Video transcript
Let's say we were to take alittle excursion to the moon. And so here we aresitting on the surface of the moon-- that's thesurface of the moon right there. And with us to our excursion tothe moon we brought two things. We brought ourselfa concrete brick. So that's my brickright over there, although it's orange-- we'll sayit's an orange concrete brick. And I also bought abird feather with us. So this is the bird feather. And then my questionto you is, if I were to hold both thebrick and the bird feather at the same time, and I wereto let go of both of them at the same time and askyou, which one of them would hit the surface of themoon first, what would you say? Well, if you based it onyour experience on Earth-- on Earth if you were totake a break and a feather, a brick would justgo straight down. A brick would justimmediately fall to the Earth, and it would doit quite quickly. It would acceleratequite quickly. While a feather wouldkind 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 makeits way down to the ground. So on Earth, at leastin the presence of air, it looks like the brickwill hit the ground first. But what wouldhappen at the moon? And what's interesting about themoon is we have no atmosphere. We have no air toprovide resistance for either the brickor the feather. So what do you thinkis going to happen? So your first temptationwould say, well, let's just use theuniversal law of gravity. So what is the force ofgravity on the brick? Well, you couldcalculate that out. The force ofgravity on the brick is going to be equal to big Gtimes the mass of the moon-- I'll say that's mfor mass and then the subscript is lowercasem for moon-- the mass of the moon times themass of the brick divided by the distancebetween the brick and the center ofthe moon squared. Fair enough. That's the force on the brick. What's going to be the forcedue to gravity on the feather? Or another way tothink about it, the weight of thefeather on the moon? We'll do the same calculation. The force on thefeather is going to be equal to big G timesthe mass of the moon times the mass of the featherdivided by the distance between the centerof this feather and the center ofthe moon squared. That's the distance,and then we square it. So if you look at bothof these expressions, they both havethis quantity right over here-- G times themass of the moon divided by the distancebetween this height and the center ofthe moon squared. So they both have thisexact expression on it. So let's replacethat expression. Let's just call thatthe gravitational field on the moon. So if you apply thisnumber by any mass, it will tell you the weightof that object on the moon, or the gravitationalforce acting downward on that object on the moon. So this is the gravitationalfield of the moon. So I'll just call it g sub m. And all it is, is all ofthese quantities combined. So if we simplifythat way, the force on the brick due tothe moon is going to be equal to that lowercaseg on the moon-- normally we use this lowercase g forthe gravitational constant on Earth, or thegravitational field on Earth, or sometimes the accelerationof gravity on Earth, but now we'rereferring to the moon. That's what this lowercasesubscript m is doing for us. So it's equal to that timesthe mass of the brick. For the case of the feather,the force on the feather is equal to allof this business. So that's the g sub m timesthe mass of the feather. So we're going to assume,which is a reasonable thing to assume, that themass of the brick is greater than themass of the feather. What's going to betheir relative forces? Well, here you have a greatermass times the same quantity. Here you have a smaller masstimes the same quantity. So if the mass ofthe brick is greater than the mass of the feather,it's completely reasonable to say that the forceof gravity on the brick is going to be greaterthan the force of gravity on the feather. So if you do allthis, and everything we've done to this pointis correct, you might say, hey, there's going to bemore force due to gravity on the brick, and that'swhy the brick will be accelerateddown more quickly. But what you need toremember is that there is more gravitationalforce on this brick. But it also has greater mass. And we remember thelarger something's mass is, the less acceleration it'llexperience for a given force. So what really determines howquickly either of these things will fall is theiraccelerations. And let's figure outtheir accelerations. I'll do this in a neutral color. We know that force is equalto mass times acceleration. So if we want to figure outthe acceleration of the brick-- or we could writeit the other way. If we divide bothsides by mass, we get acceleration is equalto force divided by mass. And acceleration isa vector quantity, and force is alsoa vector quantity. And in this situation, we'renot using any actual value. But if I were usingactual values, I would use negative numbers fordownwards and positive values for upwards. But we're not usingany signs here. But you could assume thatthe direction is implicitly being given. So what's theacceleration of the brick? That's a lowercaseb I was writing. The acceleration ofthe brick is going to be equal to the forceapplied to the brick divided by the mass of the brick. But the force applied tothe brick, we already know, is this businessright over here. It is little g on the moon,the gravitational field on the moon, timesthe mass of the brick, and we're dividing thatby the mass of the brick. So the acceleration onthe brick on the moon-- the acceleration that thebrick will experience-- is the same thing as thatgravitational field expression. It is g sub m. This is how quickly it wouldaccelerate on the moon. Now let's do the samething for the feather. I think you seewhere this is going. The accelerationof the feather is going to be the forceon the feather divided by the mass of the feather. The force on thefeather is g sub m-- g with the subscript m--times the mass of the feather, and then we'regoing to divide that by the mass of the feather. And so, once again,its acceleration is going to bethe same quantity. So they are both going toaccelerate at the same rate downwards, whichtells us that they'll both hit the groundat the same rate. They'll both accelerate fromthe same point at the same time, and they'll both havethe same velocity when they hit the ground. And they'll both hit itat the exact same time, despite one havinga larger mass. So the reality is, becauseit has a larger mass, it has a larger gravitationalattraction to the moon. But because of itsmass, that attraction gives it the same accelerationas something with a smaller mass. So any mass at the same levelon the surface of the moon would experience thesame acceleration. So now the quitenatural question is, wait, Sal, if that's trueon the moon it should also be true on Earth. And it would be true on Earth. If you did thisexact same experiment and you evacuated allthe air from the room, so that you didn'thave air resistance, and you took abrick and a feather and let them goat the same time, they will both hit the groundat the exact same time, which is a little unintuitive,to imagine a feather just plummeting the sameway a brick would. But it would if youevacuated all the air. And so the reason whywe see this over here, and I think you get thesense because I already talked aboutevacuating the air, is that the difference betweenthe brick and the feather is all due to air resistance. If you took the same brick,or if you took something that had the samemass as the brick, and you were to flatten it outso it has more air resistance-- but let's say it has the samemass, let's say this thing and this thing havethe same mass-- this thing wouldfall slower than that because it'll havemore air resistance. It has more air to collideinto, to provide resistance as it falls. And if you took a feather andif you compacted it really, really, really, really, really,really small-- the same mass as a feather, but you madeit so small that it could cut through the air-- you'll seethat it will drop a lot faster. So the real difference betweenhow things fall on Earth-- if you had no air,they would all fall at the exact same rate. It's only because of air thatthey fall at different rates. And the air does two things. For constantpressure-- so if you have two objects thathave the same shape, the object that is heavier,that has more weight, will fall fasterbecause it'll overcome-- it'll be able to provide morenet force against the air pressure. If you have somethingthat has the same weight, the object that ismore aerodynamic will fall faster-- theone that cuts through, the one that has theleast air resistance. And as a littleexperiment that you can try in the comfort ofyour own room right now, take a book like this. And you could drop it. And then you could takeanother piece of paper, or even a little postcard orsomething, and you drop it. And you'll see, ofcourse, a postcard will fall much slowerthan this book. But what you do is put thepostcard on top of the book so that the book isessentially breaking all of the air resistancefor the postcard. And what you'll see is, ifyou put it on top of the book and you were todrop it, you'll see that they fall atthe exact same rate.
