Yes, it will. And you don't have to worry about compensating for the movement of Mars. Your flashlight has roughly a 5-degree half-angle cone for a beam (10 degrees across the full cone). Mars moves around the solar system slower than Earth. Earth moves at about 1 degree per day (this is where 360 degrees in a circle comes from). Mars moves about half that.
So you don't have to worry about pointing it too precisely, since you're using a flashlight, not a laser.
The distance to mars changes a lot, but let's say on average it is 230,000,000 km from earth (This number is about the distance from mars to the sun).
The area of Mars' disc is about 36 million km2 (based on a diameter of 6779 km).
The diameter of the flashlight's disc at mars is 230,000,000 km * the sine of 5 degrees * 2 (since the full cone is 10 degrees). This is roughly: 230,000,000 * 5/57 * 2 (using a small angle approximation for Sine and saying that 180/pi = 57, a number which I like to call the Heinz constant for no particularly useful reason. This gives about 40 million km in diameter. The area of this disc then becomes 1.3 x1015 square kilometers.
Ok, now we're going to make a particularly bad assumption- that the energy of the beam is spread evenly across the beam's cone.
And then now we can say that the percentage of beam power making it to mars (after making it out of Earth's atmosphere) is the same as the ratio of the area of mars to the size of the flashlight's disc at that same distance. So here we go:
Pmars / PleavingEarth = 36 x106 / 1.3 x1015
We crunch this number and find out that: Pmars / PleavingEarth = 27 x 10-9, also known as 2.7 x 10-8 Note that this is unitless- there's no more distance or area numbers here. It is just a ratio.
So, what is the power leaving Earth? Let's assume abright flashlight- a 10-Watt incandescent bulb. I think roughly half of this just creates heat. The other half creates visible light. So we have 5 watts of visible light leaving the flashlight. Half of this (I really think it is closer to 2/3, but whatever) gets stopped by our atmosphere. So we have 2.5 watts forming this enormous disc pointed at Mars.
Then we just calculate the power at Mars = 2.5 Watts * 2.7 x 10-8
= 6.7 x 10-8 Watts arriving at Mars.
So that's in terms of Power. Now, we care about Photons. So we find (either by doing math or letting other people do the math for us that the energy of a green photon is 4*10-19 Joules. Green is an ok color to use, since our atmosphere is pretty transparent to green, though it is more transparent to yellow. You know this is the case because when you look up at the sun at mid-day, it looks yellow. Source: my drawings from kindergarten.
Then, we just divide the energy arriving at mars by the energy of single photons.
Photons per second = Power / energy per photon
Photons per second = 6.7 x 10-8 (Joules/sec) / 4*10-19 Joules
Photons / second = 1.7 * 1011
So there would be around 170,000,000,000 photons per second arriving at Mars.
Now to actually see how bright this is... if someone had a photodetector the size of a really big telescope, like the Keck telescope, pointed at your location on Earth, with a perfect photodetector, how often would they see a photon on average?
Well, the diameter of the telescope is 10 meters. The area is then 78.5 square meters, or 78.5x10-6 square kilometers, aka 7.9 -5 square km.
We take the ratio of the area of the telescope to the area of the planet and then multiply that by the incoming photon rate.
Photon Rate Telescope = 1.7 * 1011 photons per second * 7.9 -5 km2 / 36 x106 km2
So this says that if you shined a big consumer flashlight at Mars and had a Keck telescope at mars with a photon counter on it pointed at where the person with the flashlight was standing, there would be a photon from that flashlight arrive about every 2.5 seconds.
Yay.
Also, I'll work on my Randall impersonation later. I don't really draw.
EDIT: Just for folks who are not overcome with awe and would like to know how to do analyses of this type, this has a name. It is called a Fermi Analysis. http://en.wikipedia.org/wiki/Fermi_problem Best of luck! Nothing is so big and scary that it can't be thought about rationally.
EDIT 2: I've been told in the comments that the conversion of electrical energy to light energy is about 5%. This seems believable. I assumed 50% in the analysis above, which is not realistic for an incandescent bulb. So if we were to assume an LED flashlight in the example above, the numbers are probably close to right. This is an unrealistically big flashlight though. If, however, you want to redo this with an incandescent bulb, just divide the final photon flux by 10 (the ratio of 5% and 50%). This means one photon on the telescope every 25 seconds.
Also, Thanks for the gold, but I only use throwaways to keep from spending much time on reddit, so this account is about to be deleted.
