Why m g h is only a rough formula

$mgh$ treats $g$ as steady all the way up. You will make the program compare it with the exact gain in potential energy for a $1\ \text{kg}$ body lifted from the ground, and find the height at which $mgh$ is $20\%$ too big.

The idea. Exactly, the gain is $GMm\left(\dfrac{1}{R} - \dfrac{1}{R + h}\right)$, with $GM = gR^2$. Dividing, $\dfrac{mgh}{\text{exact}} = \dfrac{R + h}{R}$.

Your tasks.

1. Work out the exact gain exact for every height in h.
2. Work out quick, the gain from $mgh$.
3. Find the height at which quick is $1.2$ times exact.