I wonder if both ball 2 experiment is conducted at the same time, however one is at ground level, the other is at 10km above ground level, will they arrive at the same time?

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Based on the concept of Newton’s law of universal gravitation,

http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation

g = G m1.M2 / r^2

where:

* F is the magnitude of the gravitational force between the two point masses,

* G is the gravitational constant, G is approximately equal to 6.67 × 10^?11 N m^2 kg^-2

* m1 is the mass of the first point mass,

* m2 is the mass of the second point mass,

* r is the distance between the two point masses.

Let's assume m1 = mass of ball say = 1 kg for easy substitution and calculation of g.

mass of Earth = 5.9742 × 10^24 kilograms = M_2

http://www.google.com.sg/search?q=mass+of+earth&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a

radius of Earth = 6 378.1 kilometers

http://www.google.com.sg/search?hl=en&client=firefox-a&rls=org.mozilla:en-US:official&hs=vAF&pwst=1&sa=X&oi=spell&resnum=0&ct=result&cd=1&q=radius+of+earth&spell=1

therefore using, g = G m1.M2 / r^2

g = (6.67 × 10^?11)(1)(5.9742× 10^24) / (6378 x 10^3)^2 = 9.80 m/s^2

at a point where R' = R + 10 km =

g' = (6.67 × 10^?11)(1)(5.9742× 10^24) / ([6378+10] x 10^3)^2 = 9.77 m/s^2

so assuming the ball is on a slope tilt of angle teta,

a = g.sin (teta) = 9.80.sin (teta)

a' g'.sin(teta) = 9.77. sin (teta) where teta is say = 30 degrees

assuming motion is under constant acceleration,

equation of motion says, s = u.t + 1/2.a.t^2 and s' =u'.t + 1/2.a'.t'^2

subs in s = s' = say 1 m of simple substitution , u = u' =0

simplified........

solving which gives t = Math.sqrt[(1)(2)/9.81.sin(30^o)] = 0.639 s & t' = 0.640 s approximately.

[color=blue][b]in conclusion to answer your question, the answer should be roughly the same time unless you can conduct the experiment at a height above Earth where the g' is very different of the sea-level g .:)[/b][/color]

understand?