Escape Velocity - Video Tutorials & Practice Problems
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concept
Escape Velocity
Video duration:
7m
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Video transcript
Hey guys. So now that we've talked about velocities in gravitation, there's a very specific velocity that you're gonna need to know called the escape velocity. Let's check it out. So basically what the escape velocity is, it is the minimum launch speed that you need in order for an object to escape. And now what that word escape means, what does that mean? It means that this thing stops when it's very, very far away and it never can come back towards the earth. So let's think about something for a second. So we're used to on earth will throw an object upwards, it'll sort of reach the peak and it'll come all the way back down because of the force of gravity, right? We know that gravity pulls everything downwards if you throw this thing a little bit harder. So we've got some initial velocity right here. If you throw this thing a little bit harder, it's still gonna go up. Gravity is gonna get a little bit weaker, but it's still eventually going to come crashing back down towards the earth. The idea is that there is a launch speed. So there is a launch speed. A minimum launch speed that is so fast. So we're gonna call that V escape. Now, what happens is it gets all the way up to this magically very, very far away place. And what happens is that the force of gravity can't pull it back down towards the earth again. And so what happens is that it stops and it never returns. Now, we know, but when, when two objects get really far away from each other, the force of gravity approaches zero. So in other words, as this little R goes to infinity or gets really, really, really big, we know that the force of gravity is equal to GM little M over R squared. So if this thing in the denominator gets really, really big, then the force is goes to zero. OK. So what that means is that the object is going to stop and when it stops very far away, the final velocity is equal to zero. So in other words, the final, when it gets all the way out here to this magical place, that's really, that's infinitely far away that the final velocity is equal to zero. And the reason for that is that if it wasn't zero, if it was something that's greater than zero, then it isn't the minimum launch speed that you could have thrown this with. So I'm gonna make up a number here just for a second. Let's say you throw this up at like 1000 m per second, it gets up all the way out here and the final velocity is equal to 5 m per second. Well, then that means that you could have thrown it a little bit slower and it still would have gotten out here with 0 m per second. So that's that what that means? You just have to throw it with the minimum launch speed. All right. So this escape velocity actually comes from conservation of energy, right? We're dealing with gravitation. We have an initial position over here, a final position over here. And we know that we can't use kinematics because the accelerations are constantly changing. So we have to use conservation of. So we've got initial kinetic and potential plus any work done is equal to final kinetic and potential. Now, when you are throwing this object up with some initial velocity right here, we know that the kinetic energy is gonna be one half MV squared. So we've got one half MV squared and now the initial gravitational potential energy is not zero because you still have some distance away from the center of mass. So that's gonna be negative G big M little M over little R. Now we're talking about work, sorry, we're talking about gravity. We know that gravity is a conservative force. So that means that there's no work done by non conservative forces. It's kind of a double negative there. Now, what about these last two? What about the energies when it finally reaches up this very, very far away place. Well, we said that the energy or sorry, the velocity is equal to zero when it finally gets out all the way over here. And we know that the kinetic energy is also zero because it depends on this final velocity. Remember one half MV squared. So this thing is one half MV final squared. But we know that this thing is gonna be equal to zero. So that means that the kinetic energy is gonna be zero there. Now, what about the gravitational potential? Or again, remember the equation for gravitational potential. It depends on that little R distance. So what happens is is that as this distance gets really, really big, then the gravitational potential energy will also go to zero. So that means that both of these on the right will go to zero. And that allows us to figure out what the escape velocity is. So we're gonna go ahead and move this over to the other side. We're gonna get one half MV squared equals G big M little M over little R. Now, what we can do is we can cancel out the big MS that appear on both sides. And this then turns into uh we're gonna have to move the one half over to the other side. We're gonna get V squared is equal to two G big M over little R. And when we take the square roots, we're gonna end up with the equation for the escape velocity. So that's two GG big M over little R and square rooted. So the important thing to remember here about this equation is that the escape velocity only depends on two things. The mass that you're trying to escape and the distance your initial distance from that object that you're trying to escape your mass, the mass of the object has no difference. It makes no difference to the escape velocity. It just depends on the thing that you're trying to escape and how far you are