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If a body is at rest at Earth's surface, can we say that its kinetic energy $E_{\textrm{kin}}=0$, and its potential energy $E_{\textrm{pot}}=0$ also? Because its velocity $v$ and height above ground $h$ are zero.

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Yes, absolutely.

Potential energy is not a measurable physical quantity. What can be measured are differences in potential energy. So, if you compare the potential energies of a given mass at the Earth's surface and $100 \, \mathrm{m}$ above the surface, you cannot choose their difference, because that is governed by the laws of physics. However, you can freely choose to assign any point in space a specific potential energy (for the mass $m$). You could define that $m$'s potential energy at the surface is $0$, or $1234 \, \mathrm{J}$, or any other value.

Kinetic energy of a point particle of mass $m$ is defined by $\frac{mv^2}{2}$, where $v$ is the velocity. "The velocity with respect to what," you may ask. The choice is, again, yours to make; for different frames of references, you get different speeds and kinetic energies, and different notions of "rest". However, as long as $m$ is not accelerating (and $m\ne 0$), there is always a reference frame in which $v=0$ and thus $E_{\textrm{kin}}=0$.

So yes, for a body at rest (with regard to an inertial frame) at the surface (or any other point), you could very well claim that it has zero potential and kinetic energy.

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It all depends on where you have set your coordinate system. If it is on the earths surface then yes but if you set it say on the sun then no.

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    $\begingroup$ The choice of coordinate system does not influence the potential energy at all. The choice of a reference point and reference value for the potential energy, which is unrelated to the coordinate system is what matters. However, the choice of the inertial frame does influence the kinetic energy's value. $\endgroup$ – RQM Feb 20 '15 at 20:18
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    $\begingroup$ If you attach your coordinate system to the sun you are in fact letting the earth have a kinetic energy and thus it would not be zero. As regards potential energy this too will change since the earth distance from the sun does not remain constant. The potential energy difference from a point on the surface of the earth to another point a height h will however not be effected. Philosopher is not asking for the difference of potential energy but the exact value. $\endgroup$ – SAKhan Feb 21 '15 at 1:51
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    $\begingroup$ We do agree on the kinetic energy being dependent on the inertial frame it is measured in. However, the potential energy still does not depend on the choice of the coordinate system. If you, suddenly, include the Sun and its gravitational field into the scenario, of course a point fixed at Earth's surface will undergo a change in potential energy as that point moves towards or from the Sun. However, that would also be true if you chose a coordinate system fixed at Earth's surface. What makes the difference is you altering the problem, not you choosing a specific coordinate system. $\endgroup$ – RQM Feb 21 '15 at 12:33
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The kinetic energy of a particle is dependent on the reference frame, so if a particle is at rest in a particular reference frame, the KE is zero. If someone moving with respect to the particle calculates the KE based on the reference frame in whichthey are at rest they will say $K \neq 0$.

If you are using $$U_g = mgh$$ for gravitational potential energy, then the reference position is arbitrary and $h$ is a small deviation from that reference position. If someone else chooses a different reference position, say 2 meters above your reference position, they will say $U_g \neq 0.$

What's important for the gravitational potential energy in most simple problems is the change in $U_g$, not the actual value.

Always specify the reference frame being used and the reference height.

Also note that for large changes in vertical position one should use $$U_g = \frac{Gm_1m_2}{r_{12}}$$ which rightly recognizes that the gravitational potential energy rightly involves 2 masses. $r_{12}$ is the distance between the centers of the masses. For the simplified $mgh$ form, $m=m_2$ and $g=\frac{GM_{earth}}{r}$

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Yes of course. A stable body has potential energy... as the definition itself says that "the energy of a body at rest is called potential energy". So as the stable body which is of course in rest will contain potential energy.

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  • $\begingroup$ What do mean by "a stable body" ? $\endgroup$ – Mitchell Aug 2 '17 at 15:28
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Kinetic energy is due to motion and potential energy is due to height. There is no motion and height. So,there is only inertia in the body.its kinetic and potential is zero

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  • $\begingroup$ What does this add over the already existing answers? $\endgroup$ – ACuriousMind Aug 18 '16 at 11:27
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    $\begingroup$ Welcome on Physics SE :) While we certainly all appreciate your eagerness to contribute, posting answers on old threads should however be restricted to cases in which a substantial new contribution is made. $\endgroup$ – Sanya Aug 18 '16 at 11:36

protected by Emilio Pisanty Aug 2 '17 at 16:10

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