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Uniform Circular Motion | 27 mins | 0 completed | Learn |

Centripetal Forces | 53 mins | 0 completed | Learn |

Universal Law of Gravitation | 36 mins | 0 completed | Learn |

Gravitational Forces in 2D | 25 mins | 0 completed | Learn |

Acceleration Due to Gravity | 22 mins | 0 completed | Learn |

Satellite Motion | 45 mins | 0 completed | Learn |

Concept #1: Acceleration Due to Gravity (g)

**Transcript**

Hey guys so now that we've been talking about universal law of gravitation and how and we're taking a deeper look at how weight works the attraction forces work I want to take a deeper look into how the acceleration due to free fall works let's check it out. So first of all remember little g right little g which on earth is nine point eight meters per second squared what that actually means that what that actually refers to is the acceleration due to freefall at a planet's surface in this case the Earth OK it's the entire the entire phrase is with G. stands for. Or they could this could be on the moon or some sort of asteroid or any kind of astronomical big objects OK we got we can get that number in the way we got this number for the first time hundreds of years ago was simply most likely by grabbing objects and calculating how long it took them to the ground and you can get that experimentally but you can actually use the universal law of gravitation which gives you this equation here to derive an expression to calculate a way to find a way to calculate G. that gravity based on the properties of the planet such as the mass and the radius of the planet so let's check it out on we're going to look at two situations here the first one is the more generic case when you are at a distance far from the surface very far. So if you're over here as we know from universal of gravitation you're going to be pulled towards the earth but the earth is also going to be pulled towards you so there's sort of a mutual attraction which we're going to call F. G. but that force can also be called the weight force and weight equals mg that's sort of the old school simplistic definition but weight is also FG. and we know from our equation that F.G. is also GM1M2/r^2 where R. is the distance now instead of calling these things M1 AND M2 these problems are typically referred to the planet or the bigger mass as M the smaller mass m

So I'm going to use that here big M little m. divided by their distances Now be careful here the distance is not. Just your height so let's call this H your height the distances from your center of mass somewhere over here to the center of mass of the earth so it's your height plus the radius of the earth and I'm going to call that little r because that's not what it is right its this variable here so little r the distance is the radius plus the height okay well check this out if weight is mg. but weight is also the force of gravity or gravitational attraction then I can say that these two are the same as well right and if I set this equal to each other. You can see that the m cancels These m cancels now just to be clear these m are the same when you write weight equals mg when you write this is the mass this is your mass right so you are that whatever other object certainly not the Earth's mass and when you write this here the little m. is also the mass other than the Earth right so these are the same m and they are going to cancel and then you're left with just G. equals G. M. over r square and this equation works when you are far away from the surface so when you're far away from the surface you're G. far right here again is G. big M overr squared okay remember in physics uppercase letters big letters mean constant the G.'s a constant universal constant the mass of the earth is a constant as well it's not going to change but r could change as you move up. As you move up or farther away from the earth I should say the distance increases but this isn't the denominator so you G. will decrease the furthers the further you get from the from the earth the smaller the gravitational attraction will be therefore the smaller the acceleration due to gravity will be OK. So that's it you get this equation remember little r is big R plus H R is the radius of the planet again uppercase letters are constants of the radius and this is how it changes as you go up and down OK because you're r could change slower case and that would cause little jitter change that's why it's also lower case because it could be a different number OK now here's a little bit more specific case if you happen to be so let me start with G.far Gfar is G.M. over r square where r is bigger plus H. But if you're here on the surface your h is zero However in a more or even more generic case F. you are not just necessarily on the floor but if you are anywhere near the floor.

