What's wrong in stating that equal gravitational forces act on two identical masses at different heights, as they have the same acceleration $g$? This might be stupid but I need an answer. 
Newton's second law stated that the force exerted on a body equals the mass of the body times the acceleration of the body. And the law of gravitation states that:
$$F = G\frac{m_1m_2}{r^2}$$
My question is: if I take two identical masses and place one higher than the other then one experiences greater gravitational force than the other. Both accelerate at the same rate $g$. If this is so, according to Newton's second law the forces on the bodies must be equal. 
Where am I wrong?
 A: This has to do with the factor of $1/r^2$ in the gravitational force. In your example the quantity $r$ is the distance to the center of the Earth. So small differences in height during an experiment are completely negligible.
However, high-precision experiments do find a difference in the gravitational force (or equivalently in the value of $g$) across the Earth and hence across different distances to the center of the Earth: http://www.geophys.ac.cn/infowin/Gravity.asp
A: The answer is that falling objects do not all accelerate towards the Earth at the same rate of $9.8 \text{ m/s}^2$.
All objects, at the surface of the Earth, accelerate the same, regardless of their mass. Also, all objects at the same distance from the center of the Earth accelerate at the same rate. But objects at different heights do not accelerate exactly the same. Otherwise how could you ever escape the Earth's gravity?
The more correct way to calculate the acceleration is to do it the way you have done, using Newton's 2nd Law and Newton's Law of Universal Gravitation.
Physics teachers often teach their classes that all object's accelerate at the same rate and then don't emphasize the limits on that statement.
A: 
My question is: if I take two identical masses and place one higher than the other then one experiences greater gravitational force than the other. Both accelerate at the same rate g. If this is so, according to Newton's second law the forces on the bodies must be equal.

This link clears it up
there is 
$$F = G\frac{m_1m_2}{r^2}$$ , the law of universal gravitation
$m_1$ and $m_2$ are any two masses. In the case of $m_1$ being the earth and $m_2$ another mass
one arrives at 
$F=m_2g$ where g is the acceleration and depends on the mass of the earth, and the distance from the center of mass,  as seen in the link.
The confusion comes from ignoring orders of magnitude: the mass of the earth is so much larger than any mass on earth that the $r$ dependence in the formula can be ignored within experimental errors unless special care is taken.
A: Here the problem is that you have ignored the orders of magnitude .The radius of Earth is much larger than the difference of height considered here. As a result, we may reasonably neglect this difference with comparison to the radius of Earth and obtain the gravitational force between the identical masses and Earth to be equal. However if the measurement of forces is done with high precision in idealised situations, then we may find that the force of gravity on the lower mass is indeed greater than that on the higher mass by small amount.
A: The value of force exerted on both by the Earth is not the same, and the value of their accelerations is not the same. If you make the equations you will see that Newton's second law definitely holds true in this case.
I believe you are confused because you thought that the acceleration of a body towards the Earth is always equal to 'g', but this is not the case. The value of gravitational acceleration is g only for bodies that are close to the surface of the Earth. To get an idea of this let me give you some formulae. For an exact measurement of the acceleration varying with height we can use:
For a body of mass m at a height h from the surface of the Earth having a mass M and radius R,
$$
F = \frac{GMm}{(R+h)^2}
\\
a = \frac{F}{m}
\\
a = \frac{GM}{(R+h)^2}
$$
We also sometimes use an approximation for smaller heights (in comparison to radius of the Earth):
We have,
$$
a = \frac{GM}{(R+h)^2}\\
\implies a = (GM)(R+h)^{-2}\\
\implies a = \frac{GM}{R^2} [1 + \frac{h}{R}]^{-2}
$$
We can use binomial approximation on this, which gives us:
Since $h << R$,
$$
a = \frac{GM}{R^2} [1 - \frac{2h}{R}]
$$
since we know that $g = \frac{GM}{R^2}$, we can substitute this to get,
$$
a = g_{height=h} = g [1 - \frac{2h}{R}]
$$ 
Hope this clears your confusion!
A: Since you have placed one body higher than the other .Both will will not have the same acceleration 'g'. 
I have assumed that both the body has same mass.
g ( acceleration due to gravity ) varies with height. It decreases with height. 
And for small height  as compared to radius , one assumes the same value of 'g'.
But at the end of the day , its mere assumption. 
Two body kept at same height will have same acceleration , regardless of its mass. As acceleration is independent of mass. One can also think this case with inertia as a factor.
A: You can compute the value of $g$.  It is $g = GM_E/R_E^2$, where $M_E$ is the mass of the Earth and $R_E$ is the radius of the Earth.  For many problems near the surface of the Earth, it's adequate to use this value even though you're at some radius slightly different than $R_E$.  As long as that remains a good approximation, using $g$ is a just a handy way to group together constants in the gravitational force law.
Your question points to a spot where the approximation breaks down. In the scenario to described, either the approximation is good enough that you cannot tell the difference, or you need to use the full force law with variable radius to get higher precision and see that the forces are different.
A: Who told all objects accelerate with g irrespective of their height?Free fall acceleration g=9.8 only at heights much less than radius of the earth.Actually g is nothing but the force experienced by the object at a particular height divided by the mass of the object..Hence g is independent of the mass,but depends on the height.
