recently reunited family (Roger Stone had just been released from quarantine, and the twins had just been released from jail) is having a discussion about whether to remain until a return trip to Earth is possible, or whether to go to Venus in six days (RS, pp. 180-185). In other words, the conversation happens on 22 Aug 2149.

They decide, instead to travel to the Hallelujah Node in the asteroid belt. The departure time is said to be "six weeks" from the time of this conversation, or 3 Oct 2149. The transit time is 261 days (RS p. 185), meaning that the arrival date is 21 Jun 2150. The Hallelujah Node is said to travel "almost the same orbit" as Ceres, but "somewhat ahead" of it (RS, p. 187). The delta-vee for departure is said to have a magnitude of "twelve and a half miles per second" (RS p. 185), which is about equal to 20.1 kilometers per second.

Orbital elements of Mars.
a 1.523688 AU
e 0.093405
i 1.8497 deg
L 47.5574 deg
w 286.5016 deg
T 2452873.0

Orbital elements of Ceres.
a 2.7664122 AU
e 0.07911582
i 10.58348 deg
L 80.48630 deg
w 73.98448 deg
T 2453197.5 (here's where the problem is)

Departure from Mars at JD 2506242.2
Arrival at Hallelujah Node at JD 2506503.2

The problem is that there is no valid transfer orbit between Mars and (anywhere close to) Ceres with these times for departure and arrival. Rather, the valid transfer orbit from Mars' position at this time of departure, with the specified transit time of 261 days, has an arrival point on Ceres' orbit that is nearly in opposition to Ceres, with respect to the sun.

Here's what I think happened. Heinlein wasn't given a time of perihelion passage for Ceres; he was given a mean anomaly. While correcting the mean anomaly to the interval [0,2 pi), he mistakenly added pi instead of 2 pi to this angle. Or so I imagine. It can happen to anybody.

Other than that, Heinlein's celestial mechanics was pretty good.

Orbital elements of Hallelujah Node.
a 2.7664122 AU
e 0.07911582
i 10.58348 deg
L 80.48630 deg
w 73.98448 deg
T 2454003.8 (this time of perihelion passage makes the transfer orbit valid)

The transfer orbit from Mars at JD 2506242.2 to Hallelujah Node at JD 2506503.2 is an ellipse with aphelion at arrival. (As Heinlein predicted.)

Elements of the transfer orbit.
a = 1.551959 AU
e = 0.782885
i = 2.801 deg
L = 201.753 deg
w = 227.622 deg

The anomalies of departure:
Mean: 46.925 deg
Ecc.: 91.760 deg
True: 142.607 deg
The anomalies of arrival: all pi radians.

Heliocentric longitudes.
Mars at departure: 211.970 deg
Hallelujah at arrival: 249.341 deg
Hallelujah at departure: 189.375 deg

Magnitude of departure dV: 21.954 km/sec (fairly close to Heinlein's value)
Magnitude of arrival dV: 9.957 km/sec

Transit time: 261 days (I insisted on Heinlein's value as program input.)

Jerry Abbott
 
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  • #39
I thought that this thread had been closed. Oh well.

I've written an improved procedure. The main improvements are:

1. A more straightforward calculation of the true anomaly at the non-apsidal endpoint of the intended trajectory (i.e., either departure or arrival). In my previous paper on this subject, I missed the obvious fact that this angle can be found quite earlier in the procedure (and with much less rigmarole).

2. A more immediate solution for the calculated transit time, dt, which must be equal, or very nearly equal, to the required transit time t2-t1. This saves the user time, since he shouldn't bother with solving for the angular orbital elements if the hypothetical orbit isn't going to work out due to a mismatch in required and calculated transit times.

3. A consolidation of the four "cases" for calculating the eccentricity of the hypothetical transfer orbit into a single equation containing a sign toggle variable.

The time of departure, t1, and the time of arrival, t2, are selected by the user at the beginning. The required transit time may be found immediately, since it is simply their difference. The calculated transit time, on the other hand, is a function of the change in mean anomaly in the transfer orbit between departure and arrival, and the transfer orbit's mean motion.

Also of interest is the fact that I've found an asteroid that can be diverted into a collision with Earth with a departure delta-vee of only ~83 meters per second. The asteroid has the generic name of 2001-YB5, and I use it as my example in the new, improved procedure, which you can find at

http://jenab6.livejournal.com/12053.html

Jerry Abbott
 
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  • #41
Question... what math level, in general, is this all working with?
Hah... I feel so empowered, with all this knowledge and all, but I don't understand half the calculations in this thread.
 
  • #42
Hi Jerry

Like you I wanted to learn Astrodynamics after reading Heinlein since his figuring always seemed so precise. He wasn't always accurate though since he had to work hard with a slide-rule back in the pre-Calculator days of SF, but usually he was right. One orbit I've never been able to figure out is how the "Mayflower" in "Farmer in the Sky" could fly on a hyperbolic orbit to Jupiter and be high enough over the ecliptic plane to avoid the Asteroid Belt. Any thoughts?

In that same book he mentions the Ganymede Colony was slowly built up via a fleet of small ships flying 1,000 day orbits out to Jupiter... back in the day when we didn't know how bad cosmic ray damage could be and knew nothing of solar flares.
 
  • #43
Heinlein is my personal hero. I am a writer and I am on my first book, I am emulating him by learning my physics. Regrettably my book is almost done so I have to get some friends to help me finish up my equations lol.

