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Okay, so the formula for Gravitational Potential Energy (GPE) of satellites is: $$ U=-GmM/d$$ Mass and distance can't be negative, and $G$ is constant that is not negative.

Why is a satellite's GPE negative?

For other objects, like a ball, the GPE is calculated as: U=mgh

Correct me if I am wrong, but this is what I think:

  1. Satellite's GPE is measured relative to earth's center, while a ball's GPE is measured relative to earth's surface.

  2. The satellite was sent to its orbit with energy from the fuel, and it was launched from the earth's surface. While for the ball, it was sent to its height with energy (say, from a human body) and it was carried from the earth's surface.

So is a satellite's GPE negative because: Energy needed to bring the satellite to earth's center is greater than the energy expended when it was sent to orbit?

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marked as duplicate by John Rennie, ACuriousMind, user36790, Kyle Kanos, HDE 226868 Oct 31 '15 at 17:01

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  • $\begingroup$ Think of it like trying to climb out of an infinite well. Potential energy is just a number relative to some other number, but since a gravitational system is (to zeroth and maybe even first order) conservative, it's a zero sum game... $\endgroup$ – honeste_vivere Oct 29 '15 at 21:32
  • $\begingroup$ The reference point for potential energy is chosen for convenience. The only difference is an additive constant that has no physical consequences. For satellites we measure relative to infinity, where, by convention, the potential energy is set to be zero. For a 1/r potential the potential at the center would be infinite, so that's not a useful reference point. $\endgroup$ – CuriousOne Oct 29 '15 at 21:33
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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/64260/2451 and links therein. $\endgroup$ – Qmechanic Oct 29 '15 at 22:42
  • $\begingroup$ Noe that $U = mgh$ is in fact a limiting case of $U = -GMm/d$ where the zero point is shifted from $r = ∞$ to the ground location and where h to be small enough compared to the distance to the center such that $GM/d$ is roughly constant (1 meter compared to the radius of the earth for instance). $\endgroup$ – Cicero Oct 30 '15 at 1:19
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GPE of a system of two masses is calculated by $$GPE = U_g=\frac{-Gm_1m_2}{r_{12}}.$$

This arises from the definition of potential energy associated with a conservative force: $$U=-\int_{r_0}^{r_f} \vec{F}\cdot d\vec{r},$$

where for gravitational systems $r_o$ is $\infty$, $r_f$ is the separation distance between two masses and $\vec{F}$ is the Newtonian gravitational force, $$ \vec{F}_g = \frac{-Gm_1m_2}{r^2}\hat{r},$$

When you handle all the vectors properly, you arrive with a negative sign for the GPE of the system of the two masses. If you increase the separation, the GPE increases.

In your example of the ball where $GPE=mgh$ that is actually a change in GPE where the ball have gone from some initial $r_{12}$ with respect to the center of the Earth to a new separation from the Earth of $r_{12}+h$. When you do the maths, you get $\Delta U_g = m_{ball}gh$ where $$g=\frac{Gm_E}{r_{12}}.$$

It has nothing to do with the fuel used to send it in orbit. In fact, the ball at the surface of the Earth contributes negatively to the GPE of the system.

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