# When a star loses its mass, do the orbital radii of the planets increase?

by kd001
Tags: increase, loses, mass, orbital, planets, radii, star
 P: 43 I remember reading an article on exoplanets which suggested that when a star loses its mass, the orbital radii of the planets orbiting it increase. Apparently this makes it easier to identify such planets as they are now further away from their star. What I would like to know is, what mechanism causes this increase in orbital radius? Newton's law (r=GM/v^2) suggests that if the mass of the star decreases, then the radius should also decrease unless the velocity also decreases but why would the velocity decrease?
 P: 3,390 Sorry, read your post wrong. As the mass of the star decreases, the gravitational force decreases which allows the planet to move into a 'higher' orbit. A planet is kept at a relatively stable orbital distance between the orbital velocity and the force of gravity. If gravity decreases but the orbital velocity doesn't, the planet will spiral outwards. Newton's law of gravity is F = GM/r^2.
P: 334
A sudden disappearance or reduction of gravity is an action that demands a reaction from all things affected by it as Newton pointed out in his laws of motion.

 Newton's Three Laws of Motion http://csep10.phys.utk.edu/astr161/l...wton3laws.html

.

We have two forces at play;

1. gravity tending to decrease the distance between planet and star,
2. centrifugal force tending to increase the distance between planet and star.

As long as these two forces balance each other out, a planet's orbit remains stable. But if either is increased or decreased, then the orbit will change. Decrease gravity and the orbiting planet's angular momentum or centrifugal force will move it away from its star or sun. Increase the gravity and the planet either falls into the star or assumes a tighter orbit. The same happens with any decrease or increase in centrifugal force due to higher velocity. Just like increasing the speed of a car as it makes a narrow turn with friction between tires and road analogeus to gravity and the car's identical centrifugal force albeit on a road. Or else decreasing the friction by oil slicking the pavement. Same effect but different force [friction as gravitational force] and different scenario.

It happens in binary star systems where one star goes supernova and the other one goes flying off because its angular momentum remains intact while gravity decreases. It also happens in galactic centers when a black hole might swallow a substantial part of one of the binary's mass. This causes the other star to either assume a wider orbit or goes flying off into interstellar or intergalactic space. Some consider it to be one possible mechanism that might be producing the hypervelocity stars which seem to be leaving our galaxy.

Funny how we go calmly about our daily lives despite how precariously we are perched!
: )

