KFL@MC.LCS.MIT.EDU ("Keith F. Lynch") (02/22/86)
From: jon@csvax.caltech.edu
Ulysses is a European probe, to be launched from the Shuttle,
which will go over one the Sun's poles ...
Originally there was to have been an American probe launched at
the same time to go over the other solar pole. Funding was cut a few
years ago, making the Europeans rather mad as I recall.
How can it go over one pole but not both?
...Keithjon@CSVAX.CALTECH.EDU (02/24/86)
> From: "Keith F. Lynch" <KFL@mc.lcs.mit.edu> > > From: jon@csvax.caltech.edu > > Ulysses is a European probe, to be launched from the Shuttle, > which will go over one the Sun's poles ... > Originally there was to have been an American probe launched at > the same time to go over the other solar pole. Funding was cut a few > years ago, making the Europeans rather mad as I recall. > > How can it go over one pole but not both? > ...Keith Well, the two probes could have run into one another on the far side of the sun... (:-) I don't know if there are any plans to keep talking to Ulysses after the first polar flyby. Maybe the geometry will be wrong, or the spacecraft will lose radio lock on Earth while flying 'behind' the sun. -- Jon (jon@csvax.caltech.edu) __@/
desj@brahms.BERKELEY.EDU (David desJardins) (02/25/86)
In article <27219.509662022@csvax.caltech.edu> jon@CSVAX.CALTECH.EDU writes: >> Ulysses is a European probe, to be launched from the Shuttle, >> which will go over one the Sun's poles ... >> >> How can it go over one pole but not both? > > I don't know if there are any plans to keep talking to Ulysses >after the first polar flyby. Maybe the geometry will be wrong, or the >spacecraft will lose radio lock on Earth while flying 'behind' the >sun. I don't have any idea what these people are talking about. The whole idea is that Ulysses is being launched to rendezvous with Jupiter just like Voyager and Galileo, but it will swing around Jupiter and back over the solar pole. Presumably it will then leave the solar system; what would cause it to circle around to the other pole?? -- David desJardins
brahms@spp5.UUCP (Bradley S. Brahms) (02/26/86)
> I don't have any idea what these people are talking about. The whole >idea is that Ulysses is being launched to rendezvous with Jupiter just >like Voyager and Galileo, but it will swing around Jupiter and back over >the solar pole. Presumably it will then leave the solar system; what >would cause it to circle around to the other pole?? Well, the same force that whould cause Ulysses to go around Jupiter could cause Ulysses to orbit the sun, namly gravity. If the trajectory was correct, Ulysses would be caught by the sun and orbit over the pools. However, I have no idea if this is the plan or not. -- Brad Brahms usenet: {decvax,ucbvax,ihnp4}!trwrb!trwspp!brahms arpa: Brahms@usc-eclc
jrv@MITRE-BEDFORD.ARPA (James R. Van Zandt) (02/28/86)
>brahms!desj@ucbvax.berkeley.edu (David desJardins) >In article <27219.509662022@csvax.caltech.edu> jon@CSVAX.CALTECH.EDU writes: >> Ulysses is a European probe, to be launched from the Shuttle, >> which will go over one the Sun's poles ... >> >> How can it go over one pole but not both? It *will* go over both poles. > I don't have any idea what these people are talking about. The whole >idea is that Ulysses is being launched to rendezvous with Jupiter just >like Voyager and Galileo, but it will swing around Jupiter and back over >the solar pole. Presumably it will then leave the solar system; what >would cause it to circle around to the other pole?? It can't leave the solar system because it doesn't have enough energy. It'll be in an elliptical orbit around the sun like any planet, but the plane of the orbit will be tilted with respect to the plane of (for example) the earth's orbit. If it were easy to supply the delta V to place the probe far out of the plane of earth's orbit, we wouldn't have to use Jupiter in the first place. As it is, Jupiter is a convenient place to "turn a corner" in space. I suspect that scientists wanted two probes so they could observe both poles of the sun *at the same time*. With only one probe the long delays that make it difficult to correlate what's seen at the two poles. Does anyone know what the period and inclination of the probe's orbit will be? - Jim Van Zandt
desj@brahms.BERKELEY.EDU (David desJardins) (02/28/86)
