Radiation Pressure - Video Tutorials & Practice Problems

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1

concept

Radiation Pressure

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7m

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Hey folks. So in this video, we're gonna talk about radiation pressure, which is a kind of counterintuitive concept. But we're actually gonna see that we use this all the time in the development of space technologies and other kinds of things. And we also need to know how to solve problems involving radiation pressure. So I'm gonna show you how to do that. It's pretty straightforward. We'll leave if just a few equations. So let's check this out here. So remember like all waves, electromagnetic waves carry energy, we saw that in the last few videos where electromagnetic waves have intensities that we can calculate things like that. But additionally, electromagnetic waves also have momentum, which is a little bit of counter of a counterintuitive idea because we remember that the equation for momentum is that P equals MV. So one of the things you might be wondering is how can light have momentum if it does not have mass? It's one of those things that we just sort of discovered in the early 19 hundreds. After a bunch of experiments, we found that light can also act as if it has mass. It's one of the things that you're just gonna have to trust me about. So if light has momentum, that means it also has to obey conservation of momentum. So basically, what happens is that whenever light hits an object, for example, if you were to shine a flashlight onto a piece of paper, it actually transfers its momentum and it basically pushes an object or pushes objects with a force. All right, now, there's two basic types of situations that you'll see in your problems, one where you have absorbed light and then one where you have reflected light, the problems will almost always tell you which one you're dealing with. All right. So let's look at these two examples or these two different cases here in a case where you have a absorbed lights, imagine that you had a flashlight. So imagine that you have all these electromagnetic waves that are hitting this piece of paper over here. Now, those electromagnetic waves have momentum. So in other words, you have some initial momentum here. And what happens is that the cons the momentum has to be conserved. So when the light gets absorbed into the paper, one of the things it has to do is it has to transfer some of it so that the paper starts moving. So it's gonna have some tiny little velocity. And the only way it does that is if it accelerates the paper and that basically means that it's exerting a force on the paper. All right. So when light hits an object, it actually pushes on them with a force which is kind of counterintuitive. This is actually very similar to a completely inelastic collision. Remember a completely inelastic collision is where you have an object that hits another object, they stick together and they both move as one. The other kind of situation is where you have reflected light. This could be something like a mirror or, or something that basically just reflects all the incoming light backwards. So this is different because here you actually have light beam that's hitting the mirror and it's gonna be going in this direction. And then later on, it's actually gonna go backwards and it's gonna go in this direction. So it's basically just gonna go back the way it came in this case. What happens is the momentum change between the light is actually greater than it was for the absorbed light case. And so basically what happens is that the force that it exerts on that object is even greater. So all I'm gonna do is I'm gonna say that this is F absorbed and this is F reflected. And what we can say is that F reflected is always greater than F absorbed. All right. So this also this piece of paper, this mirror also is gonna gain some velocity, but the force is gonna be even greater. This is actually similar to an elastic collision. Remember, an elastic collision is like where you had like a, like a billiard ball where it hits something and then rebounds backwards. Now, let's go to like talk about the equations because they're actually straightforward, but they're also similar to each other for the uh case of the, of the force and the absorbed light case, it's