Can Conservation of Angular Momentum Help Estimate the Mass of Binary Asteroids?

[Beer-monster]

Homework Statement

Two asteroids (approximately the same size and mass) are orbiting each other at a given radius a. The center of mass is orbiting the sun at a distance (semimajor axis) b with an orbital period of P.

I would like to estimate the mass/size of each of the asteroids.

By analysing the motion of the orbit of the two asteroids around their center of mass I can determine what their mass is based on the period of this orbit.

However, I am stuck on trying to relate the period of the asteroids mutual orbit with the period of the center of mass's orbit around the sun.

I thought that conservation of angular momentum would play a roll, however I do not know the total momentum of the sun-asteroid-asteroid system.

Can anyone point me toward relating these two periods?

Homework Equations

[tex] L = m\omega^{2}R [/tex]

[tex] \omega = \frac{2\pi}{T} [/tex]

[tex] F_{g} = \frac{GMm}{R^{2}} [/tex]

 

[mfb]

However, I am stuck on trying to relate the period of the asteroids mutual orbit with the period of the center of mass's orbit around the sun.

There is no fixed relation, and the different orbit types are basically unrelated. For the orbit around the sun, you can treat the binary asteroid (for all reasonable binary asteroid values) as a single point-mass.

 

[Beer-monster]

Thanks for the reply mfb.

I could be wrong, but, given that the total mass of the binary asteroid is much smaller than the mass of the sun, the period of it's orbit around the sun would not dependent on it's mass. So I'm not sure how I can use the period of that orbit to determine the mass of the asteroid.

As far as I have been able to figure out so far, I can only the mass of the asteroids if I know the period of their mutual orbit. Am I missing some other way here?

Thanks again

 

[voko]

Ultimately, this depends on how accurate the data are. The masses of all the bodies in the Solar system are known from observation of their elements and Kepler's third law, so this is not impossible.

 

[ehild]

The mass of the Sun determines the period of orbit of the binary asteroid system when their mass is negligible with respect to the mass of the Sun.
As for the asteroids, you can consider the binary system as a single object with mass 1/(1/m1+1/m2) orbiting at distance a about a central mass M=(m1+m2) in the CM of the binary system. From the period of the orbit of the asteroids and from their distance, you get M, the sum of masses. If the orbits of the asteroids are identical, their mass is equal, M/2. In case the masses differ, the radii are different (see picture), and you can get the individual masses from the ratio of the radii.

ehild

 

[Beer-monster]

Thanks for the replies.

ehild.

I had tried setting up the binary asteroids into a one body problem as you have described. The problem is I don't believe I have the period of the asteroids to solve for the mass. The only period, I believe, I've been gioven is that of the asteroids center of mass around the sun. I was hoping there might be a way to determine the period of the asteroids based on that information, but I have been drawing a blank.

 

[mfb]

The only period, I believe, I've been gioven is that of the asteroids center of mass around the sun. I was hoping there might be a way to determine the period of the asteroids based on that information, but I have been drawing a blank.

That is not possible.

Ultimately, this depends on how accurate the data are. The masses of all the bodies in the Solar system are known from observation of their elements and Kepler's third law, so this is not impossible.

Many asteroid masses are poorly known, or just estimated based on their size, as their mass is so small compared to the sun. Ceres has a mass 9 orders of magnitude below the mass of the sun, an asteroid with a radius of 10km has a mass 15 orders of magnitude lower. Perturbations from other objects are much more significant than the small influence of the asteroid mass on its period.

 

[voko]

Many asteroid masses are poorly known, or just estimated based on their size, as their mass is so small compared to the sun. Ceres has a mass 9 orders of magnitude below the mass of the sun, an asteroid with a radius of 10km has a mass 15 orders of magnitude lower. Perturbations from other objects are much more significant than the small influence of the asteroid mass on its period.

I do not think this negates my statement that observation is paramount. We are not given any other bodes except the Sun, just for the record.

 

[mfb]

We are not given any other bodes except the Sun, just for the record.

That just proves that we cannot determine the mass of the asteroids based on their orbital period.

 

[voko]

mfb, you sound like we know the masses of the bodies in Solar system by sheer luck. I find this ridiculous.

 

[gneill]

mfb, you sound like we know the masses of the bodies in Solar system by sheer luck. I find this ridiculous.

We know the masses of the bodies by various methods. Bodies with moons can be "massed" by observation of period and separation. Bodies whose composition can be estimated can be massed by size. Bodies which have been visited by spacecraft have been massed most accurately by observation of the spacecraft 's perturbed motion on passage (effectively an "artificial moon" scenario).

