# Increasing mass' effect on the balance between centripetal force and centrifugal force

Okay, this is nothing more than a thought experiment which popped into my head while driving home from work today.

Take the case of a single body orbiting another, larger body, as in a planet and a star. The planet is in a stable, non-decaying orbit.

Now assume that the mass of the planet was noticeably increased instantaneously, that is, without any directional force applied to the planet's body, such as a meteor impact or other event would cause. So just take that the planet's mass increases without any force that would have an "equal and opposite" force that might alter the planet's velocity.

Given the starting stable orbit, and the instant increase in mass of the planet, I was trying to decide what would be the effect of the increased mass on the planet's orbit. I figured there were probably two possibilities...

• the increased mass would increase the pull of gravity between the planet and its star, thus causing the orbit to decay
• the increased mass would increase the planet's momentum, causing the centrifugal force to now exceed the centripetal force, and the planet would go flying off into space, breaking free of its orbit

I couldn't make up my mind though which might be more likely, so figured I'd ask people who would know a lot more about physics than I do.

I'm also completely comfortable with the fact that neither of my theories is correct, and the reaction of the system would be something else entirely.

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 It depends on how you increase,the mass. You can do it with or without modifying the velocity. Also, do you want large mass to be fixed or do you want them to orbit around the barycenter? – Manishearth♦ Mar 16 '12 at 3:41 Assume that the large mass is fixed, and the planet is orbiting strictly around the planet. The mass of the planet is increased in such a way that the process of its addition does not directly affect the planet's velocity. Also, the mass is added in uniformity, so the planet is still essentially a "perfect" sphere. – eidylon Mar 16 '12 at 3:48

As long as you add the mass in a way that does not affect its speed, then the orbit is not changed(your star must be fixed as well).

Lets say the planet(mass $M$) is orbiting at a radius $R$, about a star of mass $M_\star$. The orbital velocity is $$v_1=\sqrt{\frac{GM_\star}{R}}$$. Now, in the comments you stated that you added the mass in a way that does not affect its velocity directly. Simce momentum is conserved, the only way to do this is to give the added mass $m$ a velocity $v_1$ as well at the time it reaches the planet. As you can see, when the planet captures the mass, there is no change in angular momentum ($mv_1R+Mv_1R=(m+M)v_1R$). Now, since theres no change in angular momentum, it will orbit at the same angular velocity. If its the same angular velocity, the radius is the same as well. So it stays in stable orbit. One can get this directly from $v_1=\sqrt{\frac{GM_\star}{R}}$ as well.

What if the mass was at rest and it was captured? Well, then by conservation of linear momentum, the velocity would decrease to $v_2$. Since the velocity decreased, it will go into an elliptical orbit. If the velocity had increased, the orbit could be elliptical, but it can be hyperbolic (greater than escape velocity) and leave the system as well. This depends upon the mass ratio.

If the central mass was not fixed, then the masses orbit around the center of mass(barycenter), and the orbital angular velocity is given by$\omega=\sqrt{\frac{G\mu}{R^3}}$ (note that I'm using angular velocity in this case, as the star and planet will have different velocities). $R$ is the distance between the objects, and $\mu=\frac{MM_\star}{M+M_\star}$ is the reduced mass. One can see that a whole variety of things can happen, depending on how you add the small mass, and on the ratio between the three masses. You may want to analyse this yourself (seeing as it's not part of the question and it's a pretty interesting exercise)

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 For a body orbiting in a fixed potential, any change in velocity just kicks it into a different orbit. So if the Earth collides with a stationary blob, it will move into an elliptical orbit with a different eccentricity, but it won't spiral outwards. – David Zaslavsky♦ Mar 16 '12 at 14:14 @DavidZaslavsky Oh whoops... Didnt think about that.. Fixed.. Also added that it can leave via hyperbolic path if the velocity increased for some reason. Thanks! – Manishearth♦ Mar 16 '12 at 14:21 Thanks for the detailed answer. You covered a bunch of cases, and made the obvious assumption that the mass was significant to the overall equation (if it wasn't the whole question would be moot anyway! :) ). – eidylon Mar 21 '12 at 2:47

I will draw a picture to illustrate this later.

Motion in the central gravity field is described by the following equation: $$\frac{M\dot{r}^2}{2} - \frac{M\alpha}{r} + \frac{M\beta^2}{2r^2} = E = \text{const} \qquad (1)$$ where $$\alpha = M_\text{star}G$$ $$\beta = \frac{L_z}{M} = \frac{[\vec{r}\times\vec{v}]_z}{M} = \text{const}$$

When the mass of the planet $M$ is increased by $m$ without change of the velocity the equation (1) becomes $$\frac{(M+m)\dot{r}^2}{2} - \frac{(M+m)\alpha}{r} + \frac{(M+m)\beta^2}{2r^2} = E + \frac{mv_0^2}{2} - \frac{m\alpha}{r_0} = \text{const}$$ where $v_0$ and $r_0$ are the velocity and position of the planet at the moment when its mass is changed.

The effective potential energy $$U(r) = -\frac{M\alpha}{r} + \frac{M\beta^2}{r^2}$$ is scaled by the factor $(1+m/M)$ but is not changed qualitatively.

The total energy for finite motion is negative i.e. $E<0$.

New trajectory depends on the change of the energy $$\Delta E = \frac{mv_0^2}{2} - \frac{m\alpha}{r_0}$$

If $E + \Delta E \geq 0$ then the trajectory becomes infinite and the planet flies away. Otherwise it stays on the orbit.

One can find the change of the orbit comparing the factors $(1+m/M)$ and $(1+\Delta E / E)$.

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The orbit of a planet is independant of the mass, as long as the mass is small compared to the star it's orbiting around i.e. it doesn't significantly change the centre of mass of the star-planet system. So changing the mass by an amount small compared to the mass of the star will make little difference.

Planets obviously do move their star a bit, because that's the way the first extrasolar planets were discovered i.e. by spotting that their star moved as the planet orbited around it. However this effect is pretty small unless we're talking about a Jupiter sized planet orbiting very near the star.

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 After reading Manishearth's answer I've realised I've assumed that the mass you add is moving at the same velocity as the planet. I'm guessing this is what you mean. – John Rennie Mar 16 '12 at 12:18 Yeah, he had mentioned that in the comments.. I just wrote the other parts of the answer for fun. – Manishearth♦ Mar 16 '12 at 14:23

I assume the mass is added with the same velocity as the planet.

Consider first the case where the mass of the planet can be large. Let ${\bf r}$ be the relative position and $\mu = \frac{M m}{M+m}$ be the reduced mass. The second law takes the form $$\mu \ddot {\bf r} = {\bf F}_{\mathrm{int}} = -\frac{G M m}{r^3} {\bf r},$$ so $$\ddot {\bf r}= -\left(1+\frac{m}{M}\right) \frac{G M}{r^3} {\bf r}.$$

For the sake of argument, assume the orbit is initially circular. One can see that if $m$ increases, the bodies will tend to be closer together. (Increasing $m$ is in fact equivalent to increasing $G$.) The orbit will become elliptical.

Likewise, if $m$ decreases, the bodies will tend to be farther apart. In fact, the orbits will become more circular. The planet will never escape.

If the planet's initial and final mass are negligible, $$\ddot {\bf r}= -\frac{G M}{r^3} {\bf r}.$$ Thus, the motion of the planet is unaffected by the addition of mass. Of course, this is just the equivalence principle.

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