B Why is the momentum of a star equal to the momentum of a planet?

[Saloed]

I was wathcing a video about radial velocity method for seeking exoplanet(video) and on 3:05 author writes that momentum of a star equal momentum of a planet. Why?

 

[PeroK]

I was wathcing a video about radial velocity method for seeking exoplanet(video) and on 3:05 author writes that momentum of a star equal momentum of a planet. Why?

I assume he's using conservation of momentum of a two-body system. In the rest frame (*) of a star-planet system the total momentum is zero (by definition), hence the planet and star have equal and opposite momenta.

(*) Or, Centre of Mometum reference frame.

 

[Al_]

You mention "radial velocity" - so, did the video mean to say "radial momentum"? In a simple system of one star and one planet, would they have equal radial momentum around their shared centre of mass?

 

[etotheipi]

You mention "radial velocity" - so, did the video mean to say "radial momentum"? In a simple system of one star and one planet, would they have equal radial momentum around their shared centre of mass?

By radial momentum, do you mean the radial ($\hat{r}$) component of momentum? Then yes, the radial components of momentum must be equal in the barycentric frame

 

[stefan r]

It just sounds strange. The star is much bigger mass so a smaller change in velocity multiplies out to equal.

The gravitational force is equal too. Earth has much weaker gravity than the Sun. Earth's gravity is effecting all particles in the Sun whereas the Sun is just effecting the particles in Earth. This is also true for a comet or asteroid in an elliptical orbit. The force of gravity changes and it highest when the comet is at perihelion but at all points in the orbit the attraction of the comet to the Sun totals the same as the Sun to the comet.

 

[sysprog]

Momentum equals mass times velocity ($p=mv$) $-$ the startlingly large and immediately apprehensible difference between the mass of a star and that of a planet is equal in proportion to the velocity of the planet compared to the close-to-zero velocity of the star. If you drop a rock on Earth, it falls down until it collides with the Earth, but the Earth falls up toward the rock, too $-$ just nowhere near as much in the time from the drop to the collision, because its mass is so much larger.

 

[sophiecentaur]

If there were more than just one large planet in the system, the wobble of the star would be affected by both of the big planets according to their size. The effect of variation in red shift would presumably have more than one significant frequency component.

 

[stefan r]

If there were more than just one large planet in the system, the wobble of the star would be affected by both of the big planets according to their size. The effect of variation in red shift would presumably have more than one significant frequency component.

The momentum of the system would still add up to 0 when averaged over several orbits. Each planet is effecting the star with the same momentum change that the star is effecting the planets. At times the planets would be having an opposite effect on the star.

The measured velocity of the star includes the Sun's relative motion through space and the Earth's (telescope's) orbital and rotational motions. You have to subtract all of that out. For hot Jupiters the background will often be thousands of times higher and for rocky planet sized objects the velocity change cab be millions of times smaller than background.

 

[sophiecentaur]

The momentum of the system would still add up to 0 when averaged over several orbits.

Yes - of course - but the wobble (of the star) would produce two variations in red shift, with different periods and the two frequencies would give the clue that there are more than one large planet in the system.