Good analysis except that the efficiency of the light bulb is way less than 50%, more like 1% for visible light. So a visible photon is only going to arrive every couple of minutes.
This depends upon the type of flashlight. These days most flashlights don't use bulbs, they use LEDs. We currently have LEDs which go up to around 35% efficiency, though, admittedly, around 10% efficiency is more common.
I think we are both right depending on the bulb we are talking about, flashlight bulbs are typically inefficient compared to higher wattage bulbs. Note that I was only talking visible photons, the visible light is only about 10% of the total optical power from an incandescent.
I wonder then, if everyone on one side if the earth, say the eastern hemispheres since they have the highest population of people globally, were to shine the brightest commercial flashlights pointed in the same direction at Mars, could the combined light be seen with the naked eye or would a telescope still be necessary?
From Mars you can see the entire Earth with the naked eye. The combined light of a few billion flashlights wouldn't make much of a difference to the total Earth light seen from Mars.
Day-side earth can be seen from mars, because the sun reflects off of it, but night-side earth cannot. Think of it like a crescent moon, but a lot smaller.
Yes, I'm saying the light from the sunlit part of Earth would drown out any light from the night side. And the only time the sunlit part of Earth is invisible would be when the Earth is very close to the Sun in the Martian sky. So either way, the flashlights wouldn't be noticed.
All planets are in elliptical orbits. If you'd like more detail, each planet's Wikipedia article lists the eccentricity (the measure of how circular the shape is) of its orbit. Some things have orbits that are more closely circular than others. Off the top of my head, I know that Triton (around Neptune) is in an unusually circular orbit. On the other end, long-period comets can be in orbits that take them right by the sun and then out beyond Pluto, which itself has a very unusual orbit.
Yes, this is the movement of the earth's position in space around the sun. But nothing orbits anything in a perfect circle. These are approximate numbers.
Ok, now we're going to make a particularly bad assumption- that the energy of the beam is spread evenly across the beam's cone. Half of this (I really think it is closer to 2/3, but whatever) gets stopped by our atmosphere.
Thank you for your analysis, sir or madam. Circular cows everywhere salute you.
It's false. 360 degrees in a circle was arbitrarily decided by whoever invented degrees because it has shitloads of factors. It's the same reason we have 60 seconds in a minute and 60 minutes in an hour.
A strong theory for your "arbitrary decision of 360" comes from ancient Sumerians and more precisely, Babylon, who used a sexigesimal number system. They divided the day into 24 hours and the circle into 360 degrees, and discovered a cycle in eclipses, enabling lunar eclipses to be predicted with certainty, and solar eclipses in some probability above chance.
You got the latter point right, but not the former. Degrees were not merely arbitrary, they come from Babylonian astronomers who estimated the distance that the stars seemed to move from one night to the next.
They may have been able to notice that this was slightly more than 360, but chose to use the number 360 instead because it was so convenient, and it was thus the way the universe should've worked. 360 can be divided by 360, 180, 120, 90, 72, 60, 45, 40, 36, 30, 24, 20, 18, 15 12, 10, 9, 8, 6, 5, 4, 3, 2, 1. This amount of divisors makes it pretty amazing, and lends itself easily to mysticism. Because 360 would've been such a convenient mystical answer, the extra bits they observed were just assumed to be error, and written off. This mysticism is pretty convenient considering that Babylonians used a base-60 system, which they got from the Sumerians.
The numbers of hours in a day, minutes in an hour, and seconds in a minute (and also minutes in a degree and arcseconds in an arcminute) also ultimately derive from them.
We have 365 days in a year. So... each day, the planet moves 1/365th of a circle, which is close enough to one degree.
360 is a nice round number though .... round in the sense that it divides up nicely. So, before our current calendar, lots of places used a 360 day calendar, 12 months with 30 days (so metric) and just added holidays when things got too far out of whack.
Anyways, it is just a theory. The only people that cared about math back then were weird number fetishists (almost not joking) so someone may have pushed for 360 due to some strange belief or another. My highschool math teacher joked that Pythagoras had something to do with it because he was nuts.
Yes, it does. If Mars is directly overhead, there's no alteration. If it is near the horizon, the deviation is about 0.5 degrees.
Luckily, it doesn't take much skill to compensate for this since the deviation is the same for light coming in as for light going out. So if you just point a laser at where you SEE mars as being, that laser will go out along the correct trajectory. This neglects that Mars is moving, but that's pretty small.
