My total potential energy? The potential energy in a uniformed gravitational field is $mg \cdot \Delta h$. This assumes of course that $g$ doesn't change and only gives the difference in potential energy for $\Delta h$.
How can I calculate my total potential energy, let's say relative to the Earth's center of mass. In other words, are there any expression for $\int_0^h mgh$, where $h$ is a placeholder for every height value and $g$ is a placeholder for the local gravitational acceleration for $h$?
 A: It may help: suppose we are close to the earth and at height $h$. So
$$\Delta V=Gm_1m_E(\frac{1}{R}-\frac{1}{R+h}) $$ where $R$ is the radius of the earth and $h \ll R$. Now we approximate this relation and it's turn out that
$$\Delta V=Gm_1m_E(\frac1R-\frac1R+\frac{h}{R^2})$$
By calling $g=\dfrac{Gm_E}{R^2}$, we find $\Delta V=m_1 gh$. Even if we don't approximate, it is obvious from the first equation that by increasing $h$, $\Delta V$ will increase.
A: If we estimate the shape of the Earth to be a perfect sphere (which it isn't, but as a first approximation it will do), then you may apply the shell theorem. It states that the gravitational field of a spherically symmetric body appears as if it is concentrated in the center of mass of the body. Knowing how much Earth weighs ($5.97219\times10^{24}\,kg$), you can calculate the gravitational potential from Newton's law of gravity. Your distance from the point of mass is of course the radius of Earth ($6,371\,km$) plus your distance from the surface, which is arbitrary. This will be \begin{equation} V(x)=-\frac{G M}{x} \end{equation} where $G$ is the gravitational constant, $M$ is the mass of Earth and $x$ is your distance from its center of mass.
