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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 $h$?

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  • $\begingroup$ Right, and for different heights as well. For example, my total potential energy in relation to Earth's center of mass. $\endgroup$
    – Madde Anerson
    Commented Dec 11, 2014 at 9:29
  • $\begingroup$ Indeed, we suppose $g$ as a constant because in small distances from the earth, its attraction doesn't change so much. Otherwise, to obtain the potential energy we should use the main relation, $Gm_1m_E/r^2$, and then integrate it. Or $V=-Gm_1m_E/r$ where $m_E$ is the mass of the earth. $\endgroup$
    – MEDVIS
    Commented Dec 11, 2014 at 9:39
  • $\begingroup$ According to that expression, as $r$ gets bigger, $V$ becomes smaller. Hmm…? $\endgroup$
    – Madde Anerson
    Commented Dec 11, 2014 at 10:06
  • $\begingroup$ It's better to interpret it in this way: $mg\Delta h$ gives the difference of potentials; on the other hand $\Delta V=Gm_1m_E(\dfrac{1}{r_i}-\dfrac{1}{r_f})$ where $r_i$ and $r_f$ are the initial and final points. Now increasing $\Delta h$ is equivalent to increasing $r_f$ or decreasing $\dfrac{1}{r_f}$ which in turn is equivalent to increasing $\Delta V$. Note that in $mg\Delta h$, the origin of the potential is on the earth, but in the other formula, the origin is located at infinity. $\endgroup$
    – MEDVIS
    Commented Dec 11, 2014 at 10:30
  • $\begingroup$ That doesn't answer my objection though. $\endgroup$
    – Madde Anerson
    Commented Dec 12, 2014 at 2:44

2 Answers 2

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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.

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  • $\begingroup$ Doesn't work for negative $h$. $\endgroup$
    – ProfRob
    Commented Dec 26, 2014 at 17:58
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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.

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  • $\begingroup$ Wait… so increasing $x$ <i>decreases</i> my total potential energy ($V(x)$)? How does that make any sense? $\endgroup$
    – Madde Anerson
    Commented Dec 12, 2014 at 2:40
  • $\begingroup$ @MaddeAnerson It depends on how you define potential energy. In this case I have defined potential energy as "the work done by the gravitational field bringing a unit mass in from infinity to that point". (Using the definition of Wiki.) This way infinity has zero potential energy. $\endgroup$
    – user3237992
    Commented Dec 13, 2014 at 1:32
  • $\begingroup$ I am not sure of your definition, but my definition would be "the work done by gravity by pulling one unit mass from $h$ (or $r$, depending on what you want to call the height from the center of mass of the bigger body, for example Earth) down to the point where $h = 0$. Here I am reducing everything to point masses, of course. $\endgroup$
    – Madde Anerson
    Commented Dec 15, 2014 at 2:23
  • $\begingroup$ @MaddeAnerson The main difference between your definition and mine is that we have postulated different isopotential surfaces to be of zero potential. Your zero potential surface (thus reference point if you like because of the spherical symmetry of the gravitational field) is the surface of the Earth (approximated as a sphere). My isopotential surface (thus reference point) is a sphere with center that of the Earth's and radius of infinity. But either way you cannot calculate something as total potential energy since energy is always undeterminded up to an arbitrary additive constant. $\endgroup$
    – user3237992
    Commented Dec 26, 2014 at 11:11
  • $\begingroup$ @MaddeAnerson That is why you always have to specify an isopotential surface (or, if you project this spehrically symmetric problem into 1D, a single point) which you define to be of zero potential, and you can only calculate the potential difference between other isopotential surfaces and the zero potential surface. Your defintion and mine will yield the same results however, if you would calculate the potential difference between two points of space. $\endgroup$
    – user3237992
    Commented Dec 26, 2014 at 11:13

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