WackyDumpsite said:
And that makes no sense which is what we've been trying to tell you. Your equations don't help explain how the Sun is able to keep the Earth moving in a circular manner as opposed to flying straight. Or the Earth and the Moon, etc. To say that no energy is needed to maintain a circular orbit is nonsensical. It obviously requires energy.
Take a table. If I kick the bottom of the table, in outer space it will move in the direction it was kicked. But Earth's gravity is continuously pulling it downward. This is work being done. Work requires energy. There is no matter --> energy conversion that explains this as far as I know, and petridge pointed this out as well. So where does it come from?
I've raised this issue up in a conversation with a Stanford PhD physics professor about this. She didn't know how to answer it and said things like 'we don't know much about gravity' and 'we have this concept of potential energy.' If you have better credentials than that, by all means give a real answer, don't just refer to some equation as proof that you don't need to account for it.
Think of the simple case where the difference in mass between the two bodies is so great that the effect of the smaller mass,
m, on the greater,
M, can be ignored. Let's suppose mass
m is in a circular orbit around
M.
By definition of potential energy (and taking our "zero of potential energy" as infinity),
U = - \int_{r}^{\infty} \textbf{F} \cdot \textup{d}\textbf{r} = \frac{-GMm}{r},
where
U is potential energy due to gravity,
F gravitational force,
M the greater mass,
m the smaller mass, d
r an infinitesimal change in the position of
m (the position considered as a position vector,
r, extending from the centre of mass of the system, here effectively the centre of mass of
M, to the centre of mass of
m),
r the magnitude (length) of this position vector
r,
G the gravitational constant (which relates the units). In other words, the potential energy of the satellite
m is the work that would be done by the force of gravity if it was to move
m from infinitely far away to its current position.
By definition of kinetic energy,
T = \frac{1}{2}mv^2,
where
T is the kinetic energy of the orbiting body with mass
m, and
v its speed, defined as the magnitude of its velocity with respect to the more massive body.
By definition of circular motion,
\frac{\mathrm{d} r}{\mathrm{d} t} = 0
at all points in the orbit. This just means that the distance of the satellite from the more massive body doesn't vary over time.
Less obviously,
\frac{\mathrm{d} v}{\mathrm{d} t} = 0.
That is, the speed of the satellite is constant over time. This follows from Kepler's 2nd law in the special case of a circular orbit; the radius of the orbit doesn't change, so the only way that an equal area can be swept out over an equal time is if the speed is constant.
http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Second_law
Therefore
\frac{\mathrm{d} }{\mathrm{d} t} \left(T + U) = 0.
Kinetic energy is constant, and potential energy is constant, so their sum is constant.