We're going to say that the there are which is the radius of the earth. Is much much greater than your heights you could be on top of a ten meter building you might think that that's very high I'm sorry ten storey building I think that's very high or a hundred thousand story building but in fact it is there is negligible compared to the radius of the earth so if you're anywhere near any kind of human sized building or whatever doesn't matter if you're nearly Earth we're going to say that this number is much greater therefore age is negligible to approximately zero it's effectively zero OK so what we get is if r is r plus hand you're near the building I'm sorry Near the surface this H. is zero so your distance is simply the radius of the Earth so this equation here instead of being G.M. t little t square it's going to be G.M.. BIG R square what happens is when you near the surface your height is approximately zero therefore your radius is approximately your distance is approximately the radius of the earth OK. I'm sorry I meant to write this is a big R So notice that these are all constants that's why we're able to say that G. on earth. Is nine point eight when we made that is g on the surface of the earth OK In fact if you multiply six point six seven times since negative eleven times the mass of the earth times the radius of the earth you should get nine point eight one K. and that's a way that we can verify this so if you're far from the earth use this equation if you knew from the earth near the surface of the earth you use the second equation OK let's do an example real quick it says here the Earth has a mass of this in a radius of this calculate acceleration due to freefall at the top of Mt Everest which is this much above water OK So this is a high point there in 8500 meters. The Earth has a mass of this mass of the planet is big M five ninety eight times ten to the twenty fourth radius is bigger R six thirty seven times ten to the six OK now you have to decide are we going to use the G Far equation or the G near equation well in the five hundred meters is the highest the tallest thing on earth the reason why they give you such a big number is because they actually want you to use the G far equation. And the main idea with questions like this is to show you hey even for a very big distance like mt Everest crazy high right it's like how airplanes fly a little bit higher than that even for that distance the change in a little g. would be very small that's kind of the point of these questions is to show you that but we're going to use a real G far and let's see what we get so you're here's the here's the earth and here's Mount Everest and then you're up here some are right. Here and to find this we're going to use G far which is G.M. over r square OK now r is a distance which is a combination of h and big R so little r is big R plus h and that's going to be six point thirty seven times ten to the six plus eighty five hundred when you combine these two. You get six three seven.eight Five zero zero I'm going to I'm going to actually write everything like this because if you were to round this number you actually you probably would end up right back at. Well I guess you're in up at six thirty eight but I want to be even more precise just so we can see how much of a difference we have K. So really you know we're going to get this notice I went from six thirty seven to six thirty seven eight very small difference and now I can plug these numbers here this is just plug and chug six point six seven times ten to negative eleven the mass of the earth is five ninety eight times ten to the twenty fourth and I'm squeezing everything in here the radius is and sorry the distances this whole number here six three seven eight. Five hundred squared and if you do all of this if you do all of this very carefully you get nine. I mean I'm writing ans here it's going. Crowded in there. Nine point eight zero three meters per second square nine point eight zero three the sort of accepted standard is nine point eight or nine point eight one or if you want to be even more precise I can see the G. earth actual the average will anyway is roughly nine point eight one most people know that the next digit here is three So if you compare these two numbers if you compare these two numbers there is a zero point one percent difference which is completely negligible almost completely negligible and for most purposes it's completely likable The only as they want to say is that this number nine point eight one three is an average number the earth is not a perfect sphere so technically you do get different values of G. Even if you are different sea level with the sea level different places around the Earth OK but it's again the difference there is also going to be very small All right that said I want you to try this practice problem here real quick I'm going to keep going but hopefully you pause the video give this a shot so I have the moon has a diameter of that So remember in physics we're never going to diameters the first thing I want to do is replace that three four seven five kilometers with a radius. And then this is going to give me and I'm also changes to meters right away so we divide this part two and then change it to meters.

One seven.

One seven three eight round so that zero zero zero meters or one point seven three eight times ten to the six meters OK that's diameter of that's the radius of the moon that we now get and it says the acceleration. Due to freefall on its surface is roughly one sixth that of the earth acceleration due to freefall on its surface is what we're now calling used to call g I'm calling this g near.