away from that thing. All right guys, that's basically it. So let's go ahead and check out an example. Very common example. You guys will see. Uh So how fast must a 5 kg rock be thrown to escape the sun if it is thrown directly near, directly outwards uh near earth's orbital distance. OK. So we're gonna be looking for how fast must we throw an object to escape the sun. So we know we're gonna be dealing with an escape velocity equation that's gonna be given right here. So we know that the escape velocity is going to be equal to the square roots of two GM over little R. Now let's look at my variables. I know what this G is and I have the mass of the sun. That's the thing I'm trying to escape. So now all I have to do is figure out what is my initial distance from the sun, that thing that I'm trying to escape. So let's just go ahead and draw a little diagram right here. I've got the sun and I'm gonna have the earth that's all the way out here. Uh Something like this now. So we've got this distance over here. So if we're gonna be, if this is the earth's orbit, this little dotted line as it goes around the sun, we have to do is we have to throw this object directly outwards starting from this position right here. So that means that this is my little R and I'm gonna call this se the sun earth distance. Now, we know this is actually equal to the radius of the sun. So big R plus H because if this is my little R right? Or this is my big R, the radius of the sun and then this is my H my distance above it, then these two things will give me a little R. But what happens is that in this, uh in this particular case, the radius of the sun is much, much smaller than the actual distance in between um the sun and the earth. So what happens is that this ends up being a much, much bigger number. And so this little RSE actually just turns out to be the average distance between the earth and the sun. So that's the number we're gonna plug into there. All right. So we're gonna get the escape velocity is equal to, and let's see. So V escape is equal to square roots, two G 6.67 times 10 of the minus 11, the mass of the sun is two times 10 to the 30th. And now we're gonna have to plug in uh 1.5 times 10 to the 11th and then you got to square root there. So if you work this out, you're actually gonna get 42 or 4.21 times 10 to the fourth meters per second. So that's actually about 42.1 kilometers per second. That's how fast you would have to throw something starting from the earth's distance outwards so that it escaped the sun and can never actually come back into the solar system. All right guys, that's it for this one. Let me know if you guys have any questions.
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Problem
Problem
a) How fast does a spaceship have to go to escape Earth, starting from the launch pad on the surface? Assume it burns all its fuel very fast and then shuts off the engines.
b) One idea to make getting to space easier is to build a space elevator, a large platform high above Earth's surface where spaceships can land and take off. How high above Earth would this platform have to be for the escape velocity to be 1/5 of its surface value? (For multiple choice, select the correct answer for part (b).)
A
m
B
m
C
m
D
m
3
example
Jumping Off an Asteroid
Video duration:
7m
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Video transcript
Hey guys, let's work out this problem together. So we're landing on the surface of a large spherical asteroid. We're gonna set off walking one direction, blah blah blah, we're gonna drop some tools and eventually we have to figure out how fast do we have to jump in order to escape this asteroid. So what variable is that? What variable are we talking about here? We're talking about an escape velocity. So we can look at our escape velocity. So we've got V escape is equal to the square root of two big G big M over little R. If I take a look at this equation, if we need to figure out the escape velocity, I know that G is just a constant. I do not have the mass of the asteroid. And I also don't have the initial distance. I have any, any information about that. So this is a problem where we're gonna have to basically go it, go off in a different direction and get what these variables are. We're gonna have to go and get M and then go and get little R and then plug them back into this problem. So I like this problem because it's gonna combine a lot of different things from this chapter. Let's first start out with just drawing a diagram. So I've got a large spherical asteroid that I am gonna land on. So like that's, and the first part of this problem says that I'm gonna set off walking in one direction. And after some later time, I've realized that I've basically come back around to uh return to my spaceship. And I checked my little pedometer on my watch and I've gone 25 kilometers. So that first part of the problem um means that we are traveling around a spherical asteroid, which means that we're basically traveling around the circumference of it. So this represents the circumference. And let's check out what that circumference means. In uh in terms of an equation, we know that the circumference of a circle is related to the radius of that circle by C equals two pi R. So we can actually go ahead and figure out what the radius of this circle is. This little R distance by using this equation C equals to pi R. We know what the circumference is that 25,000 in terms of meters, all we have to do is just divide it by two par two pi and we get the radius and that's about 3980 m. So that is one part. So now we have the radius. Now, I need to go ahead and figure out what the mass is and I'm gonna go ahead and