Jump to content

Pluto-Charon combination. How two astronomical bodies orbit barycenter


Robittybob1

Recommended Posts

For two bodies to orbit a common barycenter the centrifugal forces would have to balance the centripetal forces caused by the gravitational attraction between the two masses.

 

With the Pluto-Charon pair we will need to know where the barycenter is (defined by the mass of each body and their separation) and their velocities (defined by their orbital period and orbital radius)

This is a representation of what we want to analyse: http://en.wikipedia.org/wiki/Orbit#mediaviewer/File:Orbit2.gif

 

Here is an actual photo of the two bodies orbiting each other: http://pluto.jhuapl.edu/News-Center/News-Article.php?page=20150212

Link to comment
Share on other sites

Their orbital periods are the same:

 

Their orbit s are said to be circular in shape so the circumference/ period = velocity

So the velocity is proportional to radius

and V^2 is proportional to area

 

Data from Wikipedia on Charon

Orbital period

6.3872304±0.0000011 d
(6 d, 9 h, 17 m, 36.7 ± 0.1 s)

 

Average orbital speed Charon
0.21 km/s[note 2 = Calculated on the basis of other parameters.]

 

Semi-major axis
17536±4 km to system barycenter, 19571±4 km to the center of Pluto (Barycenter closest to Pluto)

 

Mass (1.52±0.06)×1021 kg[3]
(11.6% of Pluto)

 

Data from Wikipedia on Pluto:

Mass
(1.305±0.007)×1022 kg[9]

 

The centripetal force = mV^2/r (the radius from the barycenter in this case) and that should for both objects of this binary be equal to the gravitational force between them.

Link to comment
Share on other sites

For any orbiting objects, the following formulas apply:

 

Circular orbital velocity:

 

[math] V_o= \sqrt{\frac{G m1^2}{d(m1+m2)}}[/math]

 

Where d is the distance between the centers of the objects and m2 is the mass of the object for which you want the orbital velocity.

 

In situations where one mass is much much smaller than the other, you can get an accurate enough answer by just using the mass of the larger object and assuming a value of 0 for m2.

 

Orbital period is

 

[math]T_o = 2 \pi \sqrt {\frac{a^3}{G(m1+m2)}}[/math]

 

Here a is average distance between the objects for the orbit.

 

Again, if the masses vary significantly you can get an reasonably accurate answer by ignoring the mass of the smaller of the two)

 

The barycenter will be located at a distance of

 

[math]\frac {dm1}{m1+m2}[/math]

 

from the center of m2 where d is the distance between the centers.

Edited by Janus
Link to comment
Share on other sites

Their orbital periods are the same:

 

Their orbit s are said to be circular in shape so the circumference/ period = velocity

So the velocity is proportional to radius

and V^2 is proportional to area

 

Data from Wikipedia on Charon

Orbital period

6.3872304±0.0000011 d

(6 d, 9 h, 17 m, 36.7 ± 0.1 s)

 

Average orbital speed Charon

0.21 km/s[note 2 = Calculated on the basis of other parameters.]

 

Semi-major axis

17536±4 km to system barycenter, 19571±4 km to the center of Pluto (Barycenter closest to Pluto)

 

Mass (1.52±0.06)×1021 kg[3]

(11.6% of Pluto)

 

Data from Wikipedia on Pluto:

Mass

(1.305±0.007)×1022 kg[9]

 

The centripetal force = mV^2/r (the radius from the barycenter in this case) and that should for both objects of this binary be equal to the gravitational force between them.

 

So far it looks like Charon has a centripetal force equal to the gravitational force between them but Pluto has less centripetal force than the gravitational force.

Could someone check my calculations please?

 

How could that be explained? Is Pluto still being accelerated or are the other Moons around Pluto slowing it down?

 

centripetal force for Pluto = 3.44E+18. centripetal force for Charon 3.46E+18. G force between them 3.46E+18.

 

I have yet to try Janus' formulas. I just compared the G force to the centripetal forces based on the values in Wikipedia (as above).

Trying Janus' formulas

1. velocity =sqrt((G*m^2)/(d*(m+M))) for Pluto that gave an incorrect answer of 1.74766E-09

Are the units in kg and meters and seconds?

