The MESSENGER spacecraft is currently on its way to the innermost planet, Mercury. It launched in 2004, and will arrive at the tiny planet in 2011. Why so long? Mercury orbits the Sun very quickly, much faster than the Earth. Just getting there is hard enough, but if the spacecraft isn’t moving fast enough, Mercury will whiz right past it. So MESSENGER has to change it velocity by quite a bit (how much? Well, that’s a bit complicated * (hmmm, I’m having trouble getting the anchor linking to work. Scroll down to see the footnote)). There isn’t enough fuel on board to do it, so they borrow the velocities of Earth, Venus, and even Mercury itself to get the spacecraft moving rapidly enough to catch up to the mercurial (hahaha! ha!) planet.
When MESSENGER flew by the Earth in August of 2005, the engineers back home decided to test the cameras. They got the images above (click for a higher-res version). They’re lovely! The one on the left is a "true color" image, adding red, green, and blue filtered images. You can see South America there in the center. On the right is pretty much the same view, but instead of red light, they used blue-green-infrared. Vegetation reflects IR light very well, and so the Brazilian rainforest acts almost like a mirror in IR. It shows up as very red in this image. MESSENGER detected life on Earth from more than 100,000 kilometers away!
Well, not really. Lots of things reflect IR, but it’s still pretty cool. The cameras on board are pretty hi-res, so when it gets to Mercury the images will be spectacular.
Still, for nifty images we don’t have to wait that long (as the images above show). Later on I’ll post another cool thing MESSENGER did on that flyby of Earth…
* One problem is that Mercury’s orbit is elliptical, and so it changes its velocity around the Sun. At the nearest point in its orbit (perihelion) it moves about 60 kilometers/second. At the farthest point (aphelion) it’s only moving about 40 km/sec. For reference, the Earth moves about 30 km/sec. So how much velocity the spacecraft needs depends on when it gets to Mercury. It looks to me that orbital insertion occurs at Mercury perihelion, which strikes me as odd, since that means they need to add more velocity to the spacecraft for it to enter orbit. There must be some logical reason for it, of course. Emily, if you’re reading this… 
April 19th, 2006 at 9:51 pm
Mercury is closer to the sun, and has a lower energy orbit than the Earth. You need to shed some of that energy in order to fall towards the sun. So I’d say you need to slow the spacecraft down. And it should be easier to slow down to 60km/s than to 40km/s. There is also the fact that in order to get to Mercury’s perihelion distance to the sun, you have to slow down even more than to make it go to the aphelion distance. I’m not in the mood to make the orbital calculations right now, but intuitively, I’d say you would need approximately the same amount of delta-vee change in either case. There’s also the saying that goes: it’s not the destination that’s important, it’s the journey.
April 19th, 2006 at 11:10 pm
Hmmm, I need to think about this more. It’s difficult to tell from the trajectory diagram linked in the footnote, but they did launch in the direction of Earth’s orbital motion, which would add energy to the spacecraft. In the end, the spacecraft needs to change from 30 km/sec to 60, so energy is added. Changing the shape of the orbit plays into this, but to be honest, I’m not sure exactly how. If there are any orbital mechanics reading this, pipe up! I’m always willing to learn.
April 19th, 2006 at 11:30 pm
There’s a diffence between gravitational potential energy and kinetic energy that you are forgetting about, ioresult.
True, there is a lower gravitational potential energy at the orbital distance of Mercury compared to Earth. You say “you need to shed some of that energy in order to fall towards the sun” and you are right again, but the energy shed from the potential component needs to go somewhere, so it goes into the kinetic component.
The relationship between (average) orbital velocity and (average) orbital distance is one line of algebra away from Kepler’s 3rd Law, P^2 ~ d^3 (~ means proportional to) which says that planets nearer to the Sun (smaller d) have shorter periods (smaller P), as we all know. Since v ~ d/P, then v^2 ~ 1/d, so planets nearer to the Sun have to move through space faster in order to stay in their orbits.
