If M is the mass of earth and R is the radius, the earth attracts a mass m on its surface by a force F given as:
F = GMm/R2
This force impacts acceleration to the mass m, which is known as acceleration due to gravity (g).
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By Newton's II' nd Law, we have: acceleration = force / mass
g = F/m = (GMm/R2)/m = GM/R2
On the surface of earth, g = GM/R2
Substituting the values of G, M, R we get g = 9.81 m/s2.
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Variation In Acceleration Due To Gravity:
(a) Variation of g with altitude: At an altitude h above the earth's surface,
R′ , the distance from the centre of the earth = R + h
Thus, 
( GM = gR2)
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or, 
This implies that, for h <<R,
g = g(1-2h/R)
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At a height of 100 km from the field, the acceleration due to gravity decreases by a fraction of
(b) Variation of g with depth: At a depth, x, below the earth's surface, the only change is not just due to the change in the distance from the centre of the earth, but also due to a change in the 'effective' mass of the earth that contributes to g at this point. Only that portion of the earth that is enclosed by a sphere of radius RS ( = R - x) centered at the earth's centre is effective.
If we assume the earth to be a sphere having uniform volume, then,
g′ = g(1-x/R) ( since R = RS + x)
Therefore, at a depth of 10 km below the earth's field, the fractional decrease in the acceleration due to gravity is, approximately, 0.16%
In reality, the change will be much less as the density of the earth is not constant throughout as we have assumed in fact the crust has a much lower density as compared to the core.
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(c) Variation with latitude: The effective value of the acceleration due to gravity changes with latitude owing to the rotation of the earth.
Referring to the figure, at a latitude Θ (Point Q), the effective weight,
mg′ = mg - mw2 r cosΘ . . . (i)
Where mw2 r is the "centrifugal force" and we take its component in the vertical direction and mg is the weight in the absence of rotation.
Now, r = radius of the circle rotating about Q.
= QE = R cos Θ (from ΔCQE),
Therefore, g'= g - w2 R cos2 Θ
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or, g'=
...(ii)
Now, w= 2p/86400 s-1, R Ξ 6400 km, g = 9.8 m/s2
So, w2 R/g Ξ 0.34%, which is indeed a very small effect
At poles: Θ = 90° 
g / = g
the motion of the earth has no effect on the gravity at poles.
At equator: Θ = 0° 
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(d) Effect of the surface of Earth
The equatorial radius is about 21 km longer than its polar radius. We know,  . The weight of the body increase as the body is taken from the equator to the pole.
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Work Done in Displacement of a body in gravitaional field:
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If a body of mass m is displaced in a gravitational field from point A to B, then work done by external agent can be given as
W = UB- UA
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Thus work given in displacing an object in gravitational field can be given as
Work done = mass of body * gravitational potential difference of the terminal points.
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