PF Patron
P: 10,400

## When a star loses its mass, do the orbital radii of the planets increase?

As mass is lost and gravity decreases slowly, does the planet move away slightly by converting its orbital velocity into increased distance? IE the planet moves outward a bit, which slows it down slightly into a stable orbit.
 PF Patron Sci Advisor P: 8,882 Yes, in fact the orbit of earth has increased by about one earth diameter due to mass loss of the sun over the last 4+ billion years. By the time the sun becomes a full fledged red giant, speculation is it will have moved out to around the current orbit of mars. The good news: it will be more comfortable out there than inside the sun's photosphere. The bad news: the sun will shrink down to a white dwarf a few hundred million years later. We'd better figure out interstellar travel by then or a cold day in hell will look toasty when our atmosphere starts snowing. I hope we will be able to take all our favorite animals with us. It would be a shame for all those recipes to go to waste.
 P: 3,015 If you look at the Lagrangian of a point particle in the gravitational field of a massive star: $$L = \frac{1}{2} \, m \, v^2 - \frac{G \, M \, m}{r}$$ and you scale $M \rightarrow \mu \, M$, $\mathbf{r} \rightarrow \lambda \, \mathbf{r}$, and $t \rightarrow \tau \, t$, then your Lagrangian changes as: $$L \rightarrow \left( \frac{\lambda}{\tau} \right)^2 \frac{1}{2} \, m \, v^2 - \frac{\mu}{\lambda} \, \frac{G \, M \, m}{r}$$ If you then impose: $$\left( \frac{\lambda}{\tau} \right)^2 = \frac{\mu}{\lambda} = 1 \Rightarrow \frac{\lambda^3}{\tau^2} = \mu$$ you see that the Lagrangian gets multiplied by a common factor. This means that if $\mathbf{r}(t)$ was a possible equation of a trajectory, then, so is $\lambda \, \mathbf{r}(\tau t)$ for a system with a gravitational center with a $\mu$ times bigger mass. If the system loses its mass "slowly", then the trajectory gets modified in the above scaling fashion. There is, however, another relation between the scaling factors. The total mechanical energy of the particle is not conserved, because the loop integral of the gravitational force over one period is non-zero, since it changes with time. Angular momentum, however, is conserved, because the force is always radially directed. Angular momentum scales as: $$\mathbf{M} = m (\mathbf{r} \times \mathbf{v}) \rightarrow \frac{\lambda^2}{\tau} \, \mathbf{M}$$ so, we have the further constraint: $$\frac{\lambda^2}{\tau} = 1$$ These two conditions determine $\tau$, and $\lambda$ in terms of $\mu$: $$\tau = \lambda^2 \Rightarrow \frac{\lambda^3}{\lambda^4} = \mu \Rightarrow \lambda = \mu^{-1}, \ \tau = \mu^{-2}$$ This means that if the mass of the star changes by a fractional amount $\epsilon \ll 1$, then: $$\epsilon \equiv \frac{\Delta M}{M} = \frac{ M_\mathrm{new} - M_\mathrm{old} }{ M_\mathrm{old} } = \frac{ M_\mathrm{new} }{ M_\mathrm{old} } - 1 = \mu - 1 \Rightarrow \mu = 1 + \epsilon$$ Then, notice that according to the binomial theorem: $$(1 + \epsilon)^{n} \sim 1 + n \, \epsilon, \ \epsilon \ll 1$$ Thus: $$\lambda = (1 + \epsilon)^{-1} \sim 1 - \epsilon$$ and $$\tau = (1 + \epsilon)^{-2} \sim 1 - 2 \, \epsilon$$ So, we quantify the scaling as, if the mass of the star has a fractional decrease $\epsilon$, then the linear dimensions of a planet's orbit have a fractional increase $\epsilon$, and the orbital period has a fractional increase $2 \epsilon$.
 PF Patron P: 10,400 Hmmm...what's that Lambda thing mean?
P: 3,015
 Quote by Drakkith Hmmm...what's that Lambda thing mean?
The scale of linear dimensions of geometrically similar trajectories.

$\tau$ is the scale of the periods of the orbit.

$\mu$ is the scale of the Star's mass.
 PF Patron P: 10,400 Ah ok.
P: 768
What a beautiful proof!

I was going to note that as the Earth's angular momentum is conserved and is L = Me.R.V, where Me is Earth's mass, while V = √(μ/R), thus L = Me.√(μ.R), where μ is G x the mass of Sun.

Rearrange to find R, keeping (L/Me)2 constant, as a function of μ.

R = (L/Me)2

Thus R is inversely proportional to μ.

If the Sun's mass halved instantaneously, then Earth and all the planets would be unbound, but the Sun won't lose mass suddenly. Instead the planets will very slowly spiral out as the mass declines, doing work against the Sun's gravity as they rise higher in their orbits, thus losing orbital kinetic energy while gaining potential energy.

Oddly the Earth is slowly spiralling away from the Sun faster than the Sun's mass loss via the Solar Wind implies. There's no agreed explanation presently as to why.