In article <8602280148.AA08000@mitre-bedford.ARPA> jrv@MITRE-BEDFORD.ARPA (James R. Van Zandt) writes: >>>Ulysses is a European probe, to be launched from the Shuttle, >>>which will go over one the Sun's poles ... >>>How can it go over one pole but not both? > >It *will* go over both poles. >It can't leave the solar system because it doesn't have enough energy. >It'll be in an elliptical orbit around the sun like any planet, but the >plane of the orbit will be tilted with respect to the plane of (for >example) the earth's orbit. If it were easy to supply the delta V to >place the probe far out of the plane of earth's orbit, we wouldn't have >to use Jupiter in the first place. As it is, Jupiter is a convenient >place to "turn a corner" in space. > ... Does anyone know what the period and inclination of the probe's >orbit will be? While I don't have the information necessary to give a definitive answer (hopefully someone with better information can do that), I feel I can add something to the above. The use of Jupiter is to deal with momentum problems; in order to pass over the pole of the sun one needs (1) to cancel the orbital momentum of the Earth and (2) to add momentum out of the plane of the ecliptic. Passing around Jupiter is a convenient way to change the momentum vector in almost any way desired, subject of course to certain energy constraints. Since presumably after the Jupiter encounter the spacecraft will not pass near any other planets, the path that it takes will be completely determined by its velocity after the Jupiter encounter. If its kinetic energy is greater than the depth of the solar potential well it will pass by the Sun and follow a hyperbolic path out of the solar system. If its kinetic energy is below this threshold it will enter an elliptic orbit (with aphelion at least as far from the Sun as Jupiter and thus a period of several years) and so eventually will indeed pass over both poles. A quick computation (mv^2/2 = GMm/r) gives an escape velocity of about 1.9 x 10^4 m/s, or 11.6 mi/s, which is high but possible. If Ulysses' velocity after the Jupiter encounter exceeds this (in any direction) it will escape. I think that it will leave the Jupiter encounter at a *slower* speed than it arrived due to the use of Jupiter to kill its angular momentum (the opposite of the "slingshot effect"), so it would have to arrive at a higher velocity. For comparison, the Earth's orbital velocity is 3.0 x 10^4 m/s, but of course the spacecraft will lose energy in moving from Earth's orbit to Jupiter's. If anyone does have definitive information (so we can end all of this silly speculation :-)) the numbers I would most appreciate are the actual velocity after the Jupiter encounter and the distance the spacecraft will pass from the solar pole. But any reliable data would be appreciated. Thank you... -- David desJardins
mcgeer%ji@UCBVAX.BERKELEY.EDU (Rick McGeer) (02/28/86)
When Ulysses is in its cometary orbit, what will be its closest approach to earth? Rick.
mangoe@umcp-cs.UUCP (Charley Wingate) (03/01/86)
>> Ulysses is a European probe, to be launched from the Shuttle, >> which will go over one the Sun's poles ... >> Originally there was to have been an American probe launched at >> the same time to go over the other solar pole. Funding was cut a few >> years ago, making the Europeans rather mad as I recall. Actually, there is an American instrument on Ulysses, built by JHU/APL [insert sales pitch for Space Division here]; there were to be European instruments on the American vehicle. Between this and the Halley fiasco, I doubt the ESA is going to be interested in doing much with NASA for a while. >> How can it go over one pole but not both? > I don't know if there are any plans to keep talking to Ulysses >after the first polar flyby. Maybe the geometry will be wrong, or the >spacecraft will lose radio lock on Earth while flying 'behind' the >sun. I forgot to ask my father about the orbital dynamics, but the statement that it will cross one pole but not the other indicates to me that it's going to take a hyperbolic orbit, and thus leave the solar system. THis is just aguess on my part, though. C. Wingate
KFL@MC.LCS.MIT.EDU ("Keith F. Lynch") (03/03/86)
From: brahms!desj@ucbvax.berkeley.edu (David desJardins)
I don't have any idea what these people are talking about. The whole
idea is that Ulysses is being launched to rendezvous with Jupiter just
like Voyager and Galileo, but it will swing around Jupiter and back over
the solar pole. Presumably it will then leave the solar system; what
would cause it to circle around to the other pole??
