BZZZT What consolation prize do we have for our contestant, Vanna?
An object's "weight" (force due to gravity) when it is in orbit is considerable. The force due to gravity is F=Gm1m2/r^2 where G is the universal gravitational constant, m1 and m2 are two masses, an r is the distance between them. This becomes
F2/F1=(Gm1m2/(re+oa)^2)/(Gm1m2/re^2)
where F2 is the force in orbit, F1 is the force at the earth's surface, re is the earth's radius, and oa is the orbital altitude. A little bit of algebra gives us
F2=F1*(re^2/(re+oa)^2
Plugging in the earth's radius and the space station's average altitude...
F2=0.89*F1
In other words, an object in earth orbit at the altitude of the space station weighs approximately 90% of what it weighs on the earth's surface.
However, the acceleration due to gravity is matched by the centripital acceleration, and so there is no unbalanced force, and thus "weightlessness".
Slashdot Mirror
An object's "weight" (force due to gravity) when it is in orbit is considerable. The force due to gravity is F=Gm1m2/r^2 where G is the universal gravitational constant, m1 and m2 are two masses, an r is the distance between them. This becomes
F2/F1=(Gm1m2/(re+oa)^2)/(Gm1m2/re^2)
where F2 is the force in orbit, F1 is the force at the earth's surface, re is the earth's radius, and oa is the orbital altitude. A little bit of algebra gives us
F2=F1*(re^2/(re+oa)^2
Plugging in the earth's radius and the space station's average altitude...
F2=0.89*F1
In other words, an object in earth orbit at the altitude of the space station weighs approximately 90% of what it weighs on the earth's surface.
However, the acceleration due to gravity is matched by the centripital acceleration, and so there is no unbalanced force, and thus "weightlessness".