How to prove that gravitational potential energy of a body of mass $m$ at a height $h$ is $mgh$? Many introductory physics books just write that potential energy of a body of mass $m$ at a height $h$ as $U_\text{g}=mgh$. However, they never show how this was derived.  I'm interested in knowing this derivation – if possible, avoiding calculus.
 A: At the risk of suggesting an overly simple answer, work done on a body is defined as $W = F \cdot d$.
We know that the force of gravity acting on a body is $mg$.
The perpendicular distance it travels is $h$, so
$$
W=E=F\cdot d= mgh\,.
$$
A: Suppose I let an object fall from rest for a distance $h$. Given that the gravitational acceleration is $g$ the velocity of the object will be given by the SUVAT equation:
$$ v^2 = u^2 + 2gh $$
In this case the initial velocity $u=0$ so we just get $v^2 = 2gh$. The kinetic energy of the object is given by:
$$ T = \tfrac{1}{2}mv^2 = \tfrac{1}{2}m(2gh) = mgh $$
If energy is conserved the increase in kinetic energy must be equal to the decrease in potential energy, so we get:
$$ \Delta U = -\Delta T = -mgh $$
This tells us that if we lower the object by a distance $h$ the potential energy decreases by $mgh$, and conversely that if we raise it by a distance $h$ the potential energy increases by $mgh$.
A: By definition we  know that-

Gravitational Potential Energy of a body is the work done against gravity in raising it to a certain height h.

We have$$
\text{work}~~=~~\text{force}~\times~\text{displacement}
\,.$$
Here force of the body is the weight acting vertically downwards=$mg$ and displacement is $h.$
So,$$
\text{work}
~=~mg\cdot h
~=~mgh
\,.$$
A: The form $U=mgh$ is simply an approximation to allow people to quickly calculate small changes in the gravitational potential energy of the system. The actual value of $U_g$ is usually not important in classical mechanics. $\Delta U_g$ is the important concept.
On a planet, small changes in height change $U_g$ in a space where the gravitational force is almost constant, $\vec{F}=m\vec{g}$. A change in potential energy (in classical mechanics) is calculated by the negative work done by a force while the position changes:
$$\Delta U_{y'\to y'+h}=-\int_{y'}^{y'+h} mg(-\hat{j})\mathrm\ {d}y $$
The integral is easy:
$$\Delta U_{y'\to y'+h}=\left. mgy\right| _y^{y+h}= mg(y'+h-y')=mgh$$
We see that the actual starting point is unimportant as long as the gravitational field can be considered constant. $h$ is positive if the particle is moved opposite the gravitational field (up), and negative if the particle moves down.
