Problem in understanding the concept of acceleration due to gravity?

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SUMMARY

The acceleration due to gravity is consistently 9.81 m/s² for objects falling from a height that is a small fraction of the Earth's radius. This uniform acceleration occurs because the gravitational field remains nearly constant at such heights. However, as an object moves further away from the Earth's surface, the gravitational force decreases according to the inverse square law, reaching g/4 at a distance equal to the Earth's radius. For practical purposes, the acceleration remains effectively constant for typical heights encountered in everyday scenarios.

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This question may sound quite stupid but I am having a hard time understanding acceleration due to gravity.
When a ball falls down from a specific height. Would its acceleration always even from the start be 9.8 m/s^2 or would it say initially be a different number and then come down to 9.8?
 
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Always 9.81 m/sec/sec.
 
Acceleration will be the same, numerically, as the g field. If the height the object is dropped from is a small fraction of the radius of Earth then the acceleration will be uniform because the gravitational field will not change measurably. There is very little change, even, for objects in low Earth orbit but the field drops off at an inverse square rate. So, it will be g/4 when you get to a height equal to the Earth's radius - i.e twice as far away from the centre as you are on the surface. That's a long way away, though.
 

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