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### Course: Physics library>Unit 4

Lesson 3: Newton's law of gravitation

# Acceleration due to gravity at the space station

What is the acceleration due to gravity at the space station. Created by Sal Khan.

## Want to join the conversation?

• Hypothetically, would two objects in deep space that are a few miles away from each other, with no massive objects near them within millions of miles, float towards each other due to Newton's law of gravitation?
(19 votes)
• Not necessarily. It depends on their masses and the masses of the other bodies that are "millions of miles away". It is possible that the objects in deep space would be pulled towards the other objects if the other objects' masses are much greater than the mass of the closer object.
(2 votes)
• What is a Gravitational Well? Thank You!
(5 votes)
• I disagree; you don't need to invoke the fabric of space-time to explain a gravity well.

A gravity well is simply a way of thinking of objects with mass in space, and how hard it is to pull away from those objects (i.e. how hard it is to climb out of the well).
If you are stuck in a well, you need to use enough energy to be able to escape the well. Likewise, if you are in the gravity well of a star or a black hole, you need to gain enough energy to be able to escape its pull. The more massive the object, the deeper the well.
Easy peasy.
(8 votes)
• what happens to acceleration due to gravity when we go deeper into earth ??
(5 votes)
• To clarify a bit about why exactly gravity increases and then decreases as you go from space to Earth's core (excellent figure, drdarkcheese1), let's think of the relevant equation:
a_g = G*M/r^2, where G is the gravitational constant and r is your distance from the Earth's center. Now think of M. Normally, this is just the mass of Earth when we do these calculations, because we don't normally think of gravity inside an object. If you go deep into the planet, though, the core's mass will still pull you down, but now the mass that is over your head will be pulling you up! Thus, gravity will decrease all the way to zero as we reach the center of the planet.
Pretty neat, eh?

-RNS
(4 votes)
• If you were in a space station, why would you float while the ISS is in orbit?
(3 votes)
• Because when you fall, you experience weightlessness. I recommend Sal's video on elevators, and the Normal Force in elevators. Basically, If you and, say, a platform you are on, are in freefall, there will be no normal force, as the platform isn't counteracting any pressure you are applying to it.
(3 votes)
• why does acceleration due to gravity decrease as we go into the surface of the earth <hypothetically
?>
(3 votes)
• It increases as you get closer to the mass center of Earth.

As Newton's law of universal gravitation states:
F=gravitational constant * m1 * m2 / r^2

If you decrease r (the distance of your and Earth's center of mass) you will get a greater force acting on you.
(2 votes)
• Guys, does gravity increase as we go towards the center of the Earth? Or it is maximum on the surface?
(1 vote)
• Assuming uniform density of the Earth, the gravity decreases as you go towards the center until it reaches zero at the center. The reason it is zero is because there is equal mass surrounding you in all directions so the gravity is pulling you equally in all directions causing the net force on you to be zero. In actuality, the density of the Earth is significantly higher in the core than mantle/crust, so the gravity doesn't quite decrease linearly until you reach the core, but it is zero in the center.
(4 votes)
• Well! in earth rockets pull up by the principle of Newton's 3rd law. Does it push the air molecules on the midway in the atmosphere to receive an opposite force from the air? And if it is so how does the rocket move in the space where there is nothing to be pushed or to exert force?
(1 vote)
• The rocket expels mass (rocket fuel) at very high velocity. Conservation of momentum and Newton's 3rd law explain how the rocket will move in the opposite direction of that mass expulsion.
(5 votes)
• I tried to figure out the value of g for the ISS using a different method and got 9.18m/s^2 rather than the 8.69m/s^2 that Sal got. Can someone please help me out. Here's how I did my calculation:
Centripetal acceleration is same as g in this case.
So, g = v^2/r
And, v = 2*pi*r/T (assuming a perfectly circular orbit)
==> g = 4*pi^2*r/T^2
The ISS takes 90mins or 5400s to complete one orbit around the earth. So, T = 5400s
The value of r is (4*10^5)+(6.378*10^6)
After plugging-in the values, we get g = 9.18m/s^2.

Correct me if you find a mistake somewhere.
(2 votes)
• The only thing I see that could be an issue is that the orbit of the ISS is 90 to 93 minutes depending on the height of the orbit. If you use 93 minutes you get 8.6 m/s^2.