April 20th, 2006 at 12:06 am
Another neat way to explain it is that for any point in an orbit, the kinetic energy is -1/2 the potential energy (K = -U/2, a form of the virial theorem). U = -GMm/r gets more negative the closer you are to the Sun, so K = (1/2)mv^2 gets more positive.
April 20th, 2006 at 2:17 am
Okay, you say the earth orbits at about 30km/s and mercury at its perihelion 60km/s. This is a difference in specific kinetic energy of 1/2((60km/s)^2-(30km/s)^2) or 2700MJ/kg. The difference at aphelion would be 700MJ/kg.
So, now let’s calculate the difference in potential energy: The earth orbits at 150Gm, while mercury’s perhilion is at 46Gm and aphelion at 70Gm. The difference in specific potential energy between the earth’s orbit and perihilion is -GM(1/46Gm-1/150Gm) or -2000MJ/kg. At aphelion, the difference is -1011MJ/kg. So, that means that to insert at perihelion would require the craft to speed up, gaining (2700-2000)=700MJ/kg of kinetic energy, whereas aproaching at aphelion would require it to slow down, losing (700-1000)=-300MJ/kg of kinetic energy.
It looks, however, like there would actually be some point in between the two that would require the spacecraft neither to speed up nor slow down… I don’t know why we don’t just insert at that point! It would be an awful lot easier.
April 20th, 2006 at 6:46 am
I don’t know why we don’t just insert at that point! It would be an awful lot easier.
Since the kinetic and potential energy (w/ respect to the sun) would oscillate periodically with the eccentricity of the probe’s orbit, maybe the location in space where the probe would have that particular energy never corresponds with a location from which Mercury can capture it.
I don’t know practically anything about orbital mechanics, this is just a thought.
Is the probe destined for capture, or will it just flyby (the way the Voyagers and Pioneers did with the outer planets)? Presumably, if it were simply a flyby, that wouldn’t require energies to correspond.
April 20th, 2006 at 6:51 am
Scratch that… it is a capture… I just didn’t read everything carefully enough. Sorry!
April 20th, 2006 at 7:22 am
BB, I haven’t checked your work, but don’t forget that velocity is a vector quantity, which means it requires energy to change it’s direction, as well as magnitude.
I think what’s probably been done is that Messenger is now on a solar orbit that has a syncronized (both spatially and chronologically) perhelion with mercury. Thrusting is always most efficient when you’re at maximum gravitational influence, so sometime near closest approach to mercury, they’ll go into mercury orbit (or possibly a solar orbit very similar to mercury’s). Another possibility is getting to one of the mercurian lagrangian points, but I don’t have the math of orbital mechanics learned enough to say which is cheaper.
April 20th, 2006 at 7:31 am
Regarding IR reflectance of vegetation. It depends on the health and type of vegetation as well as the IR wavelength, but generally one can expect about half of the near IR energy will be reflected and half absorbed. Ironically, green vegetation appears green to our eyes because of heavy blue and red absorption by the chlorophyll and our eyes’ higher sensitivity to green light. Vegetation reflects something like 20% or less of green light, and less than 5% red and blue.
CJSF
April 20th, 2006 at 7:58 am
Very cool images Phil. Thaks for posting.
April 20th, 2006 at 7:58 am
There appears to be a bunch of information linked here, if you know how to read it.
http://messenger.jhuapl.edu/the_mission/mission_design.html
See tables of each thruster burn delta V:
http://messenger.jhuapl.edu/missiondesignLive/index.html
Data on gravity assist:
http://messenger.jhuapl.edu/faq/faq_journey.html#7
April 20th, 2006 at 8:19 am
Ok, I’ve plugged the potential and kinetic energy formulas with the numbers mentionned above. To go from the Earth to Mercury at aphelion, you need to loose 1GJ/kg of potential gravitational energy and gain 350MJ/kg of kinetic energy, with a total of 650MJ/kg of energy to shed. Now to get from the Earth to Mercuray at perihelion, you need to loose 2GJ/kg of potential grav energy and gain 1350MJ/kg of kinetic energy, with a total of, yes, 650MJ/kg of energy to shed! See? Maths agree with my initial intuitive hypothesis, which means that wether you insert at perihelion or aphelion, the total energy differtial is exactly the same. So the real question planners need to ask themselves is, with as much time as you need (and so with as many grav assists you need), what is the trajectory you need in order to minimize the amount of delta-vee of deep space maneuvers. They obviously found a solution that required to accelerate the spacecraft at first and insert near perihelion.
