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## Homework Statement

Since an electron on a circular orbit around a proton has a centripetal acceleration, it should radiate energy according to the Larmor relation

[tex] \frac{dE}{dt} = -2/3(q^2/4\pi\epsilon_o)(a^2/c^3) [/tex]

where [itex]q, a, \epsilon_o[/itex] and [itex]c[/itex] are respectively the electron charge, its acceleration, the vacuum permittivity and the velocity of light in a vacuum. Therefore, in classical, mechanics, it should spiral and crash on the nucleus. How long would this decay take, supposing that the size of the initial orbit is 10[itex]^{-10}[/itex]m and the nucleus is a point charge (radius=0)?

## Homework Equations

[tex] a = \frac{F_{coulombic}}{m} = \frac{v^2}{r_n} = \frac{1}{m} \frac{q^2}{4 \pi \epsilon_o r^2_n} [/tex]

## The Attempt at a Solution

I can easily do this if I calculate the centripital acceleration, [itex]a[/itex], out at 10[itex]^{-10}[/itex]m, and treat it as constant in the Larmor relation.

My question is, do you think this is what the problem wants me to do? Or do you think I have to set up an integral somewhere to vary the acceleration with the distance from the nucleus?

I tried getting a function [itex]a(r)[/itex] by integrating [itex]a[/itex] with respect to [itex]r[/itex] from [itex]r=0[/itex] to [itex]r=10^{-10}[/itex]m, but then I realized that the integral would be infinite, since [itex]r[/itex] is in the denominator. Which makes me think they want me to treat [itex]a[/itex] as a constant.

Any ideas or thoughts?

Thanks!