Acceleration Due to Gravity - Video Tutorials & Practice Problems

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1

concept

Acceleration Due to Gravity

Video duration:

5m

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What's up, guys? So we took a look at the gravitational forces between objects, whether it was point masses or a planet and something else. But in a lot of problems you're gonna see you're gonna need to calculate the acceleration due to gravity. So we're gonna cover that in this video. Now that acceleration due to gravity, we can figure out from Newton's law of gravity and the the letters you'll see for that. Sometimes they're a G, but most of the times you'll see a little G. For that, we're gonna cover that, and it's got two different forms, depending on how far you are away from the planet. If you're any distance away from the planet, it's gonna be GM over little r squared. But if you're near the surface or on the surface, it's gonna be GM over big R squared. Notice the difference between those two to see how we get those equations. Let's take a look at some quick derivation. So how do we get the acceleration from a force? How do we get a from F? We always use f equals M A. So in this case, we have a forces acting on this person we know that's the gravitational force. So we write that out, the gravitational force is just m except instead of a you'll just see this become G. So now we know what the gravitational forces. It's GMM over little r squared. That's equal to M. G. So if you just cancel this EMS out that appear on both sides, we get to this expression right here. And by the way, this works for any distance away. So let's look at a special case when you're on the surface of this planet. So we're gonna start off with this equation right here. We know that little G is just big GM over little r squared. But now, when you're on the surface here, your little are is different than it was over here. We know that this little our distance is equal to big R plus H. But now, when you're on the surface, this H distance, which remember, is your height above the surface is so small compared to the big radius of the Earth. They were just gonna cancel that out and approximated to zero. So we can only do this because your heights is just so small. We're just gonna cancel it out. So what happens is your little are basically becomes your big are. So now what happens is that this equation becomes G surface. And now we're just gonna write this as big G Big M over big R squared, Alright. And that's basically the derivation. So there's a couple key points to remember about these equations. So whenever you're specifically asked or given a height, you're always gonna use this little G equation right here because it depends on the height. Whereas you're always gonna use g surface. This guy right here, whenever you're standing on the surface and in problems, you'll see the words like surface gravity or free fall due to gravity or something like that at the surface. And also also, you'll note that in these equations they only depend on Big M and not little M. So it only depends on the mass of that thing in the middle, the big planet or whatever. Finally, we know that g surface because it depends on a whole bunch of capital letters, big letters that it's a local constant, which means it's just a constant. That depends are, uh that's gonna be the same anywhere you are on the surface, whereas Giza variable and it's gonna decrease as our is increasing. So as you get farther away, your acceleration is gonna decrease. We can actually see that because as you are our increases, we know that the force of gravity gets weaker, which means that your acceleration is gonna get weaker as well. Now, finally, the last point is that your weight, what we define is the weight at any distance away from planet is just gonna be the force of gravity. We know that's GMM over r squared. But now we can actually write this as MGI with the little G Now we could actually right in terms of little G and on the surface, this speechless becomes mg at the surface. And we've actually dealt with this Ah, lot before and that's basically it, guys. So let's go ahead and take a look at an example. So we've got we're trying to figure out what the acceleration due to gravity is on Mount Everest and were specifically given what the height is. So we know we're gonna use little G for that because we're given a height. So we're just gonna start with that equation gm over little r squared. Now, I know that little are is just equal to R plus h. So I just have to make sure I have everything in that problem. I know that these air just Constance and I have with the big radius of the Earth is I could just use my table right here, and I have with the height is so I'm just gonna go ahead and plug everything in. I've got six 0.67 times, 10 to the minus 11. I've got the mass of the earth 597 times 10 of the 24th. Now I got 6.37 times 10 to the sixth, plus the height, which is 8.85 kilometers. So that's +8850 And then I've got a square that just make sure you're adding this before you square it, and yeah, so if you do that right, you should get 9 79 m per second squared. So now let's take a look at the surface gravity of the Earth because we're asked to compare the surface gravity of the earth with this guy. Right? So we have G surface. So that's the surface gravity. That's just gonna be gm over big r squared now, hopefully from forces in Quetta, Matics and stuff. You remember that this number right here, that G is just 9.8. But if you forgot for some reason, you could actually get back to it from this equation. So if you just go ahead and plug in all the numbers that we had for this except for this 8850 you're actually gonna get this is 9 81 m per second squared. So that's where this number actually comes from. It comes from this equation and so we can see that this is 9.81 m per second squared and these this number is really close. And that's because your height, even though you're standing on Mount Everest, which is the tallest mountain in the world, that thing, that height is so small compared to the huge radius of the Earth that it doesn't really have a big effect on this. So what happens is H is really small, even though it's technically the tallest mountain on our planet. Alright, guys, let me know if you have any question

2

Problem

Problem

You stand on the surface of a mysterious planet with a mass of 6×10<sup>24</sup>kg and measure the surface gravity to be 7 m/s<sup>2</sup>. What must the radius of the planet be?

