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

In summary, the mass loss of a star can cause the orbital radii of its orbiting planets to increase, making it easier to identify them. This is due to a decrease in gravitational force, which allows the planet to move into a higher orbit. This change is also reflected in the planet's angular momentum, which remains constant. The decrease in mass is gradual and leads to a slight increase in the planet's orbit and a longer orbital period.
  • #1
kd001
43
0
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?
 
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  • #2
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.
 
  • #3
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.




.

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!
: )
 
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  • #4
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.
 
  • #5
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.
 
  • #6
If you look at the Lagrangian of a point particle in the gravitational field of a massive star:

[tex]
L = \frac{1}{2} \, m \, v^2 - \frac{G \, M \, m}{r}
[/tex]

and you scale [itex]M \rightarrow \mu \, M[/itex], [itex]\mathbf{r} \rightarrow \lambda \, \mathbf{r}[/itex], and [itex]t \rightarrow \tau \, t[/itex], then your Lagrangian changes as:
[tex]
L \rightarrow \left( \frac{\lambda}{\tau} \right)^2 \frac{1}{2} \, m \, v^2 - \frac{\mu}{\lambda} \, \frac{G \, M \, m}{r}
[/tex]

If you then impose:
[tex]
\left( \frac{\lambda}{\tau} \right)^2 = \frac{\mu}{\lambda} = 1 \Rightarrow \frac{\lambda^3}{\tau^2} = \mu
[/tex]
you see that the Lagrangian gets multiplied by a common factor. This means that if [itex]\mathbf{r}(t)[/itex] was a possible equation of a trajectory, then, so is [itex]\lambda \, \mathbf{r}(\tau t)[/itex] for a system with a gravitational center with a [itex]\mu[/itex] 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:
[tex]
\mathbf{M} = m (\mathbf{r} \times \mathbf{v}) \rightarrow \frac{\lambda^2}{\tau} \, \mathbf{M}
[/tex]
so, we have the further constraint:
[tex]
\frac{\lambda^2}{\tau} = 1
[/tex]

These two conditions determine [itex]\tau[/itex], and [itex]\lambda[/itex] in terms of [itex]\mu[/itex]:
[tex]
\tau = \lambda^2 \Rightarrow \frac{\lambda^3}{\lambda^4} = \mu \Rightarrow \lambda = \mu^{-1}, \ \tau = \mu^{-2}
[/tex]

This means that if the mass of the star changes by a fractional amount [itex]\epsilon \ll 1[/itex], then:
[tex]
\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
[/tex]
Then, notice that according to the binomial theorem:
[tex]
(1 + \epsilon)^{n} \sim 1 + n \, \epsilon, \ \epsilon \ll 1
[/tex]

Thus:
[tex]
\lambda = (1 + \epsilon)^{-1} \sim 1 - \epsilon
[/tex]
and
[tex]
\tau = (1 + \epsilon)^{-2} \sim 1 - 2 \, \epsilon
[/tex]

So, we quantify the scaling as, if the mass of the star has a fractional decrease [itex]\epsilon[/itex], then the linear dimensions of a planet's orbit have a fractional increase [itex]\epsilon[/itex], and the orbital period has a fractional increase [itex]2 \epsilon[/itex].
 
  • #7
Hmmm...what's that Lambda thing mean?
 
  • #8
Drakkith said:
Hmmm...what's that Lambda thing mean?

The scale of linear dimensions of geometrically similar trajectories.

[itex]\tau[/itex] is the scale of the periods of the orbit.

[itex]\mu[/itex] is the scale of the Star's mass.
 
  • #9
Ah ok.
 
  • #10
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.