It all depends on the accuracy of the data going into the calculations.
(3 votes)
• I have two questions here:
1. If there is an astronaut between two super incredibly large stars(let say their masses are 1000000000 suns respectively). What will happen to the astronaut?
2. Is there a place where earth no longer can exert its gravitational force, or everything in the universe will have the force but just really small and the force close are to zero?
(2 votes)
• 1. That depends on where the astronaut is between the two stars. If the astronaut is at the right place, the astronaut will not accelerate at all. At any other place, the astronaut will accelerate towards one of the stars.

2. Because of Newton's Law of Gravitation, it is impossible for there to be zero gravitational force, but at ridiculously large distances, the force of gravity becomes so small that it can be neglected.
(3 votes)
• Acceleration is the rate of change of velocity of an object in time. when an object is on the earth surface how come acceleration due to gravity takes place, in which the object is stationary?
(1 vote)
• If the object is stationary then there is no acceleration due to gravity. The acceleration due to gravity can only be observed when the object is in free fall. There's still a force due to gravity, and that can be measured with a scale. But obviously if that force is offset by another force, there's not going to be acceleration, right?

If you know the acceleration due to gravity is 9.8 m/s^2 then you also know that the force due to gravity is 9.8 N/kg
(4 votes)

## Video transcript

Most physics books will tell you that the acceleration due to gravity near the surface of the Earth is 9.81 meters per second squared. And this is an approximation. And what I want to do in this video is figure out if this is the value we get when we actually use Newton's law of universal gravitation. And that tells us that the force of gravity between two objects-- and let's just talk about the magnitude of the force of gravity between two objects-- is equal to the universal gravitational constant times the mass of one of the bodies, M1, times the mass of the second body divided by the distance between the center of masses of the bodies squared. So let's use this, the universal law of gravitation to figure out what the acceleration due to gravity should be at the surface of the Earth. And I have a g right over here. I have the mass of the Earth, which I've looked up over here. And we also have the radius of the Earth. And for the sake of this, we're going to assume that the distance between the body, if we're at the the surface of the Earth, the distance between that and the center of the Earth is just going to be the radius of the Earth. And so this will give us the magnitude of the force. If we want to figure out the magnitude of the acceleration, which this really is-- I actually didn't write this is a vector. So this is just the magnitude of the acceleration. If you wanted the acceleration, which is a vector, you'd have to say downwards or towards the center of the Earth in this case. But if you want the acceleration, we just have to remember that force is equal to mass times acceleration. And if you wanted to solve for acceleration you just divide both sides times mass. So force divided by mass is equal to acceleration. Or if you take the magnitude of your force and you divide by mass, you're going to get the magnitude of your acceleration. This is a scalar quantity. This is a scalar quantity right over here. So if you want the acceleration due to gravity, you divide. Let's write this in terms of the force of gravity on Earth. So the magnitude of the force of gravity on Earth, this one right over here. So this will be in the case of Earth. I just wrote Earth really, really small. So one of these masses is going to be Earth. It's going to be this mass right over here. And so if you wanted the acceleration due to gravity at the surface of the Earth, you would just have to divide by the mass that is being accelerated due to that force. And in this case, it is the other mass. It is the mass that's sitting on the surface. So let's divide both sides by that mass. Let's divide both sides by that mass. And this will give us the magnitude of the acceleration on that mass due to gravity. So this is equal to the magnitude of acceleration, due to gravity. And the whole reason why this is actually a simplifying thing is that these two, this M2 right over here and this M2 cancels out. And so the magnitude of our acceleration due to gravity using Newton's universal law of gravitation is just going to be this expression right over here. It's going to be the gravitational constant times the mass of the Earth divided by the distance between the object's center of mass and the center of the mass of the Earth. And we're going to assume that the object is right at the surface, that its center of mass is right at the surface. So this is actually going to be the radius of the Earth squared, so divided by radius squared. Sometimes this is also viewed as the gravitational field at the surface of the Earth. Because if you multiply it by a mass, it tells you how much force is pulling on that mass. But with that out of the way, let's actually use a calculator to calculate what this value