 

(I see at least 1 error - I forgot to square the mass!)

Now I'm getting a better answer =199.6466938 m/sec but still not the same as finding the circumference and dividing by the time.

 

Previous answer using my method

velocity for Charon 199.6567905

velocity for Pluto 23.16956938

 

Using Janus' equation I thought I was calculating Pluto's velocity but the answer came out close (but not the same) to what I had previously got for Charon.

 

By Janus' method:

199.6466938 = Velocity for Charon

23.25386779 = Velocity for Pluto

 

So why were the answers different?

Edited by Robittybob1
Link to comment
Share on other sites

Using Janus' method to find the orbital period also gave surprising results.

551672.99, from using Janus method and 551856.7 from the actual listed time period in Wikipedia (I assume that is a measured time period), and the difference in the two periods = 183.71 seconds.

 

So Charon is orbiting Pluto a full 3 minutes slower than the physics equations say it should!

 

Anyone got an explanation for that?

Edited by Robittybob1
Link to comment
Share on other sites

All formulas are to good approximation, there are so many factors such as rounding errors, degree of accuracy on measurement etc etc.

 

I wouldn't worry about 3 minutes, you don't have the same precise dataset the wiki source does for that value.

Link to comment
Share on other sites

All formulas are to good approximation, there are so many factors such as rounding errors, degree of accuracy on measurement etc etc.

 

I wouldn't worry about 3 minutes, you don't have the same precise dataset the wiki source does for that value.

I used the degree of accuracy for distance +/- 4 km and added on another 4 km and it was still slow but not by much.

I think this is tending toward the proof I was looking for, trying to show that the larger mass, even the Sun, is not orbiting like a true binary .

Here we have Charon racing around Pluto with the centrifugal forces matching the gravitational force but Pluto under rating so if anything it is Pluto that is going slow not Charon but the orbital period for each is the same. I'm not familiar with the mechanics of that yet, and it could come about with Charon being captured by Pluto, so Pluto to start with was not orbiting but this binary pattern (as we see it today) developed over time. Could that be possible?

If that was the case Charon would have been further out and came in closer as momentum was given to Pluto, as Pluto gained (is gaining) more and more orbital motion.

Just like with the Earth and the Moon there are multiple solutions to their orbital period and the distance apart. The Moon was once real close to the Earth and it is being tidally accelerated further away. There is not just one distance these bodies can be apart, it is more like an infinite number of solutions.

 

If you got the time

"A True Story About Planet Pluto: | Passport to Pluto and Beyond" Edited by Robittybob1
Link to comment
Share on other sites

I used the degree of accuracy for distance +/- 4 km and added on another 4 km and it was still slow but not by much.

You have to take the degree of accuracy for distance and mass into account.

 

If you use the +/- 4km accuracy for the distance between them and the upper and lower limits of accuracy for the masses of Pluto and Charon you get maximum period of 554320.618045429 sec and a minimum period of 549059.832450658 sec. Wiki's given period of 551856.70656

sec falls right in this range. So it works out within the given degree of accuracy.

Link to comment
Share on other sites

You have to take the degree of accuracy for distance and mass into account.

 

If you use the +/- 4km accuracy for the distance between them and the upper and lower limits of accuracy for the masses of Pluto and Charon you get maximum period of 554320.618045429 sec and a minimum period of 549059.832450658 sec. Wiki's given period of 551856.70656

sec falls right in this range. So it works out within the given degree of accuracy.

The problem with that approach is that it appears the orbital period is known to a very high degree of accuracy

 

 

Orbital period

6.3872304±0.0000011 d

(6 d, 9 h, 17 m, 36.7 ± 0.1 s)

Whereas if you were to look at the difference between those two periods we get 5261 seconds or 87.68 minutes or roughly an hour and a half yet the orbit only varies by 0.1 of a second.

Link to comment
Share on other sites

According to the Wiki data the orbital speed of Charon was 210 m/sec (0.21 km/sec) but using the distance to the barycenter as a radius (2*pi*r/period. I thought that should calculate orbital velocity: Distance gone/time taken = velocity) but using that it came out to 199 m/sec.

