Friday, December 18, 2015

newtonian mechanics - Is the distance between the Sun and the Earth increasing?


$M$ = mass of the Sun


$m$ = mass of the Earth


$r$ = distance between the Earth and the Sun


The sun is converting mass into energy by nuclear fusion.


$$F = \frac{GMm}{r^2} = \frac{mv^2}{r} \rightarrow r = \frac{GM}{v^2}$$


$$\Delta E = \Delta M c^2 = (M_{t} - M_{t+\Delta t}) c^2 \rightarrow \Delta M = \Delta E / c^2$$



$$\rightarrow \frac{\Delta r}{\Delta t} = \frac{G}{v^2 c^2}.\frac{\Delta E}{\Delta t}$$


Sun radiates $3.9 × 10^{26}~\mathrm W = \Delta E/\Delta t$


Velocity of the earth $v = 29.8 \mathrm{km/s}$


There is nothing that is stopping the earth from moving with the same velocity so for centripetal force to balance gravitational force $r$ must change.


Is $r$ increasing? ($\Delta r/ \Delta t = 3.26070717 × 10^{-10} \mathrm{m/s} $)



Answer



I think the reasoning has an error. It assumes $v$ is constant, but instead we ought to assume the angular momentum is constant.


By dimensional analysis that leads to


$$r \propto \frac{L^2}{GM}$$


so as $M$ decreases, $r$ increases (the original post had $r \propto M$, not $r \propto 1/M$.



On the other hand, assuming a circular orbit seems dubious.


As the other commenters said, this effect is minute. A significant effect on the orbit of the moon around the earth is tidal evolution, which does actually push the moon further away.


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