April 20th, 2006 at 8:40 am
BB has to have done something (though I don’t know what) because kinetic + potential energy has to stay constant throughout an object’s orbit.
April 20th, 2006 at 8:40 am
CORRECTION:
BB has to have done something wrong in his calculations (though I don’t know what) because kinetic + potential energy has to stay constant throughout an object’s orbit.
April 20th, 2006 at 10:17 am
That makes sense ioresult.
April 20th, 2006 at 10:27 am
Yes, gopher65, BB forgot the (1/2) factor in the kinetic energy formula.
April 20th, 2006 at 11:23 am
You’re absolutely correct… it must have been too early in the morning, or I would have realized that it would be impossible for mercury’s total mechanical energy to change from one point in its orbit to the next.
April 20th, 2006 at 12:05 pm
The Rolling Stone article on Bush mentions his anti-science agenda:
http://www.rollingstone.com/news/profile/story/9961300/the_worst_president_in_history?rnd=1145555726696&has-player=false
April 20th, 2006 at 12:55 pm
It’s nice to see a rock magazine calling ID for pseudoscience.
Rock, really… eh… Rocks!
April 20th, 2006 at 3:19 pm
There is a contacts link on the page with the trajectory diagram linked in the footnote. If anyone here REALLY wants an eloquent explanation, then maybe one of those people can tell us.
It does seem a bit odd that the launch date was August 3rd 2004 and the first planetary swingby was with Earth on August 2nd 2005. That’s almost exactly one year later, so Earth was in almost the same position. It seems to me that launch could have been delayed until August 2nd 2005 if the weather cooperated and safety margins were still good.
April 20th, 2006 at 3:30 pm
One possible explanation for insertion at Mercury perihelion is that Earth and Venus run on their own schedules.
April 20th, 2006 at 7:44 pm
I was showing this blog to my physics students this morning since we’re talking about gravity and orbits. One of my students thought the URL at the top said “Bad Ass Astronomy”!
April 20th, 2006 at 8:42 pm
OK probably not a smart question but I have always wondered….why do most of the pictures from other planetary missions always (that I remember) have the Earth’s frame of reference pointed other than “up-down” with the poles? From this photo the poles are almost perpendicular. Does this have something to do with solar panal alignment? The old picture I remember from Ulysses was also tilted but it used RTGs right?
April 20th, 2006 at 8:53 pm
Or Cosmos teached me wrong or ALL orbits are elliptical. I think you wanted to say Mercury orbit is VERY elliptical. OR I may need to read some new theories, maybe something involving platonic solids…
April 20th, 2006 at 9:58 pm
Ah, this topic is one neat example of how I simply cannot wrap my head around physics. It seems to require more visualization (and math!) than I can muster.
To get to Mercury, you need to shed energy, or the spacecraft won’t start to fall inward towards the sun. But once you get there, you need higher velocity, since Mercury is traveling faster than the earth. That’s kind of weird!
The issue of eliptical orbits also interests me. As far as I know, all planets, and moons for that matter, have eliptical orbits of verious degrees of eccentricity.
What I wonder is this: is a perfectly circular orbit possible in theory? I would think that even if it is, such an orbit is extremely unlikely simply because any orbit must, of necessity, have a history. In order to achieve a perfectly circular orbit, you’d have to start out with exactly the right conditions of velocity and direction, and it seems to me that in a natural system, the chances are very slim that bodies condensing out of a preplanetary disk will ever happen to have just the right velocities to ensure that they end up with a perfectly cicular orbit.
