# Question #b36c5

Jun 15, 2015

It all boils down to ${E}_{g} = m \cdot g \cdot h$
$m$=mass, $g$=gravitational acceleration, $h$=height.

#### Explanation:

If the height $h$ is small enough in relation to the diameter of the earth, we may assume $g$ to be constant at $g = 9.8 m / {s}^{2}$
(actually, $g$ varies a tiny bit over the planet, second decimal)

Now we fill in what we know:
${E}_{g} = m \cdot g \cdot h \to$
$8.2 \cdot {10}^{3} = 69 \cdot 9.8 \cdot h \to$
$h = \frac{8200}{69 \cdot 9.8} \approx 12 m$