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is (F=mg) equal to (F=GmM/r^2)? And what's the difference between them?

Qmechanic
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Zheer
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    $g$ is the acceleration of something moving under gravity near the Earth’s surface, when no other forces are acting on it. It is not zero when an object is at rest. To be at rest there are additional forces opposing gravity, such as the force of the ground on your shoes, so $F$ is not only $mg$. – G. Smith Mar 11 '20 at 17:53
  • Possible duplicates: https://physics.stackexchange.com/q/286360/2451 , https://physics.stackexchange.com/q/35878/2451 and links therein. – Qmechanic Mar 11 '20 at 17:57
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    $g$ is the strength of the gravitational field, in this case, due to Earth's mass and radius. The force on a mass in a grav. field is $mg$. By Newton's 2nd Law, $ma=\sum F$. If $mg$ is the only force, then $ma=mg$, so $a=g$. – Bill N Mar 11 '20 at 18:26

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The Newton law of universal Gravitation is given by $$ F = G \frac{m M}{r^2} $$ For two bodies of mass $M$ and $m$. When describing gravitational effects near the surface of the earth, its common to use the approximate formula $F = m g$, where $g = G M/R^2$, with $R$ being the earth´s radius and $M$ the earth mass.

Alejandro Celis
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  • Isn't (g) acceleration? Can it be zero when the object is at rest? – Zheer Mar 11 '20 at 17:48
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    g is the acceleration of an object falling freely (without any other force acting, just gravity) near the surface of the Earth. For an static object near the Earth surface, lets say your door, the are other forces acting besides the gravitational one so that the object has no acceleration. – Alejandro Celis Mar 11 '20 at 17:53