I just spent several hours working out the math, and I am somewhat
confused. I assume that unlike the Voyager and Pioneer probes, that
Ulysses has some onboard fuel to be used in the vicinity of Jupiter.
If the Jupiter pass is completely passive, as with Pioneer and
Voyager, the furthest out of the ecliptic it could get would be 26
degrees, not 90 degrees as would be necessary to go over either of the
Sun's poles.
I have proven to my satisfaction that the orbit does not escape from
the solar system but will continue to pass over both poles if it
passes over one.
Can someone from JPL or NASA give us the facts?
...KeithKFL@MC.LCS.MIT.EDU ("Keith F. Lynch") (03/04/86)
From: desj@brahms.berkeley.edu (David desJardins)
The course desired is really only 1% or so out of the ecliptic;
most of the velocity is obviously toward the Sun.
Perhaps we are using words differently. I was refering to the
angle between the plane of the ecliptic and the plane of Ulysses'
orbit. If Ulysses is to pass over the Sun's north or south pole, this
angle must be 90 degrees.
But any course should be achievable by approaching Jupiter in the
right orientation (i.e. slightly above or below and to the left or
right), including 90% if this is what you want (except that turns
near 180 degrees would require orbits intersecting the planet).
I used to think so too, until I sat down and tried to analyze the
situation.
What does 26 degrees come from?
Ok, let me explain my derivation.
All units are in the Meters-Kilogram-Seconds system of units. All
velocities are relative to the Sun except when stated otherwise. The
symbols used are as follows. They represent the average values of
these quantities if the quantities vary.
RE = Radius of the Earth's orbit = 1.50E+11 meters
RJ = Radius of Jupiter's orbit = 7.78E+11 meters
VE = Velocity of Earth = 2.98E+4 meters/second
VJ = Velocity of Jupiter = 1.31E+4 meters/second
AE = Acceleration of Earth = 5.93E-3 meters/second/second
AJ = Acceleration of Jupiter = 2.21E-4 meters/second/second
VP = Perihelion velocity of Ulysses = 3.86E+4 meters/second
VA = Aphelion velocity of Ulysses = 7.45E+3 meters/second
Note that RJ/RE = (VJ/VE)**2 = (AE/AJ)**2 = 5.20
To get VP and VA, I assumed that Ulysses will follow an elliptical
path whose perihelion and starting point is on the orbit of Earth, and
whose aphelion is on the orbit of Jupiter. This is the least energy
method of getting to Jupiter. And it explains why Ulysses can't be
launched until June 1987 if it can't be launched in May 1986, i.e.
when it reaches aphelion at Jupiter's orbit, Jupiter has to be there
to meet it.
So I used equations I derived (derivation on request) for the
perihelion and aphelion velocity of an object in an elliptical orbit.
Where the perihelion is RE and the aphelion is RJ, the perihelion
velocity VP equals SQRT(2*AE*RE*RJ/(RE+RJ)) and the aphelion velocity
VA equals SQRT(2*AJ*RJ*RE/(RJ+RE)).
So Ulysses will approach Jupiter with a velocity of 7,450 meters per
second. But note that Jupiter is moving in the same direction at a
velocity of 13,100 meters per second. So Jupiter will actually
overtake Ulysses. The two will come together at a relative velocity
of VJ-VA or 5,650 meters per second.