To differentiate from some acceleration due to gravity far from the surface so and also I do what I'm supposed to use this equation here OK So g near is one six that of g near on the Earth we know that number that's one over six times nine point eight going we will go with nine point eight one just to be even more precise and this gives us one point sixty four meters per second squared that's not exactly the diameter of the moon but it's a very common thing to just remember the gravity of the moon is approximately one sixth of the earth OK so we're going to use that number that's fine and the question is let's use all these numbers to try to find the mass of the moon how might you find that well so far the only two ques I gave you are these two so it's pretty easy to figure out if you got to use one of those two equations obviously not as certain you're going to have all the equations and then you have to be able to or maybe have to memorize and remember them you have to be able to figure out which equation to use and generally speaking I like to think of it in terms of I like to think of it this way whenever you're given a variable for which there is an equation associated with Whenever you're given a value for which there's an equation associated with it you want to write that equation for example if I give you G. near and I give you that value or you're able calculate that and g near has an equation then what you should do is write G. near.

Equals one point six four and then replace g near with its equation G M over r are squared equals one point six four.

So again the idea is that if you are given a value that has an equation associated with it then you will place that value.

With its equation.

So now I can try to get out of here by solving for M okay M would be. One point six four. Times r squared divided by G. and I have all these numbers so we're done one point six four R. is the radius which is one point seven three eights times ten to the six squared divided by six point six seven times ten to the negative eleven and if you do this you get that the mass of the moon is seven points four to seven times ten to twenty second kilograms OK The nice thing about questions like this is we actually know the mass of the moon is so we can check it in fact I went to google and google the mass of the moon just to make sure that we got this right and we were Google says that we are one point zero seven percent off which is pretty neat considering that we're rounding the fact the gravity is one sixth of the earth is just an approximation OK so one percent off is pretty good let's keep going here I have one more thing to talk about i will give you tips on how to solve some of these problems and we'll do one more so protest over here if you're using G. near. Remember G. near is Gm over r square. If you use that equation to find either r or h OK so if you look were is r or h it's not in this equation not directly it's highly inside of the r okay r equals big R plus h. So you can use this equation to find r h even though they're not directly there OK Now when you do this what you want to do is you always want to find our first find or first and then use the fact that little r which you find first is r+h so you can find either r or h OK So let me just be even more clear the idea is that you're not supposed to try to or at least don't try to do this because it's going to be harder if I'm looking for if the equations like this.

OK if the questions like this by the way i actually meant I'm so sorry I meant Gfar

Remember g near is with the big R oKay I meant you far.

So if you're using G four and you're looking for r h don't do this don't try to replace this here with r+h and then solve for r Because it's going to be far more complicated in terms of just the r that we will need here so don't do this just find r first and then go to big R or h depending on what you need OK so this question says the International Space Station experience is a free for acceleration of eight point seven seven Now that's a G. freefall acceleration is a little G. is that little G. near a little G. Far well's little G. far you know this for two reasons one little G. near on the Earth should be nine point eight This is no good did a good difference from nine point eight and also the International Space Station should know it's flying way apart OK so I'm going to say that G. far

Is eight point seven seven that's my G. far and remember what I just mentioned when you are given a value so gfar that is associated with an equation gfar equals this.

Then you want to plug in that equation other words I'm going to write G. big M over r square. Because eight point seven seven and that's the starting point

For solving this so anyway it says a rotates around the earth I want to know the heights above the earth what is its height above the earth so here's the Earth here's the international space station spinning around the earth and I want to know what is h.

I am given G far so I'm going to use the G far equation you have to play with what you got right the problem is right away you notice h is not in this equation so that's what we just talked about if h is not there you have to find r first K R and then h so from looking for h I'm going to find r first and then find h So let's find r By using that equation. G.M. over r squared equals eight point seven seven that means that r I'm going to move some stuff around here r going to go up G M equals eight point seven seven r square so r will be G.M. over eight point seven seven and then we take the square root of that OK now let's plug in let's plug in these numbers.

Are will be six point six seven times ten of negative eleven big M will be was big M like.

big M is the mass of the earth remember the space station is little m and this is big M mass of the earth five point nine eight times ten to the twenty fourth we had it from before and this whole thing divided by eight point seven seven and the whole thing is square rooted when you do this when you do this you gets to approximately.

You get to approximately.

