Gravitational force distance from Earth

December 4, 2019


Tidal forces acting on Earth


The gravitational pull of the Earth is never zero; the force (and therefore the acceleration) decreases as you go further from the Earth like $1/r^2$ where $r$ is the distance from the center of the Earth. At altitudes equal to the radius of the Earth, $6378$ kilometers or so, the force drops to $1/4$ to what it is on the surface but it is not zero. It is not zero even at 400, 000 kilometers from the Earth - which is why the Moon is orbiting the Earth.

You don't feel any Earth's gravitational pull on the International Space Station or on the Moon or on any orbit because the attractive gravitational acceleration is exactly compensated by the fictitious centrifugal force.

Skydivers may fly at the same altitude as other skydivers but all of them are attracted by the Earth and all of them fall down. Because of the air resistance, the skydivers velocity doesn't increase arbitrarily high. Instead, it ultimately converges to 200 km/h or so which is approximately 55 m/s. (Of course, it's the speed before they open the parachutes - if this were the speed after they open it, that would be pretty dangerous.) In the vacuum, you would reach this speed in 5.5 seconds.

So approximately after 6 seconds, the skydivers no longer accelerate. All of them are falling at the same speed. By changing the shape of your body, you may influence the air resistance which modifies your velocity - both speed and the direction. I've tried it, too. It's quite a wind over there.

Source: physics.stackexchange.com
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