Dickfore said:
If you look at the Lagrangian of a point particle in the gravitational field of a massive star:

[tex]
L = \frac{1}{2} \, m \, v^2 - \frac{G \, M \, m}{r}
[/tex]

and you scale [itex]M \rightarrow \mu \, M[/itex], [itex]\mathbf{r} \rightarrow \lambda \, \mathbf{r}[/itex], and [itex]t \rightarrow \tau \, t[/itex], then your Lagrangian changes as:
[tex]
L \rightarrow \left( \frac{\lambda}{\tau} \right)^2 \frac{1}{2} \, m \, v^2 - \frac{\mu}{\lambda} \, \frac{G \, M \, m}{r}
[/tex]

If you then impose:
[tex]
\left( \frac{\lambda}{\tau} \right)^2 = \frac{\mu}{\lambda} = 1 \Rightarrow \frac{\lambda^3}{\tau^2} = \mu
[/tex]
you see that the Lagrangian gets multiplied by a common factor. This means that if [itex]\mathbf{r}(t)[/itex] was a possible equation of a trajectory, then, so is [itex]\lambda \, \mathbf{r}(\tau t)[/itex] for a system with a gravitational center with a [itex]\mu[/itex] 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:
[tex]
\mathbf{M} = m (\mathbf{r} \times \mathbf{v}) \rightarrow \frac{\lambda^2}{\tau} \, \mathbf{M}
[/tex]
so, we have the further constraint:
[tex]
\frac{\lambda^2}{\tau} = 1
[/tex]

These two conditions determine [itex]\tau[/itex], and [itex]\lambda[/itex] in terms of [itex]\mu[/itex]:
[tex]
\tau = \lambda^2 \Rightarrow \frac{\lambda^3}{\lambda^4} = \mu \Rightarrow \lambda = \mu^{-1}, \ \tau = \mu^{-2}
[/tex]

This means that if the mass of the star changes by a fractional amount [itex]\epsilon \ll 1[/itex], then:
[tex]
\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
[/tex]
Then, notice that according to the binomial theorem:
[tex]
(1 + \epsilon)^{n} \sim 1 + n \, \epsilon, \ \epsilon \ll 1
[/tex]

Thus:
[tex]
\lambda = (1 + \epsilon)^{-1} \sim 1 - \epsilon
[/tex]
and
[tex]
\tau = (1 + \epsilon)^{-2} \sim 1 - 2 \, \epsilon
[/tex]

So, we quantify the scaling as, if the mass of the star has a fractional decrease [itex]\epsilon[/itex], then the linear dimensions of a planet's orbit have a fractional increase [itex]\epsilon[/itex], and the orbital period has a fractional increase [itex]2 \epsilon[/itex].
 
  • #11
qraal said:
What a beautiful proof!

This analysis is due to Landau L. D., Lifschitz E. M., Mechanics vol. 1

Look up Section 10 - Mechanical similarity
 
  • #12
Drakkith said:
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.
 
  • #13
qraal said:
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?
 
  • #14
Drakkith said:
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.
 

1. How does a star losing mass affect the orbit of its planets?

When a star loses mass, the gravitational force it exerts on its planets decreases. This causes the planets to move further away from the star in their orbits, resulting in an increase in their orbital radii.

2. Can a star losing mass cause a planet to be ejected from its orbit?

In some cases, a significant loss of mass from a star can lead to a planet being ejected from its orbit. This is because the decrease in the star's gravitational pull may no longer be strong enough to keep the planet in a stable orbit.

3. Is the change in orbital radii of planets due to a star losing mass permanent?

The change in orbital radii of planets due to a star losing mass is not necessarily permanent. As the star continues to lose mass, the gravitational pull on its planets will continue to decrease, potentially causing them to move even further away. However, if the star eventually stabilizes and stops losing mass, the planets may eventually return to their original orbits.

4. How does a decrease in a star's mass affect the habitability of its planets?

The habitability of a planet depends on many factors, including its distance from its star. When a star loses mass and its planets move further away, the habitability of those planets may change. If the planets were previously too close to the star to support life, the increase in their orbital radii may make them more hospitable environments.

5. Can the orbital radii of planets decrease if a star gains mass?

Yes, the orbital radii of planets can decrease if a star gains mass. This is because the increase in the star's gravitational pull will cause the planets to move closer to the star in their orbits. However, this change in orbital radii may not be permanent if the star eventually loses mass again.

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