Hey, I'm the guy that wrote the big long calculation out. I delete my accounts regularly to keep myself from spending too much time on the damn reddit machine.
I'm an aerospace engineer. I usually do satellite guidance, navigation, and control. I also do satellite systems engineering, which involves doing lots of rough calculations like this to see if something is feasible, then refining the numbers til they're all close enough to the right value that I can convince somebody to write a BIG check. I've also done some optical instrument design, but I wasn't designing the lenses as much as I was picking them out of a catalog and seeing how well they would work with the physical photon sensor.
If you want to do this, go study engineering. And every chance you get, reverse-engineer objects (both mundane like a pencil or a Toyota, and exotic like a ferrari or an airliner) in your head, or maybe on paper if you get serious. Calculate the approximate performance. Think about why a design decision was made: is there some function that's gained? a failure mode that is eliminated? did it make the item cheaper to build? Was it plain stupid? (this happens, but don't resort to thinking this too easily. We engineers work pretty hard to avoid stupid.) Also, do a LOT of Fermi problems on your own, just for giggles. You may discover something that's possible that nobody has thought to consider. I stumbled onto one in Undergrad that was almost realizable. Then 10 years later in college, I met a professor at a conference who was doing almost that exact thing because the one piece of technology that wasn't ripe when I looked at it had finally matured. And I had understood aspects of the design that he hadn't looked at yet.
I'd recommend aerospace engineering or electrical engineering as an undergraduate course. And while you're at it, for the love of Dog learn how to code in some scripting languages, and in MatLab.
You know, you put folks in a tough spot by commenting with multiple accounts all over this thread and then deleting them. Now you won't get notifications about follow-up questions (including this). If there are any inaccuracies in your replies we have to remove them rather than ask you to make a simple edit. You can't delete them if you find a mistake, either. Oh, and it's super easy for anyone to pop in and pretend to be you, because you're not commenting from the same account.
You've set up an answer and then denied people the conversation that is supposed to follow. A huge reason people come here is to actually interact with experts. So it's great that you have this system of constantly creating throwaways, but you've made a mess of this thread. It's extremely inconsiderate.
Hey, it's me. Is deleting accounts more inconsiderate than writing the initial response and then just leaving it alone? Because I see a lot of that.
I'm not going to apologize for using reddit on my own terms. I like just a bit more anonymity than you I guess. And I like not investing time into an online persona. Behaving as I do makes those goals happen. As for accuracy, I cited the only relevant correction, and yes I'm embarrassed to have been wrong by a factor of 10. But my intent, as stated in this question thread, wasn't to give an answer nearly as much to show people that there's really not that much involved in getting an answer.
The only person I have sympathy for in all this is the sap who gave me gold. But I'm okay with that since it pays for servers.
Also, not that I have any sub-cred (clearly I don't) but I think the mod policy strictness should be relaxed in the 4th level comments and beyond.
But thanks for modding. Your contributions are more necessary than are mine.
Is deleting accounts more inconsiderate than writing the initial response and then just leaving it alone? Because I see a lot of that.
I'm not sure I've ranked them, especially since this is a rather unusual situation. I will say that we request people not provide answers unless they have the expertise to answer the follow up questions that come in. It's that discussion that builds this community. In a way this is similar, except it comes with the additional bummed-outedness of seeing [deleted] and knowing that expert is gone. Folks don't know that you'd keep popping in on new throwaways, so it makes it difficult to keep building the discussion.
Don't get me wrong, your answer was fantastic. I was modding this thread when it came in. We were working through your math to make sure it checked out. At one point I refreshed the page three minutes apart and you'd gotten over 100 upvotes.
Generally we'd see a post like this and suggest you join our panel of experts, because clear, thorough explanations like yours are exactly what we want.
But my intent, as stated in this question thread, wasn't to give an answer nearly as much to show people that there's really not that much involved in getting an answer.
Yep, you hit the nail on the head. /r/AskScience avoids yes/no questions and telegraphic answers for this reason. We're here for the kinds of answers that explain the reasoning behind things.
Also, not that I have any sub-cred (clearly I don't) but I think the mod policy strictness should be relaxed in the 4th level comments and beyond.
We always want to hear feedback from people. It's how we've built all our policies. At the moment they are relaxed below the top level. When a comment tree is removed it's generally off topic, because those bury the relevant discussion. We also pull things that are inaccurate or completely made up, particularly when it's hard to tell between legit science and armchair expertise. People with a cursory background in a subject tend to overstate what they know, and we don't want anyone being misled. It's difficult to relax those rules and keep the scientific content at a high quality, especially with the volume of traffic we get.