And BTW I resent the term "Hillbilly" especially when referring to it as an insult on intelligence. I have news for everyone, country people really aren't as portrayed on TV, believe it or not we have colleges, dentists, and we don't wear coveralls or straw hats. If you guys ever make it to NASA, AMES, or anywhere where the best and brightest are assembled you will find a hillbilly among them.
 
  • #44
Nietsnie314 said:
Question... what math level, in general, is this all working with?
Hah... I feel so empowered, with all this knowledge and all, but I don't understand half the calculations in this thread.

Algebra, trigonometry, vectors, and occasionally a little bit of calculus. The ease of doing the calculations, or in finding out which calculations are right to do, is as much dependent on being able to visualize the goings-on as it is understanding the math.
 
  • #45
qraal said:
Hi Jerry

Like you I wanted to learn Astrodynamics after reading Heinlein since his figuring always seemed so precise. He wasn't always accurate though since he had to work hard with a slide-rule back in the pre-Calculator days of SF, but usually he was right. One orbit I've never been able to figure out is how the "Mayflower" in "Farmer in the Sky" could fly on a hyperbolic orbit to Jupiter and be high enough over the ecliptic plane to avoid the Asteroid Belt. Any thoughts?

In that same book he mentions the Ganymede Colony was slowly built up via a fleet of small ships flying 1,000 day orbits out to Jupiter... back in the day when we didn't know how bad cosmic ray damage could be and knew nothing of solar flares.

If a spaceship first enters an orbit that takes it above or below the ecliptic, it can then do a course adjustment that makes the departure point (or trajectory adjustment point) the perihelion of a hyperbolic transfer orbit to Jupiter that avoids the ecliptic until arrival. It just can't begin in the ecliptic with a hyperbolic transfer orbit and reach Jupiter, avoiding the ecliptic along the way. So a course correction would be required. But I think Heinlein probably suffered a lapse of thinking that time.
 
  • #46
emc2cracker said:
Heinlein is my personal hero. I am a writer and I am on my first book, I am emulating him by learning my physics. Regrettably my book is almost done so I have to get some friends to help me finish up my equations lol.

And BTW I resent the term "Hillbilly" especially when referring to it as an insult on intelligence. I have news for everyone, country people really aren't as portrayed on TV, believe it or not we have colleges, dentists, and we don't wear coveralls or straw hats. If you guys ever make it to NASA, AMES, or anywhere where the best and brightest are assembled you will find a hillbilly among them.

Right. The reason I used the term hillbilly was to make that very point. I'm a hillbilly living in the West Virginia Allegheny Mountains. I'm a celestial mechanic. I have been employed as a defense contractor physicist. My last job, before I retired, was book editor for a small publishing company run by another hillbilly, who happened to be a retired physics professor.

I prefer an Adams cotton denim fishing hat with a 2.5 inch wide brim. I do sometimes wear overalls, but more often it's blue jeans and a blue denim cotton or tencel button shirt, with Merrell Primo Moc shoes. In the summer, that is. In the winter, I'm usually wearing Woolrich wool hunting pants (with suspenders), a Brooks Brothers cashmere sweater, a Columbia Titanium Omnitech windbreaker, a Carhartt pullover acrylic facemask hat, Woolrich "Big Woolly" merino wool socks, and Montrail Torre boots.
 
<h2>1. What is a transfer orbit?</h2><p>A transfer orbit is a path that a spacecraft takes to move from one orbit to another. It is used to change the altitude, inclination, or both of the spacecraft's orbit.</p><h2>2. How is a transfer orbit different from a regular orbit?</h2><p>A transfer orbit is different from a regular orbit because it is not a closed loop. It is an open-ended path that allows the spacecraft to reach a different orbit.</p><h2>3. What is the purpose of a transfer orbit?</h2><p>The purpose of a transfer orbit is to efficiently move a spacecraft from one orbit to another without using too much fuel. This is important for space missions as it allows the spacecraft to conserve fuel and reach its destination more efficiently.</p><h2>4. How are transfer orbits calculated?</h2><p>Transfer orbits are calculated using mathematical equations and computer simulations. Scientists and engineers use the spacecraft's current orbit and the desired orbit to determine the necessary trajectory for the transfer orbit.</p><h2>5. Are there different types of transfer orbits?</h2><p>Yes, there are different types of transfer orbits, such as Hohmann transfer orbit, bi-elliptic transfer orbit, and low-energy transfer orbit. Each type has its own advantages and is used for specific missions depending on the spacecraft's capabilities and destination.</p>

1. What is a transfer orbit?

A transfer orbit is a path that a spacecraft takes to move from one orbit to another. It is used to change the altitude, inclination, or both of the spacecraft's orbit.

2. How is a transfer orbit different from a regular orbit?

A transfer orbit is different from a regular orbit because it is not a closed loop. It is an open-ended path that allows the spacecraft to reach a different orbit.

3. What is the purpose of a transfer orbit?

The purpose of a transfer orbit is to efficiently move a spacecraft from one orbit to another without using too much fuel. This is important for space missions as it allows the spacecraft to conserve fuel and reach its destination more efficiently.

4. How are transfer orbits calculated?

Transfer orbits are calculated using mathematical equations and computer simulations. Scientists and engineers use the spacecraft's current orbit and the desired orbit to determine the necessary trajectory for the transfer orbit.

5. Are there different types of transfer orbits?

Yes, there are different types of transfer orbits, such as Hohmann transfer orbit, bi-elliptic transfer orbit, and low-energy transfer orbit. Each type has its own advantages and is used for specific missions depending on the spacecraft's capabilities and destination.

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