Of course as Ulysses approaches Jupiter it will move faster. It
will swing by Jupiter in a hyperbola, and will then slow down relative
to Jupiter and will leave the vicinity of Jupiter at 5,650 meters per
second, the same speed as it arrived. Note, however that this is
5,650 meters per second relative to Jupiter. Not relative to the Sun.
If Ulysses made a 180 degree turn around Jupiter, it would then be
going at a speed of VJ+VA or 18,750 meters per second. The solar
escape velocity in the vicinity of Jupiter is SQRT(2)*VJ or 18,500
meters per second. As such, Ulysses would have sufficient speed to
escape from the solar system (just barely).
But that will only happen if Ulysses makes a turn that is close to
180 degrees. Since the idea is to give it velocity in an out-of-
ecliptic direction, not enough velocity component will be left in the
forward direction for it to escape from the solar system. As such,
whatever trajectory it gets into will have to be a closed ellipse
about the sun. The aphelion of that ellipse will be in Jupiter's
orbit. It will keep returning to that position in space.
Fortunately, Jupiter will (I think) not be there, so its orbit will
not be interfered with.
The best way to see what velocities Ulysses can leave Jupiter at is
to construct a vector diagram. From the origin (representing zero
speed relative to the Sun) construct a horizontal line segment of
length 13,100. The end of that line represents the velocity of
Jupiter. Construct a sphere of radius 5,650 centered on the endpoint
of the line. The point where the sphere intersects the line segment
represents the velocity with which Ulysses approached Jupiter. The
surface of the sphere represents the velocities with which it can
leave Jupiter. The circle where the sphere intersects the horizontal
plane represents the velocities available if it is to remain within
the ecliptic plane. Solar escape velocity is represented by the
exterior of a sphere of radius 18,500 centered on the origin. Only a
small part of the original sphere is outside this sphere, and no part
of that small part extends far from the horizontal (ecliptic) plane.
What is wanted is a velocity represented by a point directly above
or below the origin. Only a velocity like that will bring Ulysses
over a solar pole. No such point exists on our sphere, however.
The angle between the plane of the ecliptic and the plane of Ulysses
orbit it equal to the angle between a line passing through the point
on the sphere which represents the velocity of Ulysses leaving Jupiter
and the horizontal plane. Imagine yourself sitting at the origin
studying the sphere. It should be clear that the point representing
the greatest such angle is on the top (or bottom) edge of the sphere,
as visible from where you sit. Note that the topmost (or bottommost)
point on the sphere has a lesser elevation as seen from your vantage
point.
So contruct a line tangent to the sphere at that point. The line
segment between the origin and the point on the sphere will have
length SQRT(2*VJ*VA-VA**2) or 11,800. Thus that is the aphelion
velocity of Ulysses' new orbit. The angle to the ecliptic is
ARCSIN((VJ-VA)/VJ) or 25.5 degrees. That will NOT bring Ulysses over
either pole of the Sun.
The only explanation I can think of is that fuel is expended during
the close pass to Jupiter. This would have the effect of increasing
the velocity with which Ulysses leaves Jupiter, i.e. increasing the
radius of the sphere in the vector diagram. Since a point directly
over (or under) the origin is needed, the radius must be greater than
13,100. Thus a delta-vee (change of velocity) of 7,450 meters per
second is necessary.
If the fuel is burned at or near the closest point to Jupiter, much
less delta-vee is needed, due to the fact that you are gaining energy
by expending the fuel in a deep gravity well. I haven't worked out
just how much you would gain, but it would be considerable, and the
smaller delta-vee needed is certainly within the realm of today's
technology. After all, the Galileo probe has to do the same thing,
in order to stay in orbit around Jupiter rather than flying right back
out of the Jupiter system.