But don’t know. I wonder if anyone else here can tell me. Is a circular orbit possible in principle, and real orbits elliptical simply because of historical contingencies, or must an orbit always be elliptical?
April 21st, 2006 at 3:33 am
A “perfectly circular orbit” is impossible in any n-body system in which n>2. In fact, a “perfectly elliptical orbit” is also impossible!
April 21st, 2006 at 4:25 am
All very wild and wonderful. Even for a plumber looking out at Night. Way cool.
April 21st, 2006 at 10:51 am
It is true that no real orbit is a prefect ellipse or perfect circle - no real shape is ever a perfect shape.
But don’t forget, cardoso, to say that all orbits are elliptical does not rule out the possibility that an orbit may be circular: a circle is a special-case solution of the ellipse. Just as a square is a rectangle with equal-length sides, a circle is an ellipse with equal-length semimajor and semiminor axes (and both foci in the same place).
April 21st, 2006 at 12:46 pm
Be careful, quarthinos, velocity is a vector, but that only means that a force is required to change its magnitude or direction. On the other hand, energy is a scalar, so it is not direction-dependant. This means that the force which changes the velocity vector need not necessarily do any work.
This is precisely the case in a circular orbit. Because the gravitational force is centripetal, it is perpendicular to the tangential velocity in a circular orbit (and at the instant of aphelion and perihelion in an elliptical orbit). Therefore, the force and the direction of motion are perpendicular, and no work is done.
Consider this: while work must be done to lift an object off the ground(to move it in the direction of the applied force) and get the object moving forward (to change its kinetic energy by accelerating it), no additional energy is needed to carry the object forward at a constant velocity and constant height. Because the direction of motion (forward) is perpendicular to the direction of the lifting force (upward), the work done is zero. (In mathematical terms, work is a dot-product: it takes two vector arguments and returns a scalar answer depending on the degree to which the two vectors are collinear.)
To put it another way, the fact that kinetic energy is based on the square of the velocity makes it direction-independent. In a one-dimensional system, this is a simple as saying
(-v)^2 = (v)^2
In a 2-dimesional sense, remember that the resultant velocity is the hypotenuse of the triangle made from the perpendicular component legs. So
(1/2)m(Vx)^2 + (1/2)m(Vy)^2 = (1/2)m(Vr)^2
reduces to
(Vx)^2 + (Vy)^2 = (Vr)^2
- the Pythagorean theorem!
Therefore, the resultant (i.e. overall) velocity will remain constant even as the component lengths change. (This also holds true for 3- and even n-dimensional motion.)
Of course, when an object is not at one of the “-helions” in an elliptical orbit, the force of gravity is not perpendicular to the tangential velocity, so this non-perpendicular force of gravity does change the overall velocity, and therefore the kinetic energy. But this change is caused in the component of the velocity vector which is collinear to the force, which is to say, the radial component. Therefore, the change in kinetic energy is exactly offset by the change in potential energy as the object moves to a different distance from the sun. (Indeed, you can argue that the work done by the force in that direction is “used” to change the energy form one form to the other.) Any way you look at it, the total mechanical energy of the system remains constant. (At least in a Newtonian/Keplerian an system.)
(Note: I know that “V” is not the correct abbreviation for “velocity”, but I don’t know if this blog supports the sub-script tag, and I wanted to use an upper-case letter to make my sub-scripts look small.)
April 21st, 2006 at 1:25 pm
All these equations and discussion — way too technical for me, a total lay person. So I’ll just enjoy the awesome beauty of the pictures and the brilliance of those who do understand the math! Thanks for sharing!
April 21st, 2006 at 3:18 pm
I haven’t read the related links, but possibly MESSENGER is joining Mercury at perihelion because (speculating here) that could be where the node of Mercury’s orbit hits the ecliptic. From my understanding the biggest logistical problem of sending a probe to ORBIT Mercury is because Mercury’s orbit is actually quite inclined with the ecliptic.