But the 199 m/sec gave a perfect balance with the centrifugal forces against the gravitational attraction formula.

I wonder how they calculated the higher speed with those other factors?

Can the barycenter not be the point of rotation (as I proposed in an earlier thread?)

How much larger would the RADIUS have to be to get 210 m/sec in the orbital period? 18444451.6 meters or an increase in the RADIUS of 908451.5 meters (908 km) but something orbiting a larger radius actually orbital velocity will slow, problem only gets worse.

 

My original idea was to orbit Charon around the barycenter but to move the barycenter on a circle with a radius of 908 km, so that it maintains the correct Gravitation forces, the right centrifugal forces and the calculated velocity of 220 m/sec.

 

I have not seen calculations to do such a thing as orbiting a wobbling point.

Edited by Robittybob1
Link to comment
Share on other sites

The problem with that approach is that it appears the orbital period is known to a very high degree of accuracy

Whereas if you were to look at the difference between those two periods we get 5261 seconds or 87.68 minutes or roughly an hour and a half yet the orbit only varies by 0.1 of a second.

You are missing the point. While we can directly measure the orbital period to within a certain accuracy, we also can only measure the individual components( the masses of the bodies and their distance) that determine that orbital period with certain accuracies. Different combinations of those components can all give the same answer for the period within the measured accuracy. But the trick is choosing a value for any two of them(within the known accuracy, determines the value for the third.

 

So for example, if we choose the listed median values for the mass of Pluto and Charon (1.305e22 kg and 1.52e21 kg), then to get the measured period (551856.70656 sec), the distance between them must be 19575 km.( which is within 4 km of the listed median value of 19571 km) This places the barycenter 17533 km from Charon and 2042 km from Pluto. (As an aside, I'm going to point out that the median values listed for the masses, distance. If you The gravitational attraction between the two is 3.455e18 N.

 

Charon's speed around the barycenter is 199.625 m/sec and Pluto's is 23.2513 m/sec. This gives a required centripetal force of 3.455e18 for Charon equaling the gravitational force. The required centripetal force for Pluto also works out to 3.455e18 N. So centripetal and gravitational forces are equal for both Charon and Pluto.

 

Another thing to keep in mind that any answer you get with the given values are only going to be accurate to the number of significant digits of the least accurate value. So for example, Charon's mass is only given to three significant digit accuracy, 0.06 which in actuality could be anywhere from 0.055 to 0.065. If they had given the value as 0.060 then you could assume 4 digit accuracy, 0.0600 would have been 5 digit accuracy, etc.(You cannot assume the existence of those trailing zeros)

 

The point is that any orbital velocity or period calculated using a mass that is accurate to 3 significant digits will only be accurate to the same three significant digits. In other words, that 3 min difference between calculated and measured orbital periods is well within the minimum accuracy you can get with the given parameters and are not worth fussing over. The measured accuracy of the period might be accurate to 8 significant digits, But you cannot expect the calculated period to be accurate to that degree when it relies on values which are much less accurate.

Link to comment
Share on other sites

.... The gravitational attraction between the two is 3.455e18 N.

 

Charon's speed around the barycenter is 199.625 m/sec and Pluto's is 23.2513 m/sec. This gives a required centripetal force of 3.455e18 for Charon equaling the gravitational force. The required centripetal force for Pluto also works out to 3.455e18 N. So centripetal and gravitational forces are equal for both Charon and Pluto.

 

...

Thanks for that. I must have done something wrong earlier as I calculated that Pluto didn't have the correct amount of centripetal force. I'll have double check this before going on.

 

Are these the distances you are using? These are given in Wikipedia but P-B was calculated from the other two, but seems to be where the discrepancy comes in.

full C-B P-B

19571 km

Full 19571000 meters (C-P)

Charon-barycenter 17536000 meters (C-B)

Pluto - barycenter 2035000 meters (P-B)

 

What is the P-B figure you are using? Where did you get it from?

 

You and I have different ratios of the velocities and velocity was linked to the distances to the barycenter. My ratio: 8.617199017 Yours: 8.585541454 so we must be using different figures for the radii.

