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Had physics for 2 years now on highschool, but there is a thing I am wondering about.

You know the in the height above the earth surface around where the satellites are (Or the ISS), I've calculated that there actually is a big amount of gravity-forces, even up there. (9.1 - 9.2 m/s^2) - How come that things aren't dragged down to earth, and why are you even weightless?

Why doesn't satellites fall down more frequently, and does it have something to do with their orbit-speed?

Okay, many questions here, but just a curious guy.

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marked as duplicate by John Rennie, DavePhD, Kyle Kanos, Waffle's Crazy Peanut, Jim Apr 23 at 13:49

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possible duplicate of Weightlessness for astronauts –  John Rennie Apr 23 at 7:39
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Astronauts in orbit are weightless for the same reasons that you're weightless if you went into a lift and cut the ropes holding the lift, except that with a lift you eventually hit the earth, while the space station keeps missing the earth. –  Lie Ryan Apr 23 at 10:40
    
Yes, makes sense, I never thought of it that way before :) –  denNorske Apr 23 at 10:45
    
did you mean 9.1 or 9.2 $m/s^2$? Because 1 g is the gravity at the surface; it's not larger on the ISS –  Jim Apr 23 at 13:49
    
Ahhh ofcourse, i mean m/s^2, yes! Thanks, i will do an edit ^^ –  denNorske Apr 23 at 16:45
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4 Answers 4

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These are two different effects. Satellites don't fall down because they are moving on a circular orbit. Actually, they are falling down all the time, since circular motion is accelerated (though the velocity doesn't change absolute value, it changes direction!), so it is kind of "falling around the earth".

The second question is, why doesn't an astronaut in a space station fall down to the floor? That's because, both the station and the astronaut feel the same force which holds both of them on a circular orbit. But for falling down to the floor the astronaut should feel a stronger force than the space station, otherwise there is no net force driving the astronaut and the stations floor into each other.

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So, in fact, you can say that both objects is being affected by the same forces and therefore the person acts weightless? –  denNorske Apr 23 at 8:41
    
Exactly! Since the forces are equal, there is no net acceleration towards each other. –  Photon Apr 23 at 9:10
    
okay thanks alot, that solved it for me :D –  denNorske Apr 23 at 9:12
    
That's why it's most commonly called 'free fall' –  Shadur Apr 23 at 11:20
    
But my house has the same forces on it then I do, but I am do fall down to the floor. –  Ian Ringrose Apr 23 at 13:12
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Paraphrasing Douglas Adams, "Flying is learning how to throw yourself at the ground and miss". This is pretty much how orbits work - you're actually falling all the time, but missing the body you're orbitting.

The reason you're experiencing apparent weight-lessness is because every piece of your body is accelerated the same amount (if you could achieve the same effect when accelerating in a car, you'd have the standard sci-fi "inertial dampeners") - and that means that you're not getting any information about the acceleration acting on your body. After all, you're not feeling the gravity when sitting on a chair - you're feeling the pressure the chair puts on you as it prevents you from accelerating through it. If there's nothing to oppose the acceleration, there's no way you could feel it. Accelerometers onboard would report the same thing - they can actually only measure forces that don't act uniformly on the whole body of the device. If you've got an accelerometer in your phone, you can see this quite easily - as long as you hold it in your hand, it will report a 1g of acceleration downwards. However, drop it (on a pillow or something), and during its fall, it will report 0g (ignoring air resistance - if you add air resistance to the equation, the acceleration will steadily grow from 0g to 1g as the object approaches its terminal velocity).

To be precise, the acceleration isn't actually exactly uniform on your body. The parts that are closer to Earth will be attracted slightly stronger than the parts further out. This is called the tidal force, and it's the reason we've got tides on the Earth (and going further into the past, the reason the Moon always faces us with the same side, and in the future, why the Earth will also always face the Moon with the same side). However, the sensors in your body are nowhere near precise enough to register such a tiny difference (and it really is very tiny on the two meters of the human body a hundred thousand kilometers from the Earth). However, that's only because the difference in the gravitational pull is very small over your length. The fun part starts when you get close to a black hole :)

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omg not black holes again haha! But anyway, this was a really good answer aswell, thankyou! –  denNorske Apr 23 at 11:41
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The main question has already been answered, however to cover some of the tangents raised - and give an example where two objects in orbit may not be in free fall relative to each other - there is an excellent short story by Larry Niven that actually addresses the problem of orbiting bodies and what happens to a person inside a ship as it approaches a very dense body (in this case a neutron star). The story is very old now (more than 50 years IIRC) and I haven't read it recently but so far as I know the science is still solid.

http://en.wikipedia.org/wiki/Neutron_Star_%28short_story%29

Spoiler warning

Essentially near a Neutron Star the gravitational gradient is so steep that each end of the ship receives a massively different acceleration due to being nearer or further from the star. This is enough that people inside the ship are only in free fall if they are right in the center of the ship.

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The difference is fairly negligible for most purposes, but on the ISS astronauts experience micro-gravity rather than true weightlesness as they are in a sufficiently low altitude that a thin atmosphere causes drag on the station, slowing it down and causing it to lose height. They have to boost the station back up every few weeks. See http://www.heavens-above.com/IssHeight.aspx

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