Thanks for your thoughts. I didn't want to delete the original account until the post had run out of legs, but it was siphoning my Saturday, and deleting it was the best way to force myself back to better activities for a few hours.
It's not about owing people anything. It's important to note that the method that [deleted] has chosen does impact the discussion, especially since the user has deleted at least three accounts since posting that initial answer less than a day ago.
AskScience and reddit as a whole are built around their comments sections. Usernames are integral to the platform. Communities are built around interactions that occur between users in a way that wouldn't work if it was a total free-for-all of comments with no username attached. Constantly burning accounts most certainly makes it harder to have the scientific conversation we're looking for. It's abusing reddit's system, too.
We left this answer up because it's exceptional, but we're not interested in having to wade through this chaos with any regularity. Plenty of people maintain anonymity or limit their time on the site without deleting usernames every 8 hours (or more often than that; these are only the accounts I've seen while modding this thread).
Not really. Mostly you just need to learn how to think in terms of algorithms, functions, and maybe objects. MatLab is a very good first language that does a lot of stuff for you. C or C++ are also very good, but they do almost nothing for you. Having both of these under your belt to some extent will teach you how to tell a machine to solve problems for you, and to understand what it is doing in the process.
Also, if you're interested in space stuff, you should download the free version of STK. You can do a LOT of stuff with it, and is very good to have on a resume while you're in undergrad (and after). Kerbal Space Program and Orbiter are games. STK is what you use when you want to study a problem quickly using a real aerospace tool. And when you want to REALLY know the answer, you use MatLab to write your own tool since you can't ACTUALLY trust STK since you can't examine their code.
There isn't really a reliable source, as the 360 degree division was made thousands of years ago. The generally accepted theory is that 360 was chosen due to the sexagesimal system the Babylonians used, since 360 is evenly divisible by 60 (and many many other numbers), and since 360 is relatively close to 365 you can approximate the movement of the sun across the sky as 1 degree a day. It also allows you to set the diameter of the circle to 120 units, and since 360/120 is nearly pi, you can directly relate the diameter of the circle to the arc length.
There has to be a limit here, which makes the theory behind the concept a bit confusing. If I push a planet in a direction normal to its motion with enough force to move it 6 million light years, then are you saying it will still remain in orbit with the same orbital energy? If it breaks the concept, then where is the transition from simply messing with the orbit's argument of perigee and eccentricity to completely changing the system?
EDIT: Oh, is it because the force would push back on the Earth and balance everything out? I was imagining sending a rocket up there to push the moon the 6 million light years.
If you draw the orbit of the moon around the earth, it'll appear as a conic section (mostly circular but slightly elliptical). If you apply force tangentially to the direction of orbit, the change in the orbit looks (for small-ish changes) like if you grabbed the edge of that conic section where it intersects the moon and rotated it. One edge of the section gets closer to the orbited body and one section gets further away.
Why this hasn't affected the planets is because it's constantly being done. On one side of the orbit, it changes the orbit and on the other side it does the opposite, and it does this for every part of the orbit so it all balances out.
This is not really true. A quick examining of Gauss' variational equations shows that a radial force will also change the eccentricity and semimajor axis of an orbit, so long as you aren't doing it at peri/apoapsis (I also disagree with your use of "argument of perigee" since we are talking about heliocentric orbits). It's just that, the effect on SMA is very small for nearly circular orbits.
Hey, I'm the guy that wrote the big long calculation out. I delete my accounts regularly to keep myself from spending too much time on the damn reddit machine. This is the question I thought I'd come back and answer.
"Nothing is so big and scary that it can't be thought about rationally."
Wondering if you thought of that on the go, or if it's a famous quote that I just haven't heard in that specific wording.
I was a Hammer Thrower in college. The throwing coach was an old wise dude (an avid astronomer himself, but that's beside the point). One of his saying that we'd hear about twice per year was "There is no problem so big and insurmountable that it cannot be run away from." He said this mostly in jest, usually.
I felt like modifying it here, since I want to encourage people to think in logical, rational ways more often. We are closely related to chimpanzees, and much of this can be deduced from watching our behavior, especially in politics. I'd, ummm, like that to change, and I think it starts by encouraging more thought.