 Quote by Dickfore If you look at the Lagrangian of a point particle in the gravitational field of a massive star: $$L = \frac{1}{2} \, m \, v^2 - \frac{G \, M \, m}{r}$$ and you scale $M \rightarrow \mu \, M$, $\mathbf{r} \rightarrow \lambda \, \mathbf{r}$, and $t \rightarrow \tau \, t$, then your Lagrangian changes as: $$L \rightarrow \left( \frac{\lambda}{\tau} \right)^2 \frac{1}{2} \, m \, v^2 - \frac{\mu}{\lambda} \, \frac{G \, M \, m}{r}$$ If you then impose: $$\left( \frac{\lambda}{\tau} \right)^2 = \frac{\mu}{\lambda} = 1 \Rightarrow \frac{\lambda^3}{\tau^2} = \mu$$ you see that the Lagrangian gets multiplied by a common factor. This means that if $\mathbf{r}(t)$ was a possible equation of a trajectory, then, so is $\lambda \, \mathbf{r}(\tau t)$ for a system with a gravitational center with a $\mu$ times bigger mass. If the system loses its mass "slowly", then the trajectory gets modified in the above scaling fashion. There is, however, another relation between the scaling factors. The total mechanical energy of the particle is not conserved, because the loop integral of the gravitational force over one period is non-zero, since it changes with time. Angular momentum, however, is conserved, because the force is always radially directed. Angular momentum scales as: $$\mathbf{M} = m (\mathbf{r} \times \mathbf{v}) \rightarrow \frac{\lambda^2}{\tau} \, \mathbf{M}$$ so, we have the further constraint: $$\frac{\lambda^2}{\tau} = 1$$ These two conditions determine $\tau$, and $\lambda$ in terms of $\mu$: $$\tau = \lambda^2 \Rightarrow \frac{\lambda^3}{\lambda^4} = \mu \Rightarrow \lambda = \mu^{-1}, \ \tau = \mu^{-2}$$ This means that if the mass of the star changes by a fractional amount $\epsilon \ll 1$, then: $$\epsilon \equiv \frac{\Delta M}{M} = \frac{ M_\mathrm{new} - M_\mathrm{old} }{ M_\mathrm{old} } = \frac{ M_\mathrm{new} }{ M_\mathrm{old} } - 1 = \mu - 1 \Rightarrow \mu = 1 + \epsilon$$ Then, notice that according to the binomial theorem: $$(1 + \epsilon)^{n} \sim 1 + n \, \epsilon, \ \epsilon \ll 1$$ Thus: $$\lambda = (1 + \epsilon)^{-1} \sim 1 - \epsilon$$ and $$\tau = (1 + \epsilon)^{-2} \sim 1 - 2 \, \epsilon$$ So, we quantify the scaling as, if the mass of the star has a fractional decrease $\epsilon$, then the linear dimensions of a planet's orbit have a fractional increase $\epsilon$, and the orbital period has a fractional increase $2 \epsilon$.
P: 3,015
 Quote by qraal What a beautiful proof!
This analysis is due to Landau L. D., Lifschitz E. M., Mechanics vol. 1

Look up Section 10 - Mechanical similarity
P: 768
 Quote by Drakkith As mass is lost and gravity decreases slowly, does the planet move away slightly by converting its orbital velocity into increased distance? IE the planet moves outward a bit, which slows it down slightly into a stable orbit.
Yes. If the mass loss was sudden the planets would enter elliptical or hyperbolic orbits, depending on the amount lost. But because the loss is very slow, compared to orbit times, the planets lose kinetic energy as they rise in the Sun's gravitational field.

Interestingly the motion is akin to the slow spiral of a solar-sail - the acceleration due to sunlight is equivalent to a lower effective gravity experienced by the sail.
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P: 10,400
 Quote by qraal Oddly the Earth is slowly spiralling away from the Sun faster than the Sun's mass loss via the Solar Wind implies. There's no agreed explanation presently as to why.
Really? Got a link to some information on this?
P: 768
 Quote by Drakkith Really? Got a link to some information on this?
Do a search on the ArXiv for 'Lorenzo Iorio' as author. He gives the most complete references for it and lots of other solar system astrodynamic oddities.

The rate of recession is 0.15 metres per year which is possibly low enough to be due to some kind of tidal interaction with the Sun, though no known tidal coupling produces it presently. It's a mystery. The magnitude is 1/20 of what's need to save the Earth from falling into the Sun when it hits the Red Giant Tip. We need to be ~1.15 AU out by the time the Sun swells so large, but the present recession will only put Earth at 1.0075 AU in 7.5 Gyr.

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