BTw i shouldn;t include the sun in the calculations shoudl i? becuase it doesn;t have an orbig it just woobles a bit.
Right?
Total energy of a system cycles?
Quote:Original post by rever
well thx, i think i got it now.
I get some small error when i do it for a single but the smaller the step size the less the error, si that is to be expected...and it is close to 0.05 % so i am good:D
Btw, should the sun be included in teh calculations? becuase it is the point around which the rest of them rotate.....
so i am geuessing no?
that sounds like a reasonable error. but yes you should take the sun into account. its influence will be small, since its close to the origin, but the pull of other planets will cause it to wobble a bit around the origin, so it will have some momentum.
Quote:Original post by Eelco
actually, the angular momentum should be constant about any arbitrary point. the only reason for picking the origin is because that simplifies the calculations.
Original post by rever
BTw i shouldn;t include the sun in the calculations shoudl i? becuase it doesn;t have an orbig it just woobles a bit.
Right?
This is why.
Most of the time, the center of mass will be very near, or even inside of the star in a system. The sun is wobbling because it has a small orbit of its own around the center of mass of the system. If you star is close enough to origin, and you want a simpler equation, then Eelco is right; ignoring center of mass will get you a 'good enough' answer. If you want to shave off more of that error, then account for center of mass as well.
And yes, angular momentum is a vector, you had it right.
Author
hmm i did angular momentun just of the sun it it changes by a whole lot, but the momentum of all other planet barely changes at all...
is it normal that momentum for the sun to change by a significnat amount when it is on it;s own in the calculations?
and it is negative by the way.
is it normal that momentum for the sun to change by a significnat amount when it is on it;s own in the calculations?
and it is negative by the way.
Quote:Original post by rever
k, somone here said to add the angular momentums?
That is incorrect is it not? cuz angular meomentum is a vector, what did he mean add x,y,z parts of it or something?
somthign like this?
*** Source Snippet Removed ***
this does the cross product for all the object in the system and then adds the components togeather.
Is that what was meant by adding the angular momentum?
Sorry for the pedantic remark but in your code you compute Cross( momentum, position ) rather than Cross( position, momentum ). It's not important for your current test since the only difference is a minus sign. Other than that, everything's cool.
EDIT: Just saw your last post. When talking about conservation of both linear and angular momentum, we talk about the total momentum of a closed system.
Only when you sum up all of the individual angular momentums, you should get something close to a constant vector.
Quote:Original post by erissianQuote:Original post by Eelco
actually, the angular momentum should be constant about any arbitrary point. the only reason for picking the origin is because that simplifies the calculations.
Original post by rever
BTw i shouldn;t include the sun in the calculations shoudl i? becuase it doesn;t have an orbig it just woobles a bit.
Right?
This is why.
Most of the time, the center of mass will be very near, or even inside of the star in a system. The sun is wobbling because it has a small orbit of its own around the center of mass of the system. If you star is close enough to origin, and you want a simpler equation, then Eelco is right; ignoring center of mass will get you a 'good enough' answer. If you want to shave off more of that error, then account for center of mass as well.
That is not really true. Eelco is correct that the angular momentum is conserved about any point assuming that the linear momentum is conserved too. Since in this system the linear momentum is conserved, the angular momentum can be calculated with respect to any point in the system. This is not an approximation.
Author
what about the momentum of the sun, it is really small value but it keeps chanign on me....
is that normal?
btw if i flip the sign then i get negative momentum....that is impossible....
is that normal?
btw if i flip the sign then i get negative momentum....that is impossible....
Well, take an example of a Jupiter-Sun system. We'll fudge the numbers a bit for simplicity. We'll say the Sun is at origin (r=0), and Jupiter is at r=8e11. Well call their masses 2e30 and 2e27.
The center of mass is calculated like this:
[(m1 * r1) + (m2 * r2) + (m3 * r3) + ...]/(m1 + m2 + m3 + ...)
So in this case we have (2e30 * 0 + 2e27 * 8e11) / (2e30 + 2e27)
Which is effectively 8e8 meters.
That means the orbital radius of the Sun is 8e8, which is barely outside its radius.
It moves slowly around this orbit at about 1 cm/year.
Because of its gigantic mass, its angular momentum (r*p) is ~6.5e29, which is huge.
Jupiter's angular momentum in this same system is only ~2e28 in comparison.
You can expect their total momentum to conserve, so if you ignore the Sun, then your system will appear to lose momemtum as the planets start the Sun in motion.
The center of mass is calculated like this:
[(m1 * r1) + (m2 * r2) + (m3 * r3) + ...]/(m1 + m2 + m3 + ...)
So in this case we have (2e30 * 0 + 2e27 * 8e11) / (2e30 + 2e27)
Which is effectively 8e8 meters.
That means the orbital radius of the Sun is 8e8, which is barely outside its radius.
It moves slowly around this orbit at about 1 cm/year.
Because of its gigantic mass, its angular momentum (r*p) is ~6.5e29, which is huge.
Jupiter's angular momentum in this same system is only ~2e28 in comparison.
You can expect their total momentum to conserve, so if you ignore the Sun, then your system will appear to lose momemtum as the planets start the Sun in motion.
Author
thx......
but can momentum be negative? i do the cross produc and i get a few negative numbers....because of the velocties...
but can momentum be negative? i do the cross produc and i get a few negative numbers....because of the velocties...
Quote:Original post by jjd
That is not really true. Eelco is correct that the angular momentum is conserved about any point assuming that the linear momentum is conserved too. Since in this system the linear momentum is conserved, the angular momentum can be calculated with respect to any point in the system. This is not an approximation.
If you calculate wrt a point that isn't the center of mass, you'll introduce an error term.
R' = COM - ERROR; // Center of mass minus the vector of error, in this case ERROR = COM
Ln = (Rn - R') x Pn
Ln = Rn x Pn - R' x Pn
Ln = Rn x Pn - (COM-ERROR) x Pn
Ln = Rn x Pn - COM x Pn + ERROR x Pn
L = ERROR x P
Which is not arbitrary. Touching on the earlier discussion, it will be constant but not correct.
This topic is closed to new replies.
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