What is the acceleration due to gravity if we go half of the earth's radius distance height from the earth's surface ?

dear pavankumar. 

 i think this theory will give you the answer you are looking for ...

The standard gravitational acceleration at the Earth's surface produces g-force only indirectly. The 1 g force on an object sitting on the Earth's surface is caused by mechanical force exerted in the upward direction by the ground, keeping the object from going into free-fall. An object on the Earth's surface is accelerating relative to the free-fall condition, which is the path an object would follow falling freely toward the Earth's center. It is thus experiencing proper acceleration, even without a change in velocity (which is dv/dt, the familiar "coordinate acceleration" of Newton's laws).

Objects allowed to free-fall under the influence of gravity feel no g-force, as demonstrated by the "zero-g" conditions inside a freely-falling elevator falling toward the Earth's center (in vacuum), or (to good approximation) conditions inside a spacecraft in Earth orbit. These are examples of coordinate acceleration (a change in velocity) without proper acceleration. Since the g-force felt is always a measure of proper acceleration (which, in these cases, is zero, even though the objects are freely changing velocity due to gravity) all of these conditions of free-fall produce no g-force. The experience of no g-force (zero-g), however it is produced, is synonymous with weightlessness.