# How does the Earth's orbit change as the Sun decreases in mass?

When the sun transitions into it's red giant phase it's mass is said to decrease (An article I read quoted it to go down to 67% of its mass however the number is not important). Since the orbital speed, assuming a circular orbit, is given by the equation $$v = \sqrt{\frac{GM}{r}}$$ Does this not mean that if the Mass decreases the circular orbital radius also decreases? This seems to make no intuitive sense since I would expect the radius to increase if the mass of the sun is lowered.

I have assumed the velocity of the earth in orbit will stay constant but this may be an incorrect assumption, if so can anyone explain why?

• Angular momentum remains constant, the Earth won't stay in a circular orbit. Mar 10, 2020 at 15:52
• The earth will go into an elliptic orbit Mar 11, 2020 at 14:24

Consider Kepler's Third Law: $$T^2=\Big(\frac{4 \pi^2}{GM}\Big)r^3$$ where $$T$$ is the orbital period and $$r$$ the orbit radius (for a circular orbit). So:

$$T^2 \propto r^3$$ where $$\frac{4 \pi^2}{GM}$$ is the proportionality constant.

Since as it depends on $$M$$, the ratio $$\frac{T^2}{r^3}$$ will vary accordingly.

Orbital mechanics are weird.

So, (hypothetically) we have the Earth travelling in a perfect circular orbit around the Sun. The radius of that orbit, the velocity of the earth, and the mass of the Sun are all fixed, and satisfy the relationship cited in the question.

Now the Sun loses a small amount of mass.

The Earth is still traveling at the same velocity, and requires the same centripetal force to maintain the same circular orbit. Unfortunately, the reduced mass of the Sun no longer supplies the required centripetal force.

So the Earth is now in a slightly elliptical orbit, passing perihelion. It continues along this elliptical path, climbing further out from the Sun towards aphelion, and slowing down as it climbs.

Of course, in reality, the Sun's mass loss is continuous. The Earth will repeat the preceding process over and over again. It will follow a slowly increasing spiral, moving farther and farther out and slowing down steadily.

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Let $$M(t)$$ be the solar mass and $$m$$ the Earth mass. Assuming circular orbit with radius $$r(t)$$, The gravitational force between them is $$F = \frac{\gamma mM(t)}{r^2} = ma = \frac{mv^2}{r} \Rightarrow \sqrt{\frac{\gamma M(t)}{r}} = \frac{dr}{dt},$$ where the orbital velocity is $$v = dr/dt$$. Further assume the mass of Sun is decreasing in an exponential rate, with a characteristic timescale $$\tau \sim 10^{10}$$ years (the total lifetime of Sun is about this): $$M(t) = Me^{-t/\tau},$$ where $$M$$ is the solar mass now. Plug this in, you get a separable differential equation. Separating: $$\sqrt{\gamma M}e^{-t/2\tau}dt = dr\sqrt{r}$$ Integrate and solve for $$r$$: $$r(t)^{3/2} = -3\tau\sqrt{\gamma M}e^{-t/2\tau} + C,$$ where $$C$$ is an integration constant, which can be fixed by requiring $$r(0) = r_0 = 1$$ AU (the current Earth-Sun distance). The final result for the orbit radius is $$r(t)^{3/2} = r_0^{3/2} + 3\tau\sqrt{\gamma M}\left( 1 - e^{-t/2\tau}\right),$$ from where one can easily see that the orbit radius increases as the Sun loses mass. Plugging now $$M(t)$$ and $$r(t)$$ to the orbital velocity one can see how it changes. The result is a little bit messy, but the orbital velocity $$v(t)$$ indeed decreases when the Sun loses mass and Earth is driven outward from the Sun.

• Thank you this is very helpful! Mar 10, 2020 at 20:56
• The velocity in mv^2/r is not dr/dt. For circular orbit dr/dt is zero.
– nasu
Mar 11, 2020 at 14:27