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Physics/Sounds UN-likely


Radius of earth 6,371,000 meters + 400,000 meters up to ISS orbit height = 6771000 meters distance from earth barycenter into space (weightlessness).

Fg =  G x mass earth x 1 kg  /  6771000^2

Fg = .0000000000667384 x 5972190000000000000000000  /  45846441000000  =  8.69368257 Newtons

8.69368257 Newtons  =  .886508904122242 Kilograms

I don't think a 1 kg ball up in space near the ISS feels like it weighs .88 kilograms, do you ? ?


Of course it doesn't appear that way, that would be silly, it's in orbit.  It's moving at 7,660 meters PER SECOND.  In comparison, the energy per unit mass of a bullet is laughably small (<1/50th).  The gravitational pull you mention is calculated just fine, but it applies to the space station as well.  That's enough to keep it moving in a circle, but not enough to pull it in to the center of the Earth.  When you jump from something, you feel temporarily weightless.  Gravity is pulling on you exactly as hard (or you wouldn't fall at all), you just FEEL weightless because there's no pressure being applied to you by the ground you stand on (since you're not standing on it, you're falling).  We feel contact forces, we do not directly feel the force of gravity.  We feel the force of the ground resisting gravity from pulling us into it (hence causing pressure on our feet, and in our legs, etc).  If the Earth wasn't exerting the force you calculate on the 1 kg ball, it would move in a straight line off into space and not orbit.

Your real fundamental mistake is that you tried to equate Newtons to kilograms.  Force is not mass, the units are not the same.  Force = mass*acceleration.  You can't just set different things equal.  9.8 Newtons = 1.0 kg * 9.8 m/s^2.  That is the contact force of whatever a ball sitting here on Earth applies to the ball to keep it from falling into the center of the Earth.  You similarly don't feel the gravitation of the Sun or the Moon, but calculate the force on your body by them and you'll find that it's substantial.  It's also responsible for the tides, since there you have something big enough (the Earth) that those bodies can pull on parts of our oceans more strongly (due to the difference in distances between the various oceans and the Moon or Sun) than other parts which are farther away (other side of the Earth).

Are you trying to say physics is wrong somehow?  That would be rather presumptuous, for such well-established physics.  The misconception that "weightless" means that the Earth exhibits no gravitational attraction on an object is quite common, I got that all the time as a professor from engineering and pre-med majors who were quite intelligent, but I hope this straightens that out.  


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Dr. Stephen O. Nelson


I can answer most basic physics questions, physics questions about science fiction and everyday observations of physics, etc. I'm also usually good for science fair advice (I'm the regional science fair director). I do not answer homework problems. I will occasionally point out where a homework solution went wrong, though. I'm usually good at explaining odd observations that seem counterintuitive, energy science, nuclear physics, nuclear astrophysics, and alternative theories of physics are my specialties.


I was a physics professor at the University of Texas of the Permian Basin, research in nuclear technology and nuclear astrophysics. My travelling science show saw over 20,000 students of all ages. I taught physics, nuclear chemistry, radiation safety, vacuum technology, and answer tons of questions as I tour schools encouraging students to consider careers in science. I moved on to a non-academic job with more research just recently.

Ph. D. from Duke University in physics, research in nuclear astrophysics reactions, gamma-ray astronomy technology, and advanced nuclear reactors.

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