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The International space station is constantly falling towards Earth under the pull of Earth's gravity (Just like any other object - gravity does not stop when you reach space!). However the Station is moving very fast horizontally and, as the Earth is a sphere, this means that as it falls its path takes it round the Earth in a circle - it is in "orbit". This means that if you are in the space station you are falling as fast as gravity can pull you and therefore you do not feel the pull of gravity, making you weightless.
no
Gravity
To answer this question we need to know either the height of the Earth station above the Earth or the gravitational acceleration of gravity present at the Earth station initially.Solving without these values:(1/2)Gm/R2 = Gm/r2whereG is the Gravitational Constantm is the mass of the EarthR is the radius of the Earth Station from the center of the Earth.r is the radius of the satellite where gr=1/2ge2R2 = r2SQRT(2)R=r
The international space station does not have artificial gravity. The occupants float freely and use a lot of Velcro. Don't confuse fantasy and reality.
I'm not sure if it's ever measured, but it could be approximated by calculating the surface gravity of a spherical asteroid of equal mass and dimensions.Assuming the mass of the International Space Station is 450 000 kg (M) and its mean radius is about 30 meters (r) , the surface gravity would be g = MG/r2 = about 0,00000003337 m/s2 (about 30 nanometers/square second).For comparison, Earth's surface gravity is about 9.81 m/s2, so the gravity you would experience standing on the surface of the International Space Station is about 0.3 millionth of a percent compared to earth. It's certainly too small a gravity to hold you attached to the station if you were standing on it.If you're inside the space station, in the center of the station, there is zero gravity because you are in the center of gravity because the mass of the station is situated around you.Panu, M.Sc.
In the simplest case, supposing the orbit of the space station to be a perfect circle, the centripetal acceleration would be quadrupled (F=mω2r).
because, gravity pulls ISS(Inter national Space Station)
By spinning, the centripetal force creates artificial gravity on the space station.
no.
Yes. Mass and weight are different quantities. The mass of the astronaut is always the same everywhere. The weight of the astronaut is the force on it due to gravity, which depends on the mass and the strength of gravity at the point on the planet, moon or space station the astronaut is standing on. The strength of gravity is known as "local acceleration due to gravity", and it is represented by the letter g. On the surface of the earth g is about 9.8 ms-2. On the surface of the moon g is about 1.6 ms-2 (a 6th that of the earth). On a space station, because the mass of the station is so small, g is effectively 0 ms-2. So, the weight of an object is the force on it due to gravity. The formula for weight is: W = mg Where: W = Weight (in Newtons) m = mass (of the object, in Kg) g = local acceleration due to gravity (in ms-2). Hopefully you see this formula is a restatement of F = ma. Your average man has a mass of about 70 kg. If you plug in the numbers you find the weight on the earth is 686 N, on the moon is 112 N, and on the space station it is 0 N (i.e. the astronaut is totally weightless).
As of 2010, artificial gravity technology has not been invented.