# Extra-Solar Planets and their stars wobble

[SOLVED] Extra-Solar Planets and their stars wobble

## Homework Statement

An extra-solar planet is detected by observing that its parent star had a radial velocity 'wobble' of amplitude 40 m/s. If the parent star has a mass of one Solar Mass, and the period of the 'wobble' is four days, find a lower limit for the mass of the planet in solar masses.

## Homework Equations

$$F = \frac{GMm}{r^2}$$

Kepler $$T^2 \propto R^3$$

## The Attempt at a Solution

I have calculated the radius of the planet using Keplers third law, assuming that the wobble's period will be the same as the planets period, since thewobble is due to the star/planet system orbiting a common centre of mass, but I am not sure how to fine the Mass, I think I need Netwtons Gravitational law, as stated above, but I am not sure how to find the gravitational force, F, exerted?

Any ideas?

TFM

I bit on how I found the radius,

Using:

$$\frac{a^3}{p^2} = \frac{GM_{sun}}{4 \Pi^2}$$

$$a^3 = p^2 \frac{GM_{sun}}{4 \Pi^2}$$

Inserting the mass of Star and the Period (4*86400 seconds) Gave:

$$a^3 = (4*86400)^2 \frac{(6.67*10^{-11})(2*10^{30})}{4 \Pi^2}$$

This gave the Semi-Major axis as 9.311 * 10^9 m

I am not sure now quite what to do next. My first thopught is to do:

$$F = \frac{GMm}{r^2}$$

Since I know G, M r, and m is what I am trying to find out, but I do not know a vlaue for F.

Any ideas would be very greatly appreciated,

TFM

Is $$F = \frac{GMm}{r^2}$$ the right formula, or is there a better/more useful formula?

Any ideas would be very much appreciated,

TFM

Does anyone have any ideas, because it is due in soon and I am worried that it isn't going to be finsihed in time TFM

TFM,

You're right about the period, so i'll continue from there.

You should continue by using Newton's gravitation law (as you've stated) but make this equal to the centripetal force required to make the orbit circular;

$$\frac{mv^2}{r} = \frac{GMm}{r^2}$$

and solve to find the velocity of the planet.

Now just use a certain conservation law;

$$M_{star}V_{star} = m_{planet}v_{planet}$$

being careful to use the correct units. Et voila, a lower limit for your planet. The reason it's a lower limit is to do with the fact that the measured doppler shift may not be perfectly down your line of sight, so indeed $$V_{star}$$ could be greater. If you were given an inclination of the orbit with your line of sight, you could revise this by:

$$\frac{K}{sin(i)}=V_{star}$$ where K is the observed doppler velocity.

PS. For your reference, the method you are talking about is called doppler spectroscopy.

Last edited:
Thanks for all you help, Astrorob. I get a value of 6.68x10^26 kg

Thanks,

TFM

No problem, but remember that it asks for it in solar masses so you need to divide it by ~ $$2*10^{30}$$.

Thanks, I did actually do that - I probably should havbe put that answer down in here as well. I got a value of 0.000334 Solar Masses

TFM

OUt oif interest (and not part of the question) how realistic is this question?

The radius is 0.06 AU, which, comapred to Mercury's 0.38 AU, the planet seems to close to the sun - surely it would melt away?

TFM

TFM,

Due to its nature, unless you have a massive planet orbiting close to its parent star you will not be able to detect the slightest of wobbles it produces.

Doppler spectroscopy is a method which can only used to detect large planets close to their parent stars, commonly these are called "Hot Jupiters" (because they're large and not rocky). They cause the greatest disturbances and so are consequently more likely to be detected.

Last edited:
They must be very hot.

Thanks for the Info,

TFM

Thanks for the link, Most Interseting

TFM