is. And then what I want to do is figure out, well, one, I want to compare it to the value that the textbooks give us and see, maybe, why it may or may not be different. And then think about how it changes as we get further and further away from the surface of the Earth. And in particular, if we get to an altitude that the space shuttle or the International Space Station might be at, and this is at an altitude of 400 kilometers is where it tends to hang out, give or take a little bit, depending on what it is up to. So first, let's just figure out what this value is when we use a universal law of gravitation. So let's get my calculator out. So we know what g is. It is 6.6738 times 10 to the negative 11. This EE button means, literally, times 10 to the negative 11. So this is 6.6738 times 10 to the negative 11. And then I want to multiply that times the mass of Earth, which is right over here. That is 5.9722 times 10 to the 24th. So times 10 to the 24th power. And we want to divide that by the radius of Earth squared. So divided by the radius of Earth is-- so this is in kilometers. And I just want to make sure that everything is the same units. So 6,371 kilometers-- actually, let me scroll over. Well, you can't see the kilometers right now. But this is kilometers. It is the same thing as 6,371,000 meters. If you just multiply this by 1,000. Or you could even write this as 6.371. 6.371 times 10 to the sixth meters. And we're going to square this. That's the radius of the Earth. The distance between the center of mass of Earth and the center of mass of this object, which is sitting at the surface of the Earth. And so let's get our drum roll. And we get 9.8. And if we round, we actually get something a little bit higher than what the textbooks give us. We get 9.82. Let's just round. So we get 9.82-- 9.82 meters per second squared. And so you might say, well, what's going on here? Why do we have this discrepancy between what the universal law of gravitation gives us and what the average measured acceleration due to the force of gravity at the surface of the Earth. And the discrepancy here, the discrepancy between these two numbers, is really because Earth is not a uniform sphere of uniform density. And that's what we have to assume over here when we use the universal law of gravitation. It's actually a little bit flatter than a perfect sphere. And it definitely does not have uniform density. The different layers of the Earth have different densities. You have all sorts of different interactions. And then you also, if you measure effective gravity, there's also a little bit of a buoyancy effect from the air. Very, very, very, very negligible, I don't know if it would have been enough to change this. But there's other minor, minor effects, irregularities. Earth is not a perfect sphere. It is not of uniform density. And that's what accounts for the bulk of this. Now, with that out of the way, what I'm curious about is what is the acceleration due to gravity if we go up 400 kilometers? So now, the main difference here, g will stay the same. The mass of Earth will stay the same, but the radius is now going to be different. Because now we're placing the center of mass of our object-- whether it's a space station or someone sitting in the space station, they're going to be 400 kilometers higher. And I'm going to exaggerate what 400 kilometers looks like. This is not drawn to scale. But now the radius is going to be the radius of the Earth plus 400 kilometers. So now, for the case of the space station, r is going to be not 6,371 kilometers. It's going to be 6,000-- we're going to add 400 to this-- 6,771 kilometers, which is the same thing as 6,771,000 meters, which is the same thing as 6.771 times 10 to the sixth meters. This is-- 1, 2, 3, 4, 5, 6-- 10 to the sixth meters. So let's go back to our calculator. So second entry, that's the last entry we had. And instead of 6.371 times 10 to the sixth, let's add 400 kilometers to that. So then we get 6.7. So we're adding 400 kilometers. So it was 371. Now it's 771 times 10 to the sixth. And what do we get? We get 8.69 meters per second squared. So now the acceleration here is 8.69 meters per second squared. And you can verify that the units work out. Because over here, gravity is in meters cubed per kilogram second squared. You multiply that times the mass of the Earth, which is in kilograms. The kilograms cancel out with these kilograms. And then you're dividing by meters squared. So you divide this by meters squared. You're left with meters per second squared. So the units work out as well. So there's an important thing to realize. And this is a misconception. We do a whole video on it earlier, when we talk about the universal law of gravitation, is that there is gravity when you are in orbit up here. The only reason why it feels like there's not gravity or it looks like there's not gravity is that this space station is moving so fast that it's essentially in free fall. But it's moving so fast that it keeps missing the Earth. And in the next video, we'll figure out how fast does it have to travel in order for it to stay in orbit, in order for it to not plummet to Earth due to this, due to the force of gravity, due to the acceleration that is occurring, this centripetal, this center-seeking acceleration?