The time it will take for Ulysses to rach Jupiter is
Pi*(RE+RJ)**2/(4*RE*SQRT(AE*RJ)) or 6.64E+7 seconds or 2 years and 1
month. If Galileo and Ulysses are launched in June of 1987, they
should arrive at Jupiter in July of 1989. This quite close to the
time when Voyager will start closely approaching Neputne. I hope
there are enough people to control all three probes at once!
These equations are all of my own derivation (available at request)
and there is a small chance they may not be correct. The numbers
certainly suffer from roundoff errors, and only the first two digits
should be trusted. Earth's and Jupiter's orbits are not quite
circular, so these numbers might be off by as much as ten percent or
so for the actual time of the Ulysses launch and approach to Jupiter.
Can someone tell me how I can get technical information on these
probes from JPL or NASA or wherever? All I have been able to get is
very nontechnical publications.
...Keithsteve@jplgodo.UUCP (Steve Schlaifer x3171 156/224) (03/04/86)
In article <[MC.LCS.MIT.EDU].836296.860303.KFL>, KFL@MC.LCS.MIT.EDU ("Keith F. Lynch") writes: > From: brahms!desj@ucbvax.berkeley.edu (David desJardins) > > I don't have any idea what these people are talking about. The whole > idea is that Ulysses is being launched to rendezvous with Jupiter just > like Voyager and Galileo, but it will swing around Jupiter and back over > the solar pole. Presumably it will then leave the solar system > > I just spent several hours working out the math, and I am somewhat > confused. I assume that unlike the Voyager and Pioneer probes, that > Ulysses has some onboard fuel to be used in the vicinity of Jupiter. > If the Jupiter pass is completely passive, as with Pioneer and > Voyager, the furthest out of the ecliptic it could get would be 26 > degrees, not 90 degrees as would be necessary to go over either of the > Sun's poles. The information I have about Ullyses indicates that the plan was to leave Earth in an elliptical trajectory inclined about 2 degrees from the ecliptic. A hyperbolic flyby of Jupiter with an inclination of 141 degrees (i.e. retrograde about Jupiter coming in ahead and above the planet) would then result in an elliptical orbit about the sun which was inclined about 77 degrees to the ecliptic passing over the South polar region first. Indeed, such an orbit would then pass over the North polar region at a later date but that was not of immediate interest. Also note that the mission would not pass directly over the Sun's south pole but would pass very high over the polar region. I am not directly involved with the mission but have limited access to the mission design. The information above is only approximate and was derived by means of conic orbit approximations given the planned heliocentric orbits. Opinions expressed are my own and not necessarily those of the Ulysses mission, JPL or NASA. -- ...smeagol\ Steve Schlaifer ......wlbr->!jplgodo!steve Advance Projects Group, Jet Propulsion Labs ....group3/ 4800 Oak Grove Drive, M/S 156/204 Pasadena, California, 91109 +1 818 354 3171
desj@brahms.BERKELEY.EDU (David desJardins) (03/05/86)
In article <[MC.LCS.MIT.EDU].837830.860304.KFL> KFL@MC.LCS.MIT.EDU ("Keith F. Lynch") writes: > >Note that RJ/RE = (VJ/VE)**2 = (AE/AJ)**2 = 5.20 Actually this should be RJ/RE = (VJ/VE)**2 = (AE/AJ)**.5 = 5.20. > To get VP and VA, I assumed that Ulysses will follow an elliptical >path whose perihelion and starting point is on the orbit of Earth, and >whose aphelion is on the orbit of Jupiter. This is the least energy >method of getting to Jupiter. This assumption is the problem. If we relax this only slightly we can achieve a much more satisfactory result. > ... Ulysses will approach Jupiter with a velocity of 7,450 meters per >second. But note that Jupiter is moving in the same direction at a >velocity of 13,100 meters per second. So Jupiter will actually >overtake Ulysses. The two will come together at a relative velocity >of VJ-VA