Astronomy help!

[13_CBS]

I'm doing a bit of homework for Astronomy class, and one of the questions has me utterly stumped.

The question:

Consider a planet with a mass of 10 MJupiter (10 x the mass of Jupiter) orbiting around a 3-MSun (3 x the mass of our Sun) star with a (orbital) period of 1.5 years. Calculate the distance between the center of the star and the center of mass.

I've looked all over the book, my notes, the internet, and I'm still having trouble. I can't find the formula that directly correlates the orbital period of a star, the mass of its planet, its orbital period, and the average distance of the center of the star from its center of mass.

I know that a more massive planet = a larger orbital radius for the star around the center of mass, but is there a formula or something that states the ratio of Planet-mass to Star-mass and relates that to how large the orbital radius of the star is?

Help! :smalleek:

[Egiam]

Is this an open-book assignment?

[13_CBS]

Yep. Think of it as a regular math homework assignment, except with astronomy. Open book and all that.

[Egiam]

I really am clueless. If this is a highschool assignment, I guess that this is a bad question.

Is the assignment to figure this out? If not, I would reccomend calling friends and learning if they had the same problem.

[Winterwind]

The centrifugal force experienced by that planet must be outweighed by the gravitational force between the planet and the star, for else it wouldn't be a stable orbit.

If you take the formulas for centrifugal force and gravitational force and equate them, you can solve them after the orbit's radius, giving you the distance between star and planet.

Calculating the position of the centre of mass should be easy from there. :smallwink:

[unstattedCommoner]

I'm doing a bit of homework for Astronomy class, and one of the questions has me utterly stumped.

The question:

Consider a planet with a mass of 10 MJupiter (10 x the mass of Jupiter) orbiting around a 3-MSun (3 x the mass of our Sun) star with a (orbital) period of 1.5 years. Calculate the distance between the center of the star and the center of mass.

I've looked all over the book, my notes, the internet, and I'm still having trouble. I can't find the formula that directly correlates the orbital period of a star, the mass of its planet, its orbital period, and the average distance of the center of the star from its center of mass.

I know that a more massive planet = a larger orbital radius for the star around the center of mass, but is there a formula or something that states the ratio of Planet-mass to Star-mass and relates that to how large the orbital radius of the star is?

Help! :smalleek:

I have an answer. I had to make a number of assumptions, however, which are essentially that the center of mass is fixed at the origin, and that the star and the planet follow circular orbits about the origin and are always on opposite sides of the origin.


Let n(t) be a vector with unit length which points from the star to the planet.

Suppose that the centre of mass is fixed at the origin (we can do this because the equations of motion tell us that the centre of mass moves with constant velocity, which we may assume to be zero).

Let the distance between the planet and the star be a. Let the radius of the star's orbit be L*a and the radius of the planet's orbit be (1-L)*a, for some L, 0 <= L <= 1. The star has mass M, and the planet has mass m. We will use plane polar coordinates, with the star at r_s = (L*a,theta_s) and the planet at r_p = ((1-L)*a,theta_p).

We can determine L from the condition that the centre of mass be at the origin:

M*L*a + m*(1-L)*a = 0

which gives L = m / (M + m).

The equations of motion are:

M*r_s'' = G*M*m/a^2 n
m*r_p'' = - G*M*m/a^2 n

(where ' denotes differentiation with respect to time) of which the radial components are

L*a*theta_s' = G*m/a^2
(1-L)*a*theta_p' = G*M/a^2

and the angular components reduce to

theta_s' = constant
theta_p' = constant

We need theta_s = theta_p + pi, so theta_s' = theta_p' = w.

Now we find, on substituting L, that

a^3 = G*(M+m)/w^2

Substituting w = 2*pi/T, where T is the period, gives

a^3 = G*(M+m)*T^2/(4*pi^2)

and L*a = m*a / (M+m) is easily obtained.

[13_CBS]

The centrifugal force experienced by that planet must be outweighed by the gravitational force between the planet and the star, for else it wouldn't be a stable orbit.

If you take the formulas for centrifugal force and gravitational force and equate them, you can solve them after the orbit's radius, giving you the distance between star and planet.


The formula is F = mass x (velocity)^2 / radius, right? Is the radius for the distance of the center of the planet from the center of mass or the center of the star? And wouldn't I need to figure out the velocity of the planet anyway? How would I go about doing that?

Edit: @Commoner: Thanks for the help, but...I'm having difficulty reading your notation, and the concepts you're using seem unfamiliar, which makes me suspect that your way wasn't the way we were expected to do it for this assignment...

Edit: A HA HA HA! I DID IT! VICTORY! I MIGHT NOT HAVE GOTTEN A RIGHT ANSWER, BUT AT LEAST MY ANSWER FINALLY MADE SOME GOSHDARN SENSE! A HA HA HA HA HA

[Winterwind]

The formula is F = mass x (velocity)^2 / radius, right? Is the radius for the distance of the center of the planet from the center of mass or the center of the star? And wouldn't I need to figure out the velocity of the planet anyway? How would I go about doing that?Velocity is easy - v=s/t, you have t given already (the time the planet needs for one orbit), and s is 2*pi*radius.

As for the radius, unless I have some grave error in my thinking (which is not unlikely, I'm prone to stuff like that and noticing only later :smalltongue:), it should be the distance between planet and star both times, as you can simply transform into the star's system of reference; in the centre-of-mass-system you would have to equate the gravitational force with the two opposed centrifugal forces experienced by planet and star respectively, which, I believe, would lead to the exact same result but be a more complicated calculation.

However, if you already have a solution, nevermind, congratulations, and good luck on it being right after all. :smallbiggrin:

[13_CBS]

Gotcha.

In any case, thanks for all the help folks!