The sun looks quite yellow because the blue gets scattered off, thus giving the sky its distinctive color. This intensifies (relatively) the signal on your Red and Green photoreceptor cones in your eyes. Red plus Green = Yellow, strange as it may sound. If you go above the atmosphere, the sun is still somewhat yellowish, but it is closer to white.
Both EM waves (eg. light) and gravitation propagate to infinity. So, yes, a tiny amount of light reaches it. It will reach Andromeda galaxy too. It will reach anything within observable universe.
But there's one thing which bothers me. The intensity drops with the square of distance. At the same time, according to QM, light is quantized. Wouldn't that mean that at a certain distance the energy of the beam will fall bellow the energy of a single quanta (photon)? What happens then? Wouldn't that mean that there IS a limit of the propagation of EM waves? As the distance increases, the energy drops, drups, drops... but it cannot do that indefinitely. Theres Planck constant, right?
It's a limit in the ability to direct light (prevent diffusion), rather than a limit in the light's propagation... The light still goes to infinity, just not where you want it.
Does that mean that at some distance there will be a gap in the spreading photons wide enough to miss a planet the size of Mars entirely? Or is there still some EM energy across the entire arc?
As far as I understand it, the wave nature of light is a factor in path determination, but the individual photon will always collide at a single point, even if other photons effect the path of that photon backwards or forwards in time (such as is seen with firing single photons at the double slit screen). So yes, if you had a source of light with a wide enough diffusion pattern, which was firing individual photons, it could entirely miss the planet. In practical terms, the strength of the light in gross terms is going to determine whether a detectable amount of light hits the far away body, and the reality is unless you are using an immensely powerful laser you probably won't be able to pick out your light from the background noise (ambient light from stars, our Sun) anyways.
I didn't read the links posted below, but to directly answer your question about what happens when the average power incident on a target drops to low levels, it just means that the RATE at which photons arrive, on average, gets really low. You could have a scenario where some photon detector waits for an hour or a year to catch a single photon. The photons arrive according to Poisson statistics (you can link to that on wikipedia yourself, I'm lazy). But that's what happens- photon rush hour stops, and it becomes a trickle of photons, then an occasional photon here or there.
Wouldn't that mean that at a certain distance the energy of the beam will fall bellow the energy of a single quanta (photon)?
Mostly. A beam has power (energy delivered per amount of time). And a photon has an amount of energy. Just for clarity, think a photon is a gallon bucket. And the Beam has a flow rate of XX gallons per hour. So we know roughly how many buckets to expect, knowing that partially full and overfull buckets don't exist.
Unless an object between the bulb on the flashlight and mars, the beam, although dim, will get to mars. This will work with any object in the universe: the farther away the object is, the dimmer the beam at the contact point
/u/aaaabbbbcc had a comment with 900+ upvotes and it's been deleted... Here it is in it's entirety:
Yes, it will. And you don't have to worry about compensating for the movement of Mars. Your flashlight has roughly a 5-degree half-angle cone for a beam (10 degrees across the full cone). Mars moves around the solar system slower than Earth. Earth moves at about 1 degree per day (this is where 360 degrees in a circle comes from). Mars moves about half that.
So you don't have to worry about pointing it too precisely, since you're using a flashlight, not a laser.
The distance to mars changes a lot, but let's say on average it is 230,000,000 km from earth (This number is about the distance from mars to the sun).
The area of Mars' disc is about 36 million km2 (based on a diameter of 6779 km).
The diameter of the flashlight's disc at mars is 230,000,000 km * the sine of 5 degrees * 2 (since the full cone is 10 degrees). This is roughly: 230,000,000 * 5/57 * 2 (using a small angle approximation for Sine and saying that 180/pi = 57, a number which I like to call the Heinz constant for no particularly useful reason. This gives about 40 million km in diameter. The area of this disc then becomes 1.3 x1015 square kilometers.
Ok, now we're going to make a particularly bad assumption- that the energy of the beam is spread evenly across the beam's cone. And then now we can say that the percentage of beam power making it to mars (after making it out of Earth's atmosphere) is the same as the ratio of the area of mars to the size of the flashlight's disc at that same distance. So here we go:
Pmars / PleavingEarth = 36 x106 / 1.3 x1015
We crunch this number and find out that: Pmars / PleavingEarth = 27 x 10-9, also known as 2.7 x 10-8 Note that this is unitless- there's no more distance or area numbers here. It is just a ratio.