Another possibility may be scientific. The Mariner probe in 1974-75 only imaged one side of the planet (it visited 2 or 3 times on successive orbits and amazing the same face was in daylight). A priority of the Mercury flybys (which also are used as velocity changing events) is infact to image the unseen side.
April 21st, 2006 at 3:30 pm
…And Chile is uncoloured in infrared because…? What? They have no vegetation? Have they completely slash-n-burned their entire allotment of rainforest? Surely this is some sort of poor translation of the image data, rather than a large expanse of mountains and dirt that are completely lacking in plants, and only encompasses one country, right to its political borders.
April 21st, 2006 at 5:09 pm
Those pictures are so vividly crisp, they almost look like CGI.
Really, I guess that’s a compliment to the CGI guys, huh?
April 21st, 2006 at 6:02 pm
Could it be that Chile is brown in the IR view ’cause the sun has only just cleared the Andes and the land to their west is still cool? I do know that the costal plain is arid, and Chile basically is the costal plain, so maybe that is an accurate representation of the vegetation?
In any case, I think the phenomenon is a genuine artifact of the geographic barrier of the mountains. I’d be surprised if there is anything in the image processing software which is even aware of the geopolitical boundaries.
April 22nd, 2006 at 12:32 pm
While we are on orbits and lay people, how exactly does gravity, or what ever conglomerate of forces, keep an object in orbit? I would guess there are soe opposing forces at work, I guess i just don’t see why there is a somewhat circular motion to everything, cosmos bodies spin, orbit a star, orbit the center of a galaxy, orbit the center of the universe(ok aybe thats a stretch is there a center?). Is all just a big old cosmic waltz?
April 22nd, 2006 at 12:37 pm
Phil the bad ass astronomer! Hey thats not too bad
April 23rd, 2006 at 9:38 am
Well, RAD, I can answer one of your questions: I can tell you that there is no opposing force.
I know that gravity is the only force needed to explain the circular/elliptical motion of orbits of the planets about the stars and the stars about the galactic centers. It might seem that the attractive force of gravity would just cause everything to collapse into a big ball, so a second force is needed to push the objects out from the center. But the planets are not being held away from the sun by any opposing force, they are just moving too fast to ever hit the sun. In a very real sense, the planets are trying to fall into the sun, but they are forever missing it.
It might be easier to picture this if you think of an object orbiting the Earth. If you stand on the top of a tall tower and drop a baseball off the edge, it will fall more-or-less straight down and land at the tower’s base. If you toss it horizontally from the top of the tower, it will fall a little farther away. Throw it faster, and it will land still farther away. If you could throw it fast enough, and the tower were tall enough, you could make the ball hit the ground several miles away. In fact, you can imagine that Superman might even be able to throw it so hard that it would travel halfway around the world before it landed. In this case, the baseball is just barely hitting the Earth at all… any faster, and the ball would miss the planet’s edge and fall past it to the other side. But of course, if there were no air resistance, it would miss the other side too, so it would never land at all! The ball would always be falling towards a point which is just over the horizon: always falling, but never landing - that’s an orbit!
(You might want to visit NASA’s Space Place to play the “Shoot a Cannonball Into Orbit” game. It’s intended for kids, so don’t be offended that the graphics are kinda’ juvenile, I just think the animation might be more clear than my written explanation.)
This was Newton’s great discovery in the apple orchard. He didn’t discover gravity, everyone already knew about that, he discovered that gravity was the only force needed to account for the orbital motion of the moon.
Now, as for your other questions, well, Newton himself never tried to explain why there is any gravity in the first place (though he did develop an equation to predict its magnitude). Einstein accounted for gravity as a curvature in space-time* caused by an object’s mass. He said that the moon falls towards the Earth ’cause the Earth’s mass makes a sort of dimple in the “surface” of space, and the moon is sliding down the curved surface of that dimple, like a ball rolling down a hill. Of course, like all good answers in science, this just pushes the real answer farther away by bringing up more questions. Like: why do masses curve space-time, and why does a ball roll down a hill anyway? (In fact, this is just the sort of “explanation” which Fineman used to poke fun at. Because, in an effort to explain gravity, I’ve used an analogy which only makes sense if you think you already understand gravity so, as he would say, “I’ve cheated very badly.”)
