Earth is spiraling away from Sun at rate of 1.5cm per year due to mass loss of Sun? How it was calculated? [closed]

Question

My physics teacher asked if we could calculate the rate at which Earth moves away from the Sun due to the mass loss of the Sun. It's very sensible for me to understand that Earth is spiraling away from Sun due to the nuclear fission reactions undergoing within the sun, which consumes $4.289 \times 10^{9}$ Kg every second to energy. I searched online and found that this rate is $1.5$ cm/y, but how did scientists know theoretically that Earth moves away at the rate of $1.5$ cm per year?

I tried to work out the calculations with the equation

$$m_E \frac{dv}{dt}=-G\frac{M_s(t)m_E}{r(t)^2}$$

But I found that I couldn’t work anymore using this equation. Can someone help me, please?

Answer 1

If we assume$^\dagger$ the earth always takes on a circular orbit of $r$ around the sun we get the following equation: $$\frac{mv^2}r=G\frac{Mm}{r^2}$$ We can solve this for the radius: $$r=\frac{GM}{v^2}.$$ If we now assume the velocity is constant throughout this process$^{\dagger\dagger}$ we get that the orbital radius changes at a rate of $$\dot r=\frac{G\dot M}{v^2}$$ If we now plug in the mean orbital radius for $v=29.78\text{ km/s}$ we get a rate of $$\dot r\approx1.018\text{cm/yr}$$ which is almost 50% off from the cited number but it has the right order of magnitude. So the idea behind this calculation is that we assume a circular orbit characterised by the suns mass and that losing mass just changes the radius of this orbit.

$\dagger$ This is not a good assumptions since, as you probably know, the orbit of a planet is more like an ellipse. I'm only trying to get an order of magnitude answer and taking this ellipticality into account would greatly complicate things (we would have to consider the semi-major axis and the semi-minor axis instead of the radius). For a complete/precise treatment you would probably have to solve the equation you mention and perform some analysis.

$\dagger\dagger$ This is again a simplification but if you assume a circular orbit and a slow decrease in mass this is a reasonable assumption.