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## AP®︎/College Physics 1

### Course: AP®︎/College Physics 1>Unit 3

Lesson 4: Gravitational fields and acceleration due to gravity on different planets

# 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?
• 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.
(1 vote)
• What is a Gravitational Well? Thank You!
• 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.
• what happens to acceleration due to gravity when we go deeper into earth ??
• 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
• If you were in a space station, why would you float while the ISS is in orbit?
• 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.
• why does acceleration due to gravity decrease as we go into the surface of the earth <hypothetically
?>
• 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.
• 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.
• 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.
• 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.
• 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.
• 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?
• 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.