Question

Please write legibly. Thank you.

1. Is the acceleration due to gravity a universal constant? Explain.

2. Is the acceleration due to gravity of objects affected by the Earth’s rotation? Explain.

Answer 1

1)

No.
Acceleration due to gravity is a property of a planet(For that matter any object). So it's not absolute. It differs from planet to planet. For earth, avg acceleration due to gravity is 9.8 m/s² and for moon it is 1/6th of that of earth. Also acceleration due to gravity changes within our planet since earth is not a perfect sphere. g is different at pole and different at equator. Though difference is very slight.
In general, if you consider two objects. One having quite large mass than the other. The bigger body creates a gravitational field around it. Since the bigger body cannot be moved easily because of inertia, because of the mutual attraction between the objects, small body accelerates towards bigger one. Now according to Newton's Gravitational, force between two objects is  

F=GMm/r²

Consider M as the bigger body. Rest you must be knowing.
Also

F=ma

So equate them, and from this you can find acceleration due to gravity(a) for any big inertial mass.

2)

Outward centrifugal force caused due to rotation of earth affects (decreases) the acceleration due to gravity. The change varies with the latitude.

Let us consider the earth to be a spherical ball of mass ‘M’ and radius ‘R’. An object of mass ‘m’ is at point P at latitude φ, when the earth is not rotating the weight of the object is mg. But earth is rotating with angular velocity. So, the object is moving in a circular path of radius ‘r’ as shown in the figure. The object experiences centrifugal force,

The object is being acted by two forces ‘mg’ and ‘FfFf‘. The resultant of these two forces gives the apparent weight of the object (m’g). The resultant of the two forces is given by the diagonal of the parallelogram OPAB.

From parallelogram law of vector addition:

The value of ω is 7.2921159 × 10−510−5. So, the term containing ω4ω4 is much smaller than 1 and ω2ω2 so can be neglected

At poles; φ = 90°

g’ = g.

So, at equator the effect is highest so the gravity is lowest and at the poles gravity remains unaffected of rotat