No, because the distance that determines the gravitational forces
is the distance between the CENTERS of the two objects.
Before the apple falls, its distance from the center of the Earth is
roughly 4,000 miles ... the Earth's radius. (The height of the tree
doesn't make a whole lot of difference.)
If you wanted to double the distance for gravity purposes, you'd
need to move the apple ANOTHER 4,000 MILES from the center
of the Earth ... about 15 times as high off the surface as the International
Space Station's orbit. THEN the weight of the apple would drop to
1/4 of what it weighs when it's down here.
In the given case, the force of gravity be 1/4 as strong - no, as the distance between the center of the apple and earth remain equivalent and the height of the tree is insignificant.
Gravitational force depends directly on the product of the masses of the two objects and is inversely proportional to the square of the distance between the center of mass of the two objects.
[tex]Fg = \frac{Gm1m2}{R^2}\\\\or, Fg = \frac{1}{R^2}\\[/tex]
where R = the distance between the center of mass of two objects,
In the given case,
Thus, In the given case, the force of gravity be 1/4 as strong - no, as the distance between the center of the apple and earth remain equivalent and the height of the tree is insignificant.
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