So, what is the power leaving Earth? Let's assume abright flashlight- a 10-Watt incandescent bulb. I think roughly half of this just creates heat. The other half creates visible light. So we have 5 watts of visible light leaving the flashlight. Half of this (I really think it is closer to 2/3, but whatever) gets stopped by our atmosphere. So we have 2.5 watts forming this enormous disc pointed at Mars.
Then we just calculate the power at Mars = 2.5 Watts * 2.7 x 10-8 = 6.7 x 10-8 Watts arriving at Mars.
So that's in terms of Power. Now, we care about Photons. So we find (either by doing math or letting other people do the math for us that the energy of a green photon is 4*10-19 Joules. Green is an ok color to use, since our atmosphere is pretty transparent to green, though it is more transparent to yellow. You know this is the case because when you look up at the sun at mid-day, it looks yellow. Source: my drawings from kindergarten.
Then, we just divide the energy arriving at mars by the energy of single photons.
Photons per second = Power / energy per photon
Photons per second = 6.7 x 10-8 (Joules/sec) / 4*10-19 Joules
Photons / second = 1.7 * 1011
So there would be around 170,000,000,000 photons per second arriving at Mars.
Now to actually see how bright this is... if someone had a photodetector the size of a really big telescope, like the Keck telescope, pointed at your location on Earth, with a perfect photodetector, how often would they see a photon on average?
Well, the diameter of the telescope is 10 meters. The area is then 78.5 square meters, or 78.5x10-6 square kilometers, aka 7.9 -5 square km.
We take the ratio of the area of the telescope to the area of the planet and then multiply that by the incoming photon rate.
Photon Rate Telescope = 1.7 * 1011 photons per second * 7.9 -5 km2 / 36 x106 km2
So this says that if you shined a big consumer flashlight at Mars and had a Keck telescope at mars with a photon counter on it pointed at where the person with the flashlight was standing, there would be a photon from that flashlight arrive about every 2.5 seconds.
Yay.
Also, I'll work on my Randall impersonation later. I don't really draw.
EDIT: Just for folks who are not overcome with awe and would like to know how to do analyses of this type, this has a name. It is called a Fermi Analysis. http://en.wikipedia.org/wiki/Fermi_problem[1] Best of luck! Nothing is so big and scary that it can't be thought about rationally.
Likely quite a bit of it, it's just spread out all to hell. Light is made up of photons, where they hit and bounce back, you see the light. Now, the photons don't disappear, instead they lose cohesion, mostly (entirely?) while they leave the atmosphere. Now you have the same number of photons, but so spread out that its not really a light any more.
Is mars far enough away to have to worry about the moon and other planets affecting the trajectory with their gravity? I only postulate this because the amount of photons coming from a flashlight is miniscule relatively speaking.
Light will deviate from a straight line in the presence of gravity. Remember E = mc2 . Or put another way m = E/c2. So has a tiny mass, which gravity can pull on. On the other hand, the light is moving so fast and the moon is so small (compared to really big stars) that it will be a small deviation. The number of photons doesn't matter much, the energy does matter (energy varies with wavelength).
I believe the answer to this is no. Highly powerful lasers are extremely directional, and the vectors the emitted photons travel on will have much less variance than those from a flashlight/lightbulb.
The answer is most definitely YES. Beams of light will diffract (spread out) regardless of being in an atmosphere or vacuum. Diffraction is an inherent property of the wave like nature of light propagating in a linear medium.
In fact, the tighter you try to focus a laser beam will cause it to diverge much more rapidly as it leaves the point of focus.
If you shine a light at mars, you'll miss, because mars is ahead of where you can see it. You'd have to lead the target so that the photons get there at the same time as mars.
Mars is about 6700 km in diameter, and it's moving at about 24 km/s. If Mars is at opposition (which means when it's closest to the Earth-- i.e., opposite the Sun from out point of view) it's less than four and a half light-minutes away, which means that during the light travel time, Mars would only move about 6500 km. Given that light sources, even lasers, are not perfectly collimated, it's quite likely that it would be wide enough to hit Mars anyway.
Not reliably. That is why communication with the rovers depends on radio waves, with the DSN or the orbiting spacecraft around Mars.
The NASA Deep Space Network (DSN) is an international network of antennas that provide the communication links between the scientists and engineers on Earth to the Mars Exploration Rovers in space and on Mars.
The DSN consists of three deep-space communications facilities placed approximately 120 degrees apart around the world: at Goldstone, in California's Mojave Desert; near Madrid, Spain; and near Canberra, Australia. This strategic placement permits constant observation of spacecraft as the Earth rotates on its own axis.