actually just equal to the intensity times A divided by C. And for the reflected case, it's similar, except there's gonna be a two in front of it, there's two IO A over C. So basically, the relationship between these two formulas here is that F reflected is equal to two times F absorbed. And this is because again, the momentum change of the incoming light is bigger, basically, it just rebounds. Uh And so the momentum shift of the object has to be larger in order to conserve momentum. Now, for the pressure equations, that's why we call it radiation pressure. I remember that the, the, the, the relationship between pressure force an area is that pressure is equal to F over A. So if we just take this equation here and we divide out the area, basically what you'll get here is that this is I over C and for the reflected case you'll, that this is two I over C, all right. So reflected always is gonna have a two in front of it. Otherwise the equations are pretty much the same. Let's just go ahead and take a look at an example here. So we can see how this works. So we've got a laser pointer that we're gonna be shining onto our hands and we're told that the average power output is five milliwatts. So this is our power over here. Now, the beam focuses onto a small area on the palm of our hand. So this is gonna be our area over here. That's our a and we're told that the, our hand completely absorbs the incoming light. So, problems again, are almost always gonna tell you what case you're dealing with. So we're dealing with complete absorption. Now, the first thing we wanna do is we wanna calculate the radiation pressure. So if I'm dealing with an absorption case, then I'm basically just gonna look at these formulas over here. All right. So if, if in part A I'm gonna be solving for the radiation pressure and that means that I'm gonna use the P abs uh absorbed equations. This is gonna be P absorbed is gonna be I over C. All right. Now, remember now I got the uh that C is just a constant over here, but I don't have what the intensity of light is. But remember I can always figure out intensity because intensity is related to power over area. All right. So if you actually rearrange this equation, what you'll see is that P absolute is equal to power divided by area times C. So don't get these two confused by the way, this is gonna be a lowercase P that's P, that's lowercase P for pressure. Uh Not to be confused with uppercase P for power. All right. So I'll try to make those big, so you won't get confused. The power output is five times 10 to the three watts divided by the area which is one times 10 to the minus six. And then C is gonna be three times 10 to the eight. That's the speed of light. If you go ahead and work this out, what you're gonna get is that the pressure is 1.67 times 10 to the minus six or sorry, 10 to minus fifth. And that's gonna be pascal. So this is your final answer for the radiation pressure. That's actually a very, very, very small pressure. If you think about it, let's move on to the second part here, which is, we're gonna calculate now the force that is exerted on your hands. So now we're gonna be dealing with the force absolute over here. Now, there's a couple of different ways you can actually go about this. You could just go ahead and plug it, plug this formula in which is the I A over C that would totally work or what you can do is you could just use the relationship between force pressure and area. So in other words, your force from the absorbed light is gonna be the pressure of the absorbed lights times the area. Remember we always have P equals F over A. So because we just figured out what this is in part A then we can just go ahead and make this a little bit easier on ourselves. So this is gonna be uh one times 61.67 times 10 to the minus fifth pascals times one times 10 to the minus six. That's uh meters squared. And what you're gonna get here is you're gonna get uh 1.67 times 10 to the minus 11 newtons. And that is your final answer. If you use the other equation, we would have, we would have gotten the exact same answer. All right. So if you look at this is actually an incredibly tiny force and that makes sense because the laser, when it shines onto our hand, it's not like pushing our hand away. It's an incredibly credibly small force. That's it. For this one. Folks. Let me know if you have any questions.