A

5.72×10^{13}m

B

4.43×10^{14}m

C

8.17×10^{12}m

D

7.56×10^{6}m

3

example

Find mass of planet in free fall

Video duration:

3m

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Hey, guys, let's check out this problem here. So even astronaut who's dropping a rock from rest at the surface of unknown planets and we know what the radius of the Unknown Planet is, but we're being asked for the mass. So what is that variable We're looking for what we're looking for Big M. The mass of the planet. And how do we find that? Well, there's two basic approaches we can take so far, With all of our equations, we can take either a force approach or an acceleration due to gravity approach. Now, with forces, I need to know multiple masses and I need to be told from information about the force. But I don't have any of that. I don't have multiple masses and I don't have any reference to forces, so I can't go that route. Instead, I could use the acceleration due to gravity. So now my choices are which one of my working with, um I'm working with surface gravity or some gravity at a height. My told some heights and no, I'm not. But I am told that I'm on the surface of an unknown planet, so we're actually gonna be using the G surface equation. So G surface equals big g Big M over little r squared. Now, I'm gonna try to look for this big M right here. So we go ahead, just re arrange for that. So I want to just move this r squared over to the top, and then this g is gonna biggie is gonna come down, right? And so I'm gonna get that G surface. Times are squared, divided by big G is going to be equal to the mass of the planet. Now I have the radius and I have the capital G. I'm looking for mass. So now I just need to figure out what the what the surface gravity is. How do I do that? What's what other parts of this problem can I use? Well, I have that. The rock takes 0.6 seconds to fall 1.5 m, So let's just start. Diagram will quickly. What's going on? So we've got a surface here, a rock, and we're told that that rock is dropped from rest and it falls 1.5 m in a time of 0.6 seconds. We've seen these problems before. These air actually just kinda Matics problems. So sometimes your kingdom attics problems can actually pop up on gravitation problems. We need to remember those equations, right? So how do we figure out what G surface is? Remember, we need three out of the five variables and cinematics, so let's list them. Right. So I've got v knots. I need the knots. I need t. I need Delta y and I need the final. So let's take a look. I'm told that the rock is dropped from rest, which means that the initial velocity is equal to zero. So I have that and I'm also told the time, So I t and I'm told with the delta y is, remember, that's just the distance. That's just this delta y right here. So we actually have three out of five, which means we can go ahead and ignore the VF. So I'm gonna pick the equation that doesn't involve that one. Which is the delta t one or sorry, Delta y. So I've got Delta y equals V naughty plus one half g t squared. So now what happens is if I can find out what this little G is, I can plug this thing back into this equation and solve for EMS. That's the That's what I'm doing here. So now I know that the V not term is gonna cancel because zero and then I'm just gonna start plugging everything in. Now, I got 1.5 over here, and this 1.5 is if I just choose this downward direction to be positive, right? We don't have to deal with the negatives and positives and stuff like that. So that's just equal to one half G. And then t is just the time 0.6 squared. So now we go ahead and just figure this out. I've got G equals two times 1.5. This, too, comes from the fact you move in the half over the other side and then divided by 0. squared, and you should get 8.33 meters per second squared. So now we have this thing. We're gonna plug this back into this equation and solve. So you've got 8.33 on that. We've got the radius, which is four times 10 to the six. They were gonna square that radius and then divided by 6.67 times 10 to the minus 11. That equals the mass of the planet. And if you do that, you should get two times 10 to the 24th kilograms. And that is our final answer. Let me know if you guys have any questions with this.

4

Problem

Problem

How far would you have to be above Earth's surface for g to be ½ of its surface value?

A

8.55×10^{8} m

B

2.64×10^{6} m

C

9.01×10^{6} m

D

8.13×10^{6} m

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