# Why total mechanical energy does not 'seem' to be conserved in a typical Earth-Sun system?

Considering mass of the Sun and Earth to be $$M$$ and $$m$$ respectively, we have the potential energy of the system to be equal to $$\frac{-GMm}{r}$$ when defined zero at infinite separation, $$r$$ is the separation of the two bodies. The kinetic energy of the earth(taking the sun to be frame of reference) will be $$\frac{GMm}{2r}$$ (by equating the force of gravity and the centripetal force on earth, and thereby finding the velocity and putting into equation $$\frac{1}{2}mv^2$$).

The above statement(s) seem to imply that the magnitude of the kinetic energy is half of the potential energy. Which means a change in potential energy causes half the change in kinetic energy (being opposite). Also since in this system the force of gravity being an internal conservative force, the total Mechanical energy 'should' be conserved. But the total mechanical energy is equal to $$\frac{-GMm}{2r}$$ which is a function of the separation between the two bodies. And, where is the extra half of potential energy change is going to? What am I missing here?

• How do you change potential energy in a circular orbit?
– JEB
Apr 23, 2020 at 14:48
• @JEB Both potential and kinetic energy will remain conserved in a circular orbit that is for a given radius. But when we change the orbit for example going to a lower one the decrease in potential energy will be twice than that of increase in kinetic energy. Apr 23, 2020 at 14:52
• You may find this helpful: en.wikipedia.org/wiki/Specific_orbital_energy Apr 23, 2020 at 15:10
• I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. Apr 24, 2020 at 0:40

Your argument here is based on a circular orbit, and yes, the total mechanical energy remains constant. But a change in potential energy in a gravitational circular orbit means that outside work must be done on the system, so the mechanical energy changes when the radius changes. (Energy is still conserved because $$E_{final}=E_{initial}+W_{outside}$$; you didn't create any energy. You just added or removed energy in the system.) If the radius changes the kinetic energy will change in conjunction to the work done. But, for a circular orbit, the radius cannot change without outside intervention.
If you have an elliptical orbit, the $$|U_g|=2K$$ relationship between potential ($$U_g$$) and kinetic ($$K$$) is no longer true. There will be a constant of motion, the angular momentum, as well as the mechanical energy, but the $$U_g$$ and $$K$$ will be changing, swapping back and forth, as the Earth moves from aphelion to perihelion and back again. Instead of having a constant radius, we have a constant major axis and eccentricity.
If you have radial movement the kinetical energy has two components $$K=\frac{1}{2}(\dot{r}^2+(r\dot{\theta})^2)$$. In this case, you need to consider the change of velocity in angular and radial directions.