It might be of interest to note that for centuries, while the relative distances between objects in the solar system were known quite accurately (for example the ratios of major axes of the planetary orbits), the actual scale was unknown or known only very roughly. Planetary masses were similarly ambiguous until the last century, and those of some the outer planets were "off" by several percent until visited by the Voyager spacecraft !

 

[Beer-monster]

That is not possible.

The perhaps I am thinking about the question wrong. Is 40 years really too long for the period of a binary asteroids mutual orbit.

 

[mfb]

mfb, you sound like we know the masses of the bodies in Solar system by sheer luck. I find this ridiculous.

I think you overestimate our methods to measure masses. Planets and a few big other objects influence other objects in the solar system enough to measure their mass, that is not an issue.
For all other objects: if nothing orbits the object* or comes very close to it**, we do not have a mass measurement. Our mass estimates come from the size and the estimated composition.

*like moons, binary asteroids and so on
**apart from space probes, I think we can call this "sheer luck".

Is 40 years really too long for the period of a binary asteroids mutual orbit.

It is too long. Where does your data come from?

 

[Beer-monster]

The question I am attempting gives some limited data.

The distance of the binary center of mass from the sun and it's orbital period.

Also, that the asteroids orbit with each other is circular and the maximum distance between the asteroids as measured in arcs-seconds observed from the earth.

Given the separation I can work out the gravitational force each asteroid exerts on the other, and from this the angular velocity/period. However, without the values of that period I do not know how to get a value of the mass or the radius.

One thought, since I know the orbit of the center of mass I can determine the acceleration of the center of mass. If I know that, can I then relate the movement of the asteroids (treating them as one body orbiting the CoM) to the acceleration of the CoM using Newton's laws?

 

[mfb]

Given the separation I can work out the gravitational force each asteroid exerts on the other, and from this the angular velocity/period.

This is not possible without the mass.Please post the full and exact problem statement, otherwise we just keep guessing what exactly you have.

 

[Beer-monster]

Here's the full question:


A binary asteroid consisting of two components that are separated by 5.0 arcseconds on the sky at maximum separation. The binary’s center of mass orbits the Sun with semimajor axis a =60.0 AU. The orbital period is 40 years. Estimate the physical radii of each of the two components, stating any assumptions that you make.

 

[ehild]

The binary asteroid is far enough from the Sun to consider it a single body. If the semimajor axis of the CM is 60 AU the period can not be 40 years, according to Kepler's Third Law. So the 40 years is the period of the components. From these data you can estimate the distance between the asteroids, and the sum of the masses.

ehild

 

[mfb]

Hmm, my estimate was wrong. With that significant separation, 40 years as period for the binary asteroid is possible.

From these data you can estimate the distance between the asteroids, and the sum of the masses.

And from the masses, you can find some estimate for the radius.

 

[ehild]

The two-body approximation can be used for the two asteroids, if the the force of the Sun on them is much smaller than their mutual attraction. If you find out the total mass of the asteroids, neglecting the effect of Sun, you get that the force of Sun is not much smaller, so it influences the shape of the orbit and the time period.
It is interesting that the "tug of war" value, the ratio attraction between the asteroids over to that of the Sun is much greater than 1, (over 100) in case of the other planets and their moons, but its is about 0.5 between the Earth and Moon.

The Moon is unique among natural satellites in that it experiences a stronger gravitational attraction to the Sun than to its primary, the Earth. As a consequence, its path is always concave to the Sun. It can be argued that this makes the Moon a planet, orbiting the Sun, rather than a satellite of the Earth. Usually, it is considered to be in orbit around the Earth, but its orbit is substantially distorted from a simple elliptical shape by the gravity of the Sun

http://en.wikipedia.org/wiki/Moon#Orbit

ehild

 

[mfb]

The two-body approximation can be used for the two asteroids, if the the force of the Sun on them is much smaller than their mutual attraction.

That is a sufficient, but not a necessary condition.
The approximation is good if the binary asteroids are within their Hill sphere.

The Hill sphere of an asteroid with a diameter of 10km and roughly the density of earth, orbiting at 60 AU from the sun, has a radius of ~70000km. At 1/10 this distance, deep within the Hill sphere, the gravitational attraction on an object from the sun is roughly 500 times stronger than the gravitational attraction from the asteroid. And yet you can ignore the sun for the orbital dynamics of the asteroid system, as the asteroid feels nearly the same force.