If there is a better explanation, or an answer to those two questions, they are past my depth. Maybe someone else could help us both there.
Also, the whole “falling past the horizon” explanation can only account for the orbit of the planets if they were already moving fast enough to miss the sun the first time. The question as to why the matter in the planets was moving in the first place is a tricky one. The short answer is “the big bang got them going”, but that’s not a very good answer, really. In point of fact, in the early days of our solar system, most of the matter in this area was not moving fast enough, so it did fall in towards the center, where it coalesced to form our sun. (Can you have “days”, or even a “solar system” before your have a sun?) Anyway, most of the solar system did hit the “sun” - that’s why the sun is so big today. But the real question isn’t “why didn’t the matter that made up our planet fall into the center also.” The real question, as you’ve implied, is “what do we mean by ‘center.’”
If all there were to get the system going was a big bang, all the matter in the universe would be evenly distributed into a smooth, uniform, bubble expanding out* from the original explosion. No area would be any more dense than any other area, so none of the matter would have been attracted in one direction any more than the other, and there wouldn’t be a sun to fall past. The only point which could really be considered a center is the spot/moment* at which the “bang” occurred. If the bang weren’t big enough, everything might eventually slow down, stop, and fall back towards the place/time* where it came from (the so-called “Big Crunch”). But that wouldn’t explain why some matter clumped together to form our sun (or any of the other big, glowing lumps of stuff we see when we look at the night sky).
So the three big questions in cosmology are:
- “Why isn’t the universe uniformly dense?”
- “Will it ever stop expanding?”
and
- “If it does, what will happen after* that?”
‘Last I heard, no one really knows the answer to any of these, though there are a few very promising theories in the works.
Again, maybe someone more in the know than I will chime in at this point.
* Einstein’s theory of relativity holds that the universe is not just a collection of points, but also a collection of moments, and that the two are inextricably linked. This is what is meant by “space-time”. It is also why, when you talk about things like the Big Bang and “End of the Universe”, you cannot make a meaningful distinction between a spatial place and a temporal moment. I think I kind’ get relativity, in a general sort of way (no pun intended), but I wouldn’t presume to try to explain it to you. Here again, maybe the blogosphere will help us both out.
April 24th, 2006 at 11:01 am
Thanks for the info I am going to check this out more
April 24th, 2006 at 1:44 pm
SFWriter, take a look at this:
http://worldatlas.com/webimage/countrys/sa.htm
Note the color photo (composite image) of South America at the bottom. Note the color variations, where the vegetation shown green maps to the infrared in the MESSENGER image (especially the higher res version). Note also the low infrared matches the brown colorations on the South America photo, especially along the Western coast, but also in Argentina, Paraguay, and Brazil. Finally note the political map higher on that page, and see that the boundary does NOT align with the Chilean border, but rather with the geographical feature of the Andes Mountains. In fact, lower Chile is green (covered by clouds on the MESSENGER image), and the brown arid territory in the north extends into Bolivia and Peru.
Also, if you look at the MESSENGER image in “true color” as opposed to infrared, you will note that the coloration pattern falls along the exact same borders, with the “red” of the infrared mapping to dark coloration on the Amazon forest.
May 5th, 2006 at 1:07 pm
Um, if that messenger sequence is from real photos of the earth rotating, then why aren’t the weather patterns moving, or at least changing some? It would appear the cloud patterns are staying exactly the same throughout the rotation. Shouldn’t there be at least SOME variation in the cloud cover? Otherwise it just looks like one picture texture mapped on a sphere with simulated shadows.
March 1st, 2008 at 1:20 am
scientists are the best!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!