How do they make International space station to orbit the earth beyond Earth's gravity acting on it? We all know that ISS is rotating at an altitude of just 350km away. How could the ISS escape Earth's gravitational pull?
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asked Mar 1, 2014 at 18:29
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$\begingroup$ And what is the speed of ISS? I think that is the answer - the closer you are the faster is your speed (while angular velocity remains the same). $\endgroup$– Tomáš ZatoMar 1, 2014 at 18:50
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$\begingroup$ The question might to be migrated to the Space Exploration site. $\endgroup$– GeraldMar 1, 2014 at 21:03
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1$\begingroup$ Praveen - it is not rotating. It is orbiting - which means it is entirely within the action of Earth's gravity. It is not escaping - as you can see by the fact it keeps orbiting. $\endgroup$– Rory AlsopMar 3, 2014 at 14:04
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3 Answers
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Your question presumes that the ISS is beyond Earth's gravity, that it has escaped earth's gravitational pull. This is not correct. All objects with mass in the universe affect all other things with mass in the universe, the effect just gets weaker with distance. So the ISS is feeling the effect of gravity from Earth significantly more than the moon is.
The reason the ISS doesn't just fall to the earth, either directly or gradually spiralling towards the earth, is that it is travelling fast enough around the earth that it is continually "missing" earth. It is sometimes described as 'falling' constantly around earth.
If I am to be properly correct though, the reality is that ISS is in fact falling towards the earth, getting closer and closer to Earth all the time. It needs occasional boosts to push it further back out in it's orbit.
Just to blow your mind a little bit: The ISS is pulling on the Earth with the same force that the Earth is pulling on ISS.
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$\begingroup$ That the orbit is not stable and gradually decreasing is not an effect of gravity, but of the drag of earth's atmosphere even in 400km height $\endgroup$ Jan 20, 2021 at 3:09
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ISS is travelling with about 8 km/s around Earth. At that velocity the centrifugal force on the circular orbit is strong enough to cancel out gravity. The parts ISS is made of, needed to be accelerated to that speed by rockets.
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$\begingroup$ It doesn't really annihilate gravity, it just provides an opposite and equal force (almost) that keeps the satellite at a near constant altitude. $\endgroup$ Mar 3, 2014 at 15:33
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2$\begingroup$ @ChrisLovell Now used 'cancel out' instead; hope that's less misleading. $\endgroup$– GeraldMar 3, 2014 at 20:30
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$\begingroup$ Centrifugal force is completely out of place here. In a centrifuge, the objects inside it are driven against the outside walls. No object onboard the ISS is being driven against the outside walls -- they are orbiting with the station. In other words, they and the ISS are falling around earth, entirely subject to gravity (which is NOT cancelled out). $\endgroup$ Sep 22, 2014 at 6:01
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$\begingroup$ @Cyberherbalist The fictitious centrifugal force (the apparent force in the non-inertial reference frame rotating with the ISS around approximately the center of Earth) plays a role here. It is balanced by the gravitational force of Earth which acts as a centripetal force here. That is the reason why the objects inside ISS and the whole ISS do not fly away from Earth. $\endgroup$ Sep 22, 2014 at 12:59
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Centrifugal force is too small to cancel gravity. The ISS, as all the satellites, are actually "always falling", but the horizontal component of their speed will always take them over the horizon line, keeping the distance to Earth constant. You have a very good explanation here
http://www.lasalle.edu/~smithsc/Astronomy/Orbits/orbits.html
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1$\begingroup$ Why do you say that the "centrifugal force is too small"? The centrifugal force is a so-called fictitious force that is only defined in rotating reference frames, but from the perspective of such a frame, it is perfectly correct to say that the centrifugal force is equal and opposite to the gravitational force for a circular orbit. But of course, it is also correct (and more fundamental) to explain the orbit in the context of an inertial frame where there is no centrifugal force, invoking the horizontal component of velocity as you did. $\endgroup$ Sep 22, 2014 at 14:32