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The context is orbital mechanics (moving objects away/toward planets, for example). The following graph confuses me:

enter image description here

How come total energy isn't a straight, horizontal line? Due to conservation of energy? I would think that E total = E kinetic + E potential = some constant value.

I've having quite the conceptual confusion.

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    $\begingroup$ I would guess this graph shows the energy for the same object in different orbits around an object. (e.g. circular orbits at different radii). The total energy in one orbit is constant (even if the orbit is not circular), but you need to change the energy to move to a different orbit. That is why spacecraft have engines! $\endgroup$ – alephzero Apr 16 at 11:37
  • $\begingroup$ Thanks for the reply. You say "you need to change the energy to move to a different orbit" --> as far as I understand, this is true insofar that you change the TYPE of energy. But how come the total energy is changing? E.g. in class we learnt that if we move a satellite to a higher orbit, the satellite's total energy is the same (it lost kinetic energy, gained potential energy, E total unchanged). $\endgroup$ – Lolo123 Apr 16 at 12:31
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In class we learnt that if we move a satellite to a higher orbit, the satellite's total energy is the same (it lost kinetic energy, gained potential energy, E total unchanged).

It sounds like you misunderstood what was said in your class. Assuming circular orbits for simplicity, larger orbits have more total energy than smaller orbits.

Since the orbit is circular and no forces are acting on the satellite except gravity, the gravitational force provides the necessary centripetal force, which allows us to equate $$ \frac{mv^2}{r} = \frac{GMm}{r^2}$$ Multiplying by $\frac{r}{2}$, this implies that the kinetic energy is equal to $\frac{GMm}{2r}$. We can add this to the potential energy to obtain the total energy $$E = -\frac{GMm}{2r}$$ So we see that increasing the orbital radius causes the energy to become less negative, i.e. it increases.


This result can be generalized by the virial theorem, which says that the average kinetic energy of an object in a gravitational orbit is minus one-half of its average potential energy. This result holds for general orbits and reduces to our result for circular orbits, in which the kinetic and potential energies are actually constant.

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For a orbital motion with a periodic motion, radial $r(t)$, the potential $$ V(r) = -\frac{GMm}{r}. $$

Using Virial theory, the average kinetic energy in a cycle of motion $\langle E_k \rangle = -\langle V(r) \rangle / 2$: $$ \langle E_k \rangle = \frac{1}{2} \langle \frac{GMm}{r} \rangle. $$

The total energy $E$: $$ E = \langle E_k \rangle + \frac{1}{2} \langle \frac{GMm}{r} \rangle = -\langle \frac{1}{2}\frac{GMm}{r} \rangle. $$

As alephzero mentioned, for a given average radius $\langle r \rangle$, $ E = \langle E_k \rangle + \langle V(r) \rangle$ is a constant. But the post ploted $E$ for different average radius $r$.

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  • $\begingroup$ @ ytlu What is a "sematic radial"? and where does $E_k=V/2$ come from? Could you explain? $\endgroup$ – mike stone Apr 16 at 12:38
  • $\begingroup$ Yes. I edited to a more precise terms. $\endgroup$ – ytlu Apr 16 at 13:27

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