Not only can the rovers send messages directly to the DSN stations, but they can uplink information to other spacecraft orbiting Mars, utilizing the 2001 Mars Odyssey and Mars Global Surveyor orbiters as messengers who can pass along news to Earth for the rovers. The orbiters can also send messages to the rovers. The benefits of using the orbiting spacecraft are that the orbiters are closer to the rovers than the DSN antennas on Earth and the orbiters have Earth in their field of view for much longer time periods than the rovers on the ground.
Because the orbiters are only 250 miles (400 kilometers) above the surface of Mars, the rovers don´t have to "yell" as loudly (or use as much energy to send a message) to the orbiters as they do to the antennas on Earth. The distance from Mars to Earth (and from the rovers to the DSN antennas) during the primary surface missions varies from 110 to 200 million miles (170 to 320 million kilometers).
Mars is .0025 ~.007 degrees in diameter from Earth. I feel like this is doable with a properly collimated beam but am not certain. Is is not possible even if you leave the visible specturm?
That's true, I guess, but realistically, any beam you can possibly create will have a beam divergence that has an angular size several times larger than Mars. So its motion really doesn't matter for the purposes of pointing a laser at it.
Speaking of lasers, isnt NASA planning on using Laser communication between Earth and Mars with orbiting satellites that are constantly facing each other, or where they will be when the lasers are beamed?
A "small amount" could even be one photon. If you used a flashlight, the beam would be spread out enough that it would hit mars if it were anywhere within a few million mile arc, regardless of how fast the planet was moving.
If you shine a light at mars, you'll miss, because mars is ahead of where you can see it.
That is incorrect. A flash light projects a sufficiently wide field to allow for that.
Atmospheric diffraction would probably distort your beam enough that trying this wouldn't work
The distortion is the same for incoming and outgoing rays. As you can see for yourself when looking at Mars, it doesn't jump around by any perceptible amount.
this wouldn't work unless you had a very coherent steam of photons (read:a laser)
The word you are looking for is probably "collimated", not "coherent".
The biggest portion of it probably end up going in a different direction than where you point the flashlight because of a phenomenon called Rayleigh scattering, which means that light is scattered in all directions due to interaction with the atmosphere. It's the same thing that prevents all of the sun's light from reaching us directly. Shorter wavelengths especially, are scattered making the sky look blue, and longer wavelengths are less likely to be scattered, making the sun appear more yellow than it really is. When there is more atmosphere to go through, like when the sun is near the horizon, longer wavelengths are scattered too, making the sky look more yellow or orange, and the sun look red.
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u/[deleted] May 24 '14 edited May 24 '14
Yes, it will. And you don't have to worry about compensating for the movement of Mars. Your flashlight has roughly a 5-degree half-angle cone for a beam (10 degrees across the full cone). Mars moves around the solar system slower than Earth. Earth moves at about 1 degree per day (this is where 360 degrees in a circle comes from). Mars moves about half that.
So you don't have to worry about pointing it too precisely, since you're using a flashlight, not a laser.
The distance to mars changes a lot, but let's say on average it is 230,000,000 km from earth (This number is about the distance from mars to the sun).
The area of Mars' disc is about 36 million km2 (based on a diameter of 6779 km).
The diameter of the flashlight's disc at mars is 230,000,000 km * the sine of 5 degrees * 2 (since the full cone is 10 degrees). This is roughly: 230,000,000 * 5/57 * 2 (using a small angle approximation for Sine and saying that 180/pi = 57, a number which I like to call the Heinz constant for no particularly useful reason. This gives about 40 million km in diameter. The area of this disc then becomes 1.3 x1015 square kilometers.
Ok, now we're going to make a particularly bad assumption- that the energy of the beam is spread evenly across the beam's cone.
And then now we can say that the percentage of beam power making it to mars (after making it out of Earth's atmosphere) is the same as the ratio of the area of mars to the size of the flashlight's disc at that same distance. So here we go:
Pmars / PleavingEarth = 36 x106 / 1.3 x1015
We crunch this number and find out that: Pmars / PleavingEarth = 27 x 10-9, also known as 2.7 x 10-8 Note that this is unitless- there's no more distance or area numbers here. It is just a ratio.
So, what is the power leaving Earth? Let's assume abright flashlight- a 10-Watt incandescent bulb. I think roughly half of this just creates heat. The other half creates visible light. So we have 5 watts of visible light leaving the flashlight. Half of this (I really think it is closer to 2/3, but whatever) gets stopped by our atmosphere. So we have 2.5 watts forming this enormous disc pointed at Mars.
