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Would a brick or feather fall faster?

What would fall faster on the moon, a brick or a feather? Created by Sal Khan.

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  • blobby green style avatar for user Kevin Kwok
    I'm still convinced the brick would still, albeit imperceptably, "hit the surface of the moon" (the question posed in ) before the feather (even assuming that the moon is a true vacuum). F_f and F_b aren't only the forces of the moon pulling on the objects, but also the same force that the object pulls on the moon, so the moon is accelerating at G*m_f and G*m_b respectively, where the latter is larger and thus the brick collides first. My question is: am I wrong?
    (69 votes)
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    • leafers sapling style avatar for user Peter Collingridge
      Technically, you're correct.

      When you drop the brick, it accelerates towards the moon at the same rate as a feather would, but the moon also accelerates towards the brick a tiny amount. This tiny amount is a slightly bigger tiny amount than the moon would accelerate toward a feather.

      Of course, if you drop the brick and the feather at the same time, the moon will accelerate towards them both, so they will collide at the same time.
      (111 votes)
  • mr pants teal style avatar for user Sumaiyah Ihatexams
    the last experiment was ToTaLlY cool but why is it that the paper didnt float up but fell with the book???!!!
    (25 votes)
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  • hopper cool style avatar for user Riya
    why does the moon have no atmosphere
    (21 votes)
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  • blobby green style avatar for user eseraserra
    Why is r(or distance) squared? Isn't it just distance from one object to another?
    (9 votes)
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  • blobby green style avatar for user anthonymendoza132
    can you explain air friction to me? for example, when a skydiver is falling and reaches a terminal velocity of 120MPH after falling for 15 sec, what is the force of air friction on the skydiver as he falls at terminal speed?
    (7 votes)
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    • male robot hal style avatar for user Reinhard Grünwald
      The reasoning for AndrewM's absolutely correct answer is this:

      Terminal speed means that this speed no longer changes. In other words there is no accelleration happening. This means there is no net force acting on the diver. This means that all forces on the diver have to balance each other. There are only two relevant forces: friction and gravitational pull.
      (9 votes)
  • male robot donald style avatar for user Mohammad Umar
    what is the ratio g(earth)andg(moon)?
    (4 votes)
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  • leafers ultimate style avatar for user walker.jonathon
    If you would shoot a gun and drop a book would the bullet and book hit the ground at the same time
    (3 votes)
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    • leaf green style avatar for user tankballs22
      it depends entirely on the direction you fired the gun. if you fired it perpendicular to the surface, and the bullet had enough velocity, it could potentially reach orbit. if you fired it straight into the ground, obviously it would be moving much faster than pure gravity could pull it, and therefore hit the ground before the book.
      (4 votes)
  • primosaur ultimate style avatar for user Tomislav Peran
    a) We could calculate time by using kinematic formula: Vo=0 -> t=2X/g. From here we can see that the mass is irrelevant.

    b) Air resistance is: F=p*A. Bigger the surface bigger the resistance force.

    Is this correct?
    (3 votes)
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  • leaf green style avatar for user Hadia
    How do we prove that the acceleration due to free fall is the same as Earth's gravitational field strength?
    (1 vote)
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  • old spice man blue style avatar for user TheMarioGomez10
    Hello! I have a question regarding Newton's second law and law of gravitation. If, through reconciling F=ma and F=G(Mm)/r^2, we see that the acceleration due to gravity on Earth is equivalent to the mass of the earth multiplied by the universal gravitational constant, divided by the square of the distance between the object and the Earth, shouldn't we see a massive decrease in acceleration as distance increases? Why don't we see it in real-life? Please let me know if I am mistaken. Thank you very much!
    (2 votes)
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Video transcript