2

Problem

Problem

A radio transmits a wave with intensity 27.0 W/m^{2} towards a flat surface (perfectly reflecting) with area 2m^{2}. Calculate the force and radiation pressure on the surface.

A

Pressure = $1.8\times10^{-7}Pa$ ; Force = $3.6\times10^{-7}N$

B

Pressure = $9.0\times10^{-8}Pa$ ; Force = $1.8\times10^{-7}N$

C

Pressure = $1.8\times10^{-7}Pa$ ; Force = $9.0\times10^{-8}N$

D

Pressure = $9.0\times10^{-8}Pa$ ; Force = $3.6\times10^{-7}N$

3

example

Example 1

Video duration:

7m

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All right folks. So hopefully you get a chance to tackle this problem on your own. If you got a little stuck, I'm here to help you. Uh Let's just go ahead and get started. We'll work it out together. So we want to make a laser beam or we want to use a laser beam to make an object float in the air. All right, we're told that the laser beam is pointing straight upwards. Uh And it exerts a radiation pressure on a horizontal disk and we're told what the mass and radius of that disc is. All right. So this problem is very visual. Uh So I actually sort of want to draw this out here. So it kind of makes sense. So we've got this disk here and we're told that the radius of this disc is equal to 100 millimeters. All right. So the radius of this disk here, uh I'm gonna call this, I'm gonna call this, our dis is equal to 100 millimeters. And I'm also told that the mass of the disc is 7.8 g. All right. So the disc which completely absorbs light um is 2.75 m from the laser, the radius of this beam is 50 millimeters, basically what's going on, right. You've got this horizontal disk here, the laser beam is pointing straight upwards and, and it's gonna sort of try and make this object float by just using radiation pressure. So you've got this laser beam that's pointing straight up at this thing. Here's like the laser, here's all the light and stuff like this. All right. And we want to make this object float. All right. So we're told that the distance from the disk to the laser is 2.75 m. I'm gonna call this Y 2.75. Uh And let's see, the radius of the beam itself is 50 millimeters. So, whereas the radius of the disk was 100 the radius of the beam is 50. So it's smaller than the disk, right? So this distance over here. So I'm gonna call this, our laser is equal to 50 50 millimeters. OK. So we want to figure out what's the laser power in order to be able to balance this disc here. All right. So there's a lot of variables going on here, but we really just want to figure out what is P, what is the P, the power of the laser? All right. So then let's get started here. How do we figure this out? Well, first of all, we have to figure out what floats even means, right? The object wants wants to float in the air, one with the power so that the disc is balanced and sort of suspended in the air like this. So remember what objects float? What that means is that the forces will cancel the, basically the sum of all forces is zero. So in other words, there is a force that is being uh produced from radiation pressure and there's gonna be a downward force because of gravity. So that's just gonna be FG force of gravity. When those two things are equal to each other, the disc will float. So in other words, what happens is that float? Really just means that F radiation is equal to FG, they're equal and opposite. All right. So that's the first thing you need to know, right? And uh we'll figure out which type of radiation pressure this is whether it's reflecting or absorbing. Uh Basically, you can actually figure this out right here. This completely absorbs like this disc. So that means that we're not gonna be using reflecting equations and instead we're gonna be using our absorption equations over here. So actually we do know what type of pressure or sorry, what what type of surface this is, this is actually gonna be the force of absorbing radiation pressure. All right. So when F absorbed is equal to FG, you have this object here that floats. All right. So basically, what I want to do is I want to set these two equations. Equal to each other. All right. So let's start with that through our f absorbed equation. Remember uh just has involves I A over C there's no factor of two in there. So basically what I have here is that what I have is I times the area uh divided by C and this equals FG over here. Remember the force of gravity is really just M times G and this is actually just gonna be the mass of the disk times G. All right. So we have with the mass of the disk is G is just a constant 9.8 C is a constant. So now what we need is we're gonna need the intensity and the area. But notice how in this problem, what happens is I actually don't have intensity. Um And that's actually the, our target variable isn't really in this equation here. Remember our target variable is power. So in other words, we want to start with our F absorbed equation in FG. But ultimately, we want to figure out P, how do we do that? Well, remember that intensity and power are always related by this equation over here. I equals P over A. All right. So what happens here is I'm gonna take this intensity and actually gonna sort of rewrite this as power divided by area. All right. So when I do that, what ends up happening is I end up getting power times area divided by C uh top sorry, power over area times area over uh over C equals M disc over or times G. All right. So in other words, what you can see here is that the areas are actually gonna cancel. All right, notice how, what happens here is that the force is absorbed also depends on the area. But there's actually two different types of areas here. There's the area of the disk itself and there's the area of the laser beam itself. Really. What happens is that these areas, um the area that is going to be used in the force calculation really just has to do with the area that the laser beam is hitting, right? Because what happens is it's not like the laser beam is hitting the entire surface area of the disc, it's hitting a smaller portion of it. So the only thing that actually exerts pressure is the part of the area that the laser beam is hitting. So in other words, this area inside here, everything else that's outside of that actually doesn't matter, doesn't contribute to the force that's being generated. So the area here this laser, but the area from the intensity equation is also the area of the laser. So this is basically just a a laser on top and bottom all that's just to say that that actually just cancels out. All right. So uh basically what happens here was actually you don't have to figure out what the area is this P is just gonna equal M dis times G and then we move the C over to the other side as well. So this is gonna be the power of our laser. So now we're actually just ready to start plugging in some numbers. So we have the mass of the disk which is 7.8 g. Uh The 7.8 g is equal to 0.0078 kg. To be very careful with your units, you have to plug stuff in, in SI units. So this is just gonna be 0.0078. This is gonna be 9.8 and this is gonna be three times 10 to the 8 m per second. Now, when you work this out, you should get it for a power is you should get 2.29 times 10 to the seven and that's in watts, that's actually equal to 2 22.9 megawatts. Uh And so far, at least currently, I don't think there's a laser that can provide that much power, but that is your final answer. That's how much power you would need to make a tiny little disk suspend in the air just from radiation pressure. So again, we can see that the forces involved in the radiation pressure are very, very, very small, but they are measurable. Uh And it takes a lot of light to do that. All right, to actually make something float. All right. Folks. So that's it for this one. Let me know if you have any questions.

