Consider an apple at the top of a tree. the apple is pulled by earth's gravity with a force of 1n. if the tree where twice as tall, would the force of gravity be 1/4 as strong?

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.

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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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.

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2