do that over here somewhere. So I need to figure out what the mass of this asteroid is. So we're done with sort of this first part of the problem. Now, we got to look at the second part. We're gonna grab a tool from the toolbox and drop it and then noting that again, using our watch or something like that, it takes about 30 seconds to hit the ground from 1.4 m high. So if we're looking for the big mass of the planet, we have two approaches. We could use, well, you could use um forces, gravitational forces or we can use gravitational accelerations. Hopefully you guys recognize that this part of the problem right here, this dropping 30 seconds and hitting the ground, sort of looks like a kinematics problem, the vertical kinematics, which we've done a lot of before. So I wanna just go ahead and draw a quick diagram of what's going on in that part. So you've got this little uh this ground right here and I'm gonna be dropping a rock and I'm told that the distance that it takes uh that the distance that it falls is equal to 1.4 it takes a time of 30 seconds and we know that it is dropped, which means that the initial velocity is equal to zero. So what, what is my target variable here? Well, I'm looking for uh let's see if I can keep going. This is gonna be my delta Y, right? It's gonna be the distance that this thing falls. I have the final velocity when it gets to the surface. But I don't know what that is and I have the gravitational acceleration that is gonna be G at the surface, right? Because this is gonna be gravitational acceleration, but we're standing really close to the surface of the asteroid. And I also don't know what that is. So if I take a look at my equations for M, my M equations involve forces and gravitational accelerations forces. I don't have any information about masses and forces. Uh But I do have some sort of kinematics in which I have this acceleration. So I'm gonna grab G surface as the equation I'm gonna use because I'm on the surface and that's gonna be big G big M over big R squared. So now what happens is if I can figure out what this uh mass is. Um So let me go ahead and rearrange for that, right? So I've got G surface then I've got R big R squared divided by big G equals M. So I just need to figure out what this G surface is by looking at this part of the problem. So I basically got to go somewhere else to go and find out what G surface is. Let's go ahead and do that. So if I'm looking for G surface here. I need, that's a kinematic variable. I need three of the five other ones, right? So I need, I've got the velocity, initial velocity final, I've got the T and the delta Y, which one of those do I? Which ones, which one of the variables do I have? I have the initial velocity, I don't have the final velocity I, the time and I have the delta Y. So I'm gonna pick the equation that doesn't have that V final. Hopefully, you guys remember your kinematic formulas. This is gonna be delta Y equals the initial velocity T plus one half G times T squared. Now, we know that the initial velocity is equal to zero. So that means that this is going to cancel out. And then we basically can choose this downward direction to be positive because in everything turns, turns out to be positive. So the delta Y is gonna be 1.4 that's the meters. And then we've got one half of G and this is gonna be G surface and then the T is just equal to 30. Then we have that squared. So let me go ahead and move down. So basically, I'm gonna move this one half over to the other side and then the 30 squared goes down. So we basically end up with two times 1.4 divided by 30 squared. Let me go ahead and write right then uh D divided by 30 squared. Is equal to G surface. So you go ahead and plug all that stuff into your calculator. You should get that G surface is equal to 3.11 times 10 to the minus three. So now we have G surface, so now we can plug this thing back into this equation. And if we go ahead and sol for M, we're gonna get 3.11 times 10 to the minus three. Now we've got R squared, big R squared. What is that number right there? So now what we've got to do is this big R squared corresponds to the radius at the surface. In other words, this distance right here, which is actually this R that we calculated in this part. So this little R is actually big R and that's 3980. So now we can go ahead and plug that in. So we've got 3980 that's gonna be squared. Now, we've just got to divide by 6.67 times 10 to the minus 11. That's capital G. And that's gonna be the mass of the uh spherical asteroid, which is equal to, we got 7.39 times 10 to the 14th that's in kilograms. Now that I have this mass right here, I can plug this all the way back into that equation and I can figure out what the final escape velocity is going to be. So it's basically just like working out words to find all your variables and then plugging them all back into your original equation. That's what we've done. So we've got the escape is equal to. Then I've got the square roots of two times 6.67 times 10 to the minus 11. Now, I've got the mass of the asteroid which is uh 7.39 times 10 to the 14th. So let me go ahead and actually move all of this stuff down. I'm gonna move this stuff down a little bit there. There we go. And then I've got uh divided by r the distance which is gonna be 3980. If you go ahead and plug all this stuff in, you're gonna get, the escape velocity is equal to 4.98 that's meters per second. So this is how fast you'd have to be jumping, which is actually probably doable if you were actually uh like in a spacesuit. So this is the final answer for how fast you'd have to be traveling to escape this asteroid. Let me know if you guys have any questions.