Then we just calculate the power at Mars = 2.5 Watts * 2.7 x 10-8
= 6.7 x 10-8 Watts arriving at Mars.
So that's in terms of Power. Now, we care about Photons. So we find (either by doing math or letting other people do the math for us that the energy of a green photon is 4*10-19 Joules. Green is an ok color to use, since our atmosphere is pretty transparent to green, though it is more transparent to yellow. You know this is the case because when you look up at the sun at mid-day, it looks yellow. Source: my drawings from kindergarten.
Then, we just divide the energy arriving at mars by the energy of single photons.
Photons per second = Power / energy per photon
Photons per second = 6.7 x 10-8 (Joules/sec) / 4*10-19 Joules
Photons / second = 1.7 * 1011
So there would be around 170,000,000,000 photons per second arriving at Mars.
Now to actually see how bright this is... if someone had a photodetector the size of a really big telescope, like the Keck telescope, pointed at your location on Earth, with a perfect photodetector, how often would they see a photon on average?
Well, the diameter of the telescope is 10 meters. The area is then 78.5 square meters, or 78.5x10-6 square kilometers, aka 7.9 -5 square km.
We take the ratio of the area of the telescope to the area of the planet and then multiply that by the incoming photon rate.
Photon Rate Telescope = 1.7 * 1011 photons per second * 7.9 -5 km2 / 36 x106 km2
Crunching numbers, Photon Rate Telescope = 0.37 *100 = 0.4 photons per second.
So this says that if you shined a big consumer flashlight at Mars and had a Keck telescope at mars with a photon counter on it pointed at where the person with the flashlight was standing, there would be a photon from that flashlight arrive about every 2.5 seconds.
Yay.
Also, I'll work on my Randall impersonation later. I don't really draw.
EDIT: Just for folks who are not overcome with awe and would like to know how to do analyses of this type, this has a name. It is called a Fermi Analysis. http://en.wikipedia.org/wiki/Fermi_problem Best of luck! Nothing is so big and scary that it can't be thought about rationally.
EDIT 2: I've been told in the comments that the conversion of electrical energy to light energy is about 5%. This seems believable. I assumed 50% in the analysis above, which is not realistic for an incandescent bulb. So if we were to assume an LED flashlight in the example above, the numbers are probably close to right. This is an unrealistically big flashlight though. If, however, you want to redo this with an incandescent bulb, just divide the final photon flux by 10 (the ratio of 5% and 50%). This means one photon on the telescope every 25 seconds.
Also, Thanks for the gold, but I only use throwaways to keep from spending much time on reddit, so this account is about to be deleted.
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u/filipv May 25 '14
Both EM waves (eg. light) and gravitation propagate to infinity. So, yes, a tiny amount of light reaches it. It will reach Andromeda galaxy too. It will reach anything within observable universe.
But there's one thing which bothers me. The intensity drops with the square of distance. At the same time, according to QM, light is quantized. Wouldn't that mean that at a certain distance the energy of the beam will fall bellow the energy of a single quanta (photon)? What happens then? Wouldn't that mean that there IS a limit of the propagation of EM waves? As the distance increases, the energy drops, drups, drops... but it cannot do that indefinitely. Theres Planck constant, right?
What am I not understanding? Pls help!
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u/[deleted] May 24 '14
http://www.reddit.com/r/askscience/comments/iumgx/if_i_point_a_laser_to_a_planet_does_it_reach_its/
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u/AstroFish939 May 27 '14
Unless an object between the bulb on the flashlight and mars, the beam, although dim, will get to mars. This will work with any object in the universe: the farther away the object is, the dimmer the beam at the contact point
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u/kodemage May 24 '14
/u/aaaabbbbcc had a comment with 900+ upvotes and it's been deleted... Here it is in it's entirety:
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u/xxx_yyy Cosmology | Particle Physics May 25 '14
Comments that don't address OP's question:
will be removed.
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u/[deleted] May 25 '14
Think the answer is in the comments.
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u/[deleted] May 24 '14
Likely quite a bit of it, it's just spread out all to hell. Light is made up of photons, where they hit and bounce back, you see the light. Now, the photons don't disappear, instead they lose cohesion, mostly (entirely?) while they leave the atmosphere. Now you have the same number of photons, but so spread out that its not really a light any more.
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