4

example

Example 2

Video duration:

6m

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Welcome back, everyone, hopefully had a chance to work this problem now on your own. So we're told this problem that is spacecraft with a reflective sail like material may actually eventually used for low cost space travel. This is a real technology that's actually being used. Um And the science is sound. So we're told that a 400 kg satellite near earth is equipped with these super reflective sails. So I kind of want to draw this out here. And basically what happens is those sails are gonna capture sunlight from the sun and it's gonna generate a radiation pressure. So let me just draw this out here. So you got the sun like this. Uh sorry about my sun. And then at some distance where the earth is, you're gonna have this satellite over here, the satellite like this and this satellite has these big big reflective sails like this. So what happens is the sunlight that travels is gonna hit those reflective sails and it's eventually going to push them to the right. So this is gonna be an f from radiation pressure. All right. So eventually what happens is that, that that object, that satellite is gonna pick up speed and that's actually one way that we could generate large amounts of velocity for spacecraft with very little fuel. You're only just using the light from the sun. All right. So let's take a look here. The area of each of these reflective sails is 5000 m squared. So, in other words, the area is equal to 5000. Uh but actually what happens is the total amount of area is going to be twice of 5000 because there's two of them. So there's two of them uh each are completely, all right. So this area total is gonna equal 10,000. And also we know that we're dealing with completely reflective sales, which means we're gonna be using our reflective radiation pressure equations and not our absorbing radiation equations. All right. So the intensity of sunlight is approximately 1350. In the words, the eye from the sun is about 1350. So now what we want to do is we want to figure out what is in the first part, what's the force that's calculated? Or what's the first force that's exerted on the satellites? All right. So that's F, so we actually know also what type of F we're dealing with again because we know that it is reflective. So we basically want to figure out what is F reflective. All right. So let's get started here with our equation. Our equation says that F reflected is equal to two times I A over C. All right. So do we have everything we need, we have the intensity, we also have the area and we have C that's just a constant. So in other words, what happens is you get two times the intensity which is 1350 in the area, which area we're gonna use here? We're gonna use 5000 or 10,000. Remember we're gonna use the total area of both of the reflective sales. So that means we're actually gonna use the 10,000. All right. So 10,000 is our total area over here. Then we divide by speed of light three times 10 of the eighth. And when you get, uh when you calculate this, what you're gonna get for F reflect is you're gonna get a force of about 0.09 newtons. So this doesn't seem like a very, very big force. But remember this is a constant 0.09 force that will be pushing this satellite. All right. So now let's take a look at the second part here. That's the first part. The second problem asks us assuming that the satellite starts from rest. Then how is it moving? How fast is it moving after one year? So how fast is it moving? That's actually gonna be a velocity. So we're actually gonna look at this force here and we're gonna see how fast it's moving after a certain amount of time. So hopefully, you folks realize that, that has nothing to do with intensity or radiation pressure or anything like that. That's just straight up kinematics. So what, what's basically asking us here is what is the final after one year? So the delta T is gonna be one year. All right. So let's just look at our kinematics equations. This is a constant force that will be pushing this. So it's gonna be a constant acceleration. All right. So in order to figure out the final uh we have, we need to have the initial plus the acceleration times time. That's the most simple velocity formula um of the kinematics equations, right? So we're really looking for what is the final? Now, let's take a look at what the initial is, right? We're told actually what the initial is because we're assuming that satellite starts from rest. So that basically means that V knot is equal to zero. All right. So now we just need acceleration and time. All right. So how do we figure out the acceleration? Well, we can figure out acceleration because we have the force we have 0.09 newtons. So let's just go over here real quick and figure out the acceleration A is equal to remember just F equals ma uh F is um A is equal to F divided by M. So in other words, we have uh f reflect over mass which is 0.09 divided by the 400 kg mass. And what you should get is you should get a acceleration of 2.25 times 10 to the negative or, and that's gonna be meters per second squared. All right. So now this equation, this calculation, sorry, this acceleration but get plugged into the velocity formula. And now the other thing we have to figure out is what is the time. So in other words, this t is equal to one year. And basically what you could do is you could multiply a bunch of stuff out on your calculator. You can multiply this by 365 days times 24 hours times 60 minutes times 60 seconds. And you can see that all the, the units will cancel. Uh But basically what you should get is three points about 15 or 33 million, 153,600. Um That's, you know, uh you can calculate that out yourself. And let's see. Now we plug this into our equation for V final. You get V finals equal to zero plus and then we have two points 25 times 10 to the minus four and then we have seconds, which is 3 million, 153,000. You have to convert it because you can't plug in years into this formula. You have to have everything in si units and what you should get out of here folks is you should get about um you should get about 7100 m per second. All right. So this is actually your final answer for that second part here. So even though we can see that the force is actually very small, the force is only about 0.09 newtons. After it acts for a long time, you actually can get some pretty significant velocity changes. All right. So this is one of those things where we would send the satellite out and years later it would be traveling super, super fast out of the solar system. Anyway, that's it. For this one. Folks, let me know if you have any questions, I'll see you in the next video.

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