Where I have that six point seven four times ten to the six six seven four times ten to the six meters that's my little or but I need h well r=R+h therefore h is little r minus big R.

Something that's curious here is that little or even little kids letter is actually a bigger number always than h OK then the I'm sorry Little r is a bigger number than big R.

h have to subtract little r here which I just got six point seven four times ten to the six minus big r big r is six point thirty seven times ten to the six and if you do this I get zero point three seven times ten to the six just to keep the powers consistent and if you further convert that you see that h is three seven zero zero zero zero. Kilometers or meters sorry let me meters and once again we can Google this and I google it and that is actually exactly. What Google shows OK So let's start this answer is correct I want to make one final comment here when it comes to the mass of the earth and the radius of the earth different books and different professors will use different numbers for the mass of the earth is five point nine is seven sometimes so this is the mass of the Earth sometimes five point ninety eight and sometimes it just runs to six all of these obviously times ten to the twenty fourth the radius of the Earth sometimes. You see six thirty seven sometimes you see six thirty eight and sometimes you see it rounded to six point four times ten to the six meters OK So if you're slightly off it's OK I'm obviously depends on what values are given to you by your professor or what he what he prefers to use Ok that said let me know if you guys have any questions.

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Concept #1: Acceleration Due to Gravity (g)

Five homogeneous planets have relative masses and sizes as shown in the figure. A body of mass m would weigh least on which planet?
A. 1
B. 2
C. 3
D. 4
E. 5

Some planet X has twice the mass of Earth and half the radius. What would the gravitational acceleration be at the surface of planet X?

What is the gravitational acceleration due to Earth at an altitude of 3 Earth radii?

The mass of Earth is ME = 6.0 × 10 24 kg, and the mass of the Sun is 2.0 × 10 30 kg. They are a distance of 1.5 × 10 8 km apart. Calculate the acceleration of Earth due to the gravitational attraction of the Sun. (G = 6.67 × 10–11 N·m2/kg2)A. 6.0 × 10–3 m/s2B. 6.0 m/s2C. 9.8 m/s2D. 6.0 × 103 m/s2E. 1.8 × 10–8 m/s2

The radius of the moon is R. A satellite orbiting the moon in a circular orbit has an acceleration due to the moon's gravity of 0.14 m/s2. The acceleration due to gravity at the moon's surface is 1.62 m/s2. The height of the satellite above the moon's surface is
A. 0.42R
B. 3.4R
C. 2.4R
D. 11R
E. 1.7R

The strange planet Zztop has mass 8.00 x 1025 kg. Zztop is spherical and has uniform density. If an object of mass 0.600 kg is projected straight upward from the surface of Zztop with an initial speed of 16.0 m/s, it returns to the surface in a time of 8.00 s after it is projected upward. What is the magnitude of the acceleration due to gravity near the surface of Zztop?

The strange planet Zztop has mass 8.00 x 1025 kg. Zztop is spherical and has uniform density. If an object of mass 0.600 kg is projected straight upward from the surface of Zztop with an initial speed of 16.0 m/s, it returns to the surface in a time of 8.00 s after it is projected upward. What is the radius of planet Zztop?

Planet Sec is spherical and has uniform density. Its radius is R = 5.00 x 106 m and its mass is M = 4.00x1022 kg. A rock is released from rest from a height of 20.0 m above the surface of the planet. How many seconds does it take the rock to reach the surface of the planet? Air resistance can be neglected.

Planet X has radius 4.00 x 106 m and mass 5.00 x 1024 kg. You stand on the surface of Planet X and throw a rock vertically upward with a speed of 12.0 m/s. What is the maximum height above your hand reached by the rock?

You are a member of a group of scientists who travel to planet Bubba, a planet that orbits a bright star in our southern sky. The radius of the planet is Rb = 5.0 x 106 m. You land on the planet. While you are standing on the surface of the planet you throw a small rock straight up with an initial speed of 6.0 m/s. The rock reaches a maximum height of 12.0 m above the point from which it was thrown. What is the mass of planet Bubba?

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