Let's say we were to take a little excursion to the moon. And so here we are sitting on the surface of the moon-- that's the surface of the moon right there. And with us to our excursion to the moon we brought two things. We brought ourself a concrete brick. So that's my brick right over there, although it's orange-- we'll say it's an orange concrete brick. And I also bought a bird feather with us. So this is the bird feather. And then my question to you is, if I were to hold both the brick and the bird feather at the same time, and I were to let go of both of them at the same time and ask you, which one of them would hit the surface of the moon first, what would you say? Well, if you based it on your experience on Earth-- on Earth if you were to take a break and a feather, a brick would just go straight down. A brick would just immediately fall to the Earth, and it would do it quite quickly. It would accelerate quite quickly. While a feather would kind of float around. If you had a feather on Earth, it would just float around. It would go that way, then it would go that way, and it would slowly make its way down to the ground. So on Earth, at least in the presence of air, it looks like the brick will hit the ground first. But what would happen at the moon? And what's interesting about the moon is we have no atmosphere. We have no air to provide resistance for either the brick or the feather. So what do you think is going to happen? So your first temptation would say, well, let's just use the universal law of gravity. So what is the force of gravity on the brick? Well, you could calculate that out. The force of gravity on the brick is going to be equal to big G times the mass of the moon-- I'll say that's m for mass and then the subscript is lowercase m for moon-- the mass of the moon times the mass of the brick divided by the distance between the brick and the center of the moon squared. Fair enough. That's the force on the brick. What's going to be the force due to gravity on the feather? Or another way to think about it, the weight of the feather on the moon? We'll do the same calculation. The force on the feather is going to be equal to big G times the mass of the moon times the mass of the feather divided by the distance between the center of this feather and the center of the moon squared. That's the distance, and then we square it. So if you look at both of these expressions, they both have this quantity right over here-- G times the mass of the moon divided by the distance between this height and the center of the moon squared. So they both have this exact expression on it. So let's replace that expression. Let's just call that the gravitational field on the moon. So if you apply this number by any mass, it will tell you the weight of that object on the moon, or the gravitational force acting downward on that object on the moon. So this is the gravitational field of the moon. So I'll just call it g sub m. And all it is, is all of these quantities combined. So if we simplify that way, the force on the brick due to the moon is going to be equal to that lowercase g on the moon-- normally we use this lowercase g for the gravitational constant on Earth, or the gravitational field on Earth, or sometimes the acceleration of gravity on Earth, but now we're referring to the moon. That's what this lowercase subscript m is doing for us. So it's equal to that times the mass of the brick. For the case of the feather, the force on the feather is equal to all of this business. So that's the g sub m times the mass of the feather. So we're going to assume, which is a reasonable thing to assume, that the mass of the brick is greater than the mass of the feather. What's going to be their relative forces? Well, here you have a greater mass times the same quantity. Here you have a smaller mass times the same quantity. So if the mass of the brick is greater than the mass of the feather, it's completely reasonable to say that the force of gravity on the brick is going to be greater than the force of gravity on the feather. So if you do all this, and everything we've done to this point is correct, you might say, hey, there's going to be more force due to gravity on the brick, and that's why the brick will be accelerated down more quickly. But what you need to remember is that there is more gravitational force on this brick. But it also has greater mass. And we remember the larger something's mass is, the less acceleration it'll experience for a given force. So what really determines how quickly either of these things will fall is their accelerations. And let's figure out their accelerations. I'll do this in a neutral color. We know that force is equal to mass times acceleration. So if we want to figure out the acceleration of the brick-- or we could write it the other way. If we divide both sides by mass, we get acceleration is equal to force divided by mass. And acceleration is a vector quantity, and force is also a vector quantity. And in this situation, we're not using any actual value. But if I were using actual values, I would use negative numbers for downwards and positive values for upwards. But we're not using any signs here. But you could assume that the direction is implicitly being given. So what's the acceleration of the brick? That's a lowercase b I was writing. The acceleration of the brick is going to be equal to the force applied to the brick divided by the mass of the brick. But the force applied to the brick, we already know, is this business right over here. It is little g on the moon, the gravitational field on the moon, times the mass of the brick, and we're dividing that by the mass of the brick. So the acceleration on the brick on the moon-- the acceleration that the brick will experience-- is the same thing as that gravitational field expression. It is g sub m. This is how quickly it would accelerate on the moon. Now let's do the same thing for the feather. I think you see where this is going. The acceleration of the feather is going to be the force on the feather divided by the mass of the feather. The force on the feather is g sub m-- g with the subscript m-- times the mass of the feather, and then we're going to divide that by the mass of the feather. And so, once again, its acceleration is going to be the same quantity. So they are both going to accelerate at the same rate downwards, which tells us that they'll both hit the ground at the same rate. They'll both accelerate from the same point at the same time, and they'll both have the same velocity when they hit the ground. And they'll both hit it at the exact same time, despite one having a larger mass. So the reality is, because it has a larger mass, it has a larger gravitational attraction to the moon. But because of its mass, that attraction gives it the same acceleration as something with a smaller mass. So any mass at the same level on the surface of the moon would experience the same acceleration. So now the quite natural question is, wait, Sal, if that's true on the moon it should also be true on Earth. And it would be true on Earth. If you did this exact same experiment and you evacuated all the air from the room, so that you didn't have air resistance, and you took a brick and a feather and let them go at the same time, they will both hit the ground at the exact same time, which is a little unintuitive, to imagine a feather just plummeting the same way a brick would. But it would if you evacuated all the air. And so the reason why we see this over here, and I think you get the sense because I already talked about evacuating the air, is that the difference between the brick and the feather is all due to air resistance. If you took the same brick, or if you took something that had the same mass as the brick, and you were to flatten it out so it has more air resistance-- but let's say it has the same mass, let's say this thing and this thing have the same mass-- this thing would fall slower than that because it'll have more air resistance. It has more air to collide into, to provide resistance as it falls. And if you took a feather and if you compacted it really, really, really, really, really, really small-- the same mass as a feather, but you made it so small that it could cut through the air-- you'll see that it will drop a lot faster. So the real difference between how things fall on Earth-- if you had no air, they would all fall at the exact same rate. It's only because of air that they fall at different rates. And the air does two things. For constant pressure-- so if you have two objects that have the same shape, the object that is heavier, that has more weight, will fall faster because it'll overcome-- it'll be able to provide more net force against the air pressure. If you have something that has the same weight, the object that is more aerodynamic will fall faster-- the one that cuts through, the one that has the least air resistance. And as a little experiment that you can try in the comfort of your own room right now, take a book like this. And you could drop it. And then you could take another piece of paper, or even a little postcard or something, and you drop it. And you'll see, of course, a postcard will fall much slower than this book. But what you do is put the postcard on top of the book so that the book is essentially breaking all of the air resistance for the postcard. And what you'll see is, if you put it on top of the book and you were to drop it, you'll see that they fall at the exact same rate.