or 5,650 meters per second. > ... and will leave the vicinity of Jupiter at 5,650 meters per >second, the same speed as it arrived. Note, however that this is >5,650 meters per second relative to Jupiter. Not relative to the Sun. A very good analysis. We desire to leave the Jupiter encounter with an absolute (relative to Sun) velocity directly out of the ecliptic. Its magnitude should be relatively small, depending on how close we wish to pass to the solar pole. Thus we need to leave with a relative (to Jupiter) velocity slightly greater than that that of Jupiter (with a component equal to Jupiter's velocity but in the opposite direction, and a smaller component out of the plane of the ecliptic). Since I don't know how close Ulysses is supposed to pass to the Sun, I am simply going to note that it should be much less than the orbital velocity of Jupiter. Thus we desire to leave Jupiter with a *relative* velocity slightly greater than the orbital velocity of Jupiter; we must arrive at Jupiter with this same relative velocity. I will use the following notation: Vir = radial component of initial velocity (after leaving Earth) Vit = theta component of initial velocity Vfr = radial component of final velocity (when arriving at Jupiter) Vft = theta component of final velocity These quantities are related by the following equations: RE * Vit = RJ * Vft (1) 1/2 (Vir^2 + Vit^2) - AE*RE = 1/2 (Vfr^2 + Vft^2) - AJ*RJ (2) Here (1) represents conservation of angular momentum; (2) represents conservation of energy. If we solve (1) and (2) with Keith Lynch's assumption that Vir = Vfr = 0, we get his solution of: >VP [ = Vit ] = Perihelion velocity of Ulysses = 3.86E+4 meters/second >VA [ = Vft ] = Aphelion velocity of Ulysses = 7.45E+3 meters/second But let us not make this assumption. The *relative* velocity on arrival at Jupiter is the vector (Vfr,Vft-VJ). If we want its magnitude to be approximately equal to VJ, we have the constraint equation: Vfr^2 + (Vft - VJ)^2 = VJ^2 (3) We can now eliminate Vfr and Vft from (1), (2), and (3) with the following result: 1/2 Vir^2 + 1/2 Vit^2 - (RE/RJ)*VJ*Vit = AE*RE - AJ*RJ (4) (4) is a constraint on the *initial* velocity which will cause Ulysses to arrive at Jupiter with the correct *final* velocity. The relative velocity with which we leave the Earth is the vector (Vir,Vit-VE). In order to minimize the thrust requirement we can minimize the magnitude of this vector: Vir^2 + (Vit - VE)^2 (5) Minimizing (5) with respect to the constraint (4), using Lagrange multipliers, yields the solution: Vir = 0 1/2 Vit^2 - (RE/RJ)*VJ*Vit - (AE*RE - AJ*RJ) = 0 Using the quadratic formula with the actual values yields: Vir = 0 Vit = 4.05E+4 m/s Note that an initial velocity of 3.86E+4 m/s was required for the minimum-energy path to Jupiter, so we see that a relatively small addition to the initial delta-V will suffice to encounter Jupiter with the required velocity. > The only explanation I can think of is that fuel is expended during >the close pass to Jupiter. This would have the effect of increasing >the velocity with which Ulysses leaves Jupiter, i.e. increasing the >radius of the sphere in the vector diagram. Since a point directly >over (or under) the origin is needed, the radius must be greater than >13,100. Thus a delta-vee (change of velocity) of 7,450 meters per >second is necessary. We see that a much smaller delta-V will suffice if provided upon leaving Earth orbit. > Can someone tell me how I can get technical information on these >probes from JPL or NASA or wherever? All I have been able to get is >very nontechnical publications. Ditto. Note that my analysis seems to confirm that I was wrong when I said that Ulysses would follow a hyperbolic path around the Sun and leave the solar system. It appears that additional launch energy would be necessary to cause this to happen. It should indeed go into an eccentric orbit with perihelion near the site of its Jupiter encounter. -- David desJardins