 

NASA site

 

 

Charon (P1)

Mean distance from Pluto (km) 19,600

 

So what does that mean? So roughly full distance in thousand kilometers = 19.6

C-B = 17.6 and P-B = 2.0

Edited by Robittybob1
Link to comment
Share on other sites

Thanks for that. I must have done something wrong earlier as I calculated that Pluto didn't have the correct amount of centripetal force. I'll have double check this before going on.

 

Are these the distances you are using? These are given in Wikipedia but P-B was calculated from the other two, but seems to be where the discrepancy comes in.

full C-B P-B

19571 km

Full 19571000 meters (C-P)

Charon-barycenter 17536000 meters (C-B)

Pluto - barycenter 2035000 meters (P-B)

 

What is the P-B figure you are using? Where did you get it from?

You have to remember that the values given are only within certain measurement tolerances.

 

Thus if you assume a Pluto mass of exactly 1.305e22 kg and a Charon mass of exactly 1.52e21 kg and a Pluto-Charon distance of exactly 19571000 m then the Charon-Barycenter distance works out to 17529276 m. not only is this not equal to 17536000 m but it is outside the measured tolerance for that distance. What this means is that these three exact values don't work together. But since these values are just possible choices from a range of possible values, this isn't really a problem.

 

Thus if you keep the distance and Pluto's mass the same and you adjust The mass of Charon to 1.515e21 kg (well within the accepted values for Charon's mass), then you get 17535294 m for the Charon-Barycenter distance, which is well within the excepted tolerance. Which gives you a Pluto-barycenter distance of 2035706 meters.

 

The point is that we are dealing with a range of values and not exact ones. The values given just happen to be those that fall in the middle for the range of possible values for each parameter. Some of them are arrived at through direct measurement and others are worked out indirectly. There is no requirement that taking the median values of three parameters and plugging them into an equation will result in an answer that is equal to the median value of a forth parameter.

Link to comment
Share on other sites

...

The point is that we are dealing with a range of values and not exact ones. The values given just happen to be those that fall in the middle for the range of possible values for each parameter. Some of them are arrived at through direct measurement and others are worked out indirectly. There is no requirement that taking the median values of three parameters and plugging them into an equation will result in an answer that is equal to the median value of a forth parameter.

Since Pluto has 4 moons if you account for all the distance and motion just with Charon you aren't leaving yourself any centrifugal/

centripetal forces to deal with the other Moons. So I am happy that they don't quite fit as that is what I'd expect.

The other moons may tend to fix Pluto in its central location giving it a lower value of centrifugal force compared to Charon, so Pluto orbits closer to the barycenter without having to be more massive.

Edited by Robittybob1
Link to comment
Share on other sites

It's good to see some real science and descoverary being quoted here.. the numbers are unraviling as we speak..
Pluto is a long way from us and until very recantly little was known of the other bodies and respective masses.. Orbital periods and the precision of information required is not at our fingertips as readily as you seem to think. Hours of study and months of observation.. Not one of the telescopes involved is a back yard observer.. Gaining information with such precision is diffacult and time consuming.. Understanding what is seen just as.. I will wait a little longer, for the most recant data which will expand our knowledge greatly... Waiting..

Link to comment
Share on other sites

It's good to see some real science and descoverary being quoted here.. the numbers are unraviling as we speak..

Pluto is a long way from us and until very recantly little was known of the other bodies and respective masses.. Orbital periods and the precision of information required is not at our fingertips as readily as you seem to think. Hours of study and months of observation.. Not one of the telescopes involved is a back yard observer.. Gaining information with such precision is diffacult and time consuming.. Understanding what is seen just as.. I will wait a little longer, for the most recant data which will expand our knowledge greatly... Waiting..

Thanks Mark, and thanks to Janus to for his knowledge of formulas. I just can't wait to see what New Horizons will reveal about Pluto and its moons.

I found it weird to see the moon and the "planet" orbiting each other tidally locked it really amazed me and you would wonder what sets this off.

If my theory about what causes a magnetic field is correct I predict that Pluto will have no strong magnetic field if any. There may be some magnetism locked into the rock core but that will be all.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.