DimaMazin Posted May 20, 2015 Posted May 20, 2015 Is the momentum=gamma *speed of center of their masses*(mass of the Earth+mass of the Moon) Or the momentum=gamma *speed of center of their masses *(mass of the Earth+mass of the Moon+ + kinetic energy of the Moon relatively of the Earth / c2)
Enthalpy Posted May 21, 2015 Posted May 21, 2015 I vote for the internal kinetic energy having a mass and a global kinetic energy and a global momentum. BUT the gravitational interaction energy, too - and this one has the opposite sign and twice the value of the kinetic energy. My reasons - not very educated - is that at composite particles like atom nuclei and protons, the attractive energy and internal kinetic energy are big (as compared with the constituents), not always well known, and though we can compute the momentum and global kinetic energy taking the global mass of the nucleus or proton, which does include the internal kinetic energy and the attraction. One "experiment" is under observation and modelling with pairs of stars, a neutron and a white dwarf, whose individual mass comprises different amounts of internal interactions and internal kinetic energy. An other experiment, still under planning, wants to orbit a piece of titanium concentric with one of tungsten, and observe if any drift appears. A more accurate version of this http://www.scienceforums.net/topic/74526-q-on-general-relativity/#entry740748 Though, I have still to think at http://www.scienceforums.net/topic/85377-relativistic-corrections-to-hydrogen-like-atoms/ which may tell a different story. Hope someone easier with these questions will bring a better justified answer.
pavelcherepan Posted May 28, 2015 Posted May 28, 2015 Is the momentum=gamma *speed of center of their masses*(mass of the Earth+mass of the Moon) Or the momentum=gamma *speed of center of their masses *(mass of the Earth+mass of the Moon+ + kinetic energy of the Moon relatively of the Earth / c2) I think you need to define the frame of reference first. Is it sun-centred, Earth-centred or uses some other point of reference? Because in different reference frames velocities of Earth and Moon will be different and the momentum will be different as a result.
DimaMazin Posted May 29, 2015 Author Posted May 29, 2015 I think you need to define the frame of reference first. Is it sun-centred, Earth-centred or uses some other point of reference? Because in different reference frames velocities of Earth and Moon will be different and the momentum will be different as a result. Different result doesn't change equation. The question about equation.
Strange Posted May 29, 2015 Posted May 29, 2015 (edited) The frame of reference can change the equation. Anyway, momentum is [math]\gamma m_0 v[/math]; where [math]m_0[/math] is the total (rest) mass of the system. Edited May 29, 2015 by Strange
pavelcherepan Posted May 29, 2015 Posted May 29, 2015 (edited) Different result doesn't change equation. The question about equation. Sorry, Dima. The way question was phrased I thought you were after a more practical solution rather than the general formula, which Strange has given in the post above. Still, if you use this formula in Earth-Moon barycenter FoR and use [latex]v[/latex] for the velocity of the barycenter the momentum will be zero. Hence in this FoR the total momentum of the system is better calculated as simply the momentum of the Moon in it's orbit around the barycenter. If you get to Sun FoR, you can use the formula and get a generalised result or you can get the total momentum of the system as a vector sum of momenta of the Earth and the Moon in their orbits around the Sun. This would give a slightly different result and it will be changing over time due to the shape of Moon orbit around the Sun. There will be some slight differences both to the magnitude and the direction of total momentum compared to the generalised formula for the barycenter. So, depending on what level of accuracy you want to achieve and the FoR you want to use it in, the formula may change. Edited May 29, 2015 by pavelcherepan
Robittybob1 Posted May 30, 2015 Posted May 30, 2015 (edited) .... Still, if you use this formula in Earth-Moon barycenter FoR and use [latex]v[/latex] for the velocity of the barycenter the momentum will be zero. Hence in this FoR the total momentum of the system is better calculated as simply the momentum of the Moon in it's orbit around the barycenter. .... There will be some slight differences both to the magnitude and the direction of total momentum compared to the generalised formula for the barycenter. So, depending on what level of accuracy you want to achieve and the FoR you want to use it in, the formula may change. Your diagrams don't show Earth and Moon orbit around the Sun and the E-M barycenter very accurately. The barycenter line would be the curve of the circle and Earth and Moon swapping sides across this line. Edited May 30, 2015 by Robittybob1
pavelcherepan Posted May 30, 2015 Posted May 30, 2015 Your diagrams don't show Earth and Moon orbit around the Sun and the E-M barycenter very accurately. The barycenter line would be the curve of the circle and Earth and Moon swapping sides across this line. Yes, but it's rather hard to show it on a diagram given than the Earth would oscillate ~4700 km from the barycenter and the Moon ~380,000 km.
Robittybob1 Posted May 30, 2015 Posted May 30, 2015 Yes, but it's rather hard to show it on a diagram given than the Earth would oscillate ~4700 km from the barycenter and the Moon ~380,000 km. Did you draw the diagrams? (There is no indication of source.) It would be not much more difficult just to have the barycenter as the point orbiting the Sun and the both the Moon and the Earth doing a wobbling curve around that, criss-crossing each other during a lunar month.
pavelcherepan Posted May 30, 2015 Posted May 30, 2015 Did you draw the diagrams? (There is no indication of source.) Sorry about that. The diagrams (only the lower one is to scale) are by a French astronomer Lucien Rudaux and taken from the below website: http://io9.com/is-the-moon-a-planet-1064356920
DimaMazin Posted May 30, 2015 Author Posted May 30, 2015 The frame of reference can change the equation. Anyway, momentum is [math]\gamma m_0 v[/math]; where [math]m_0[/math] is the total (rest) mass of the system. What is the formula we can use for definition of m0 of Earth+Moon ?
pavelcherepan Posted May 30, 2015 Posted May 30, 2015 What is the formula we can use for definition of m0 of Earth+Moon ? [latex]m_0[/latex] is just the sum of rest masses of Earth and Moon. Mass of the Earth is [latex]5.97*10^{24}kg[/latex] and mass of the Moon is [latex]0.073*10^{24}kg[/latex] so the [latex]m_0 = 6.043*10^{24}kg[/latex].
Mordred Posted May 30, 2015 Posted May 30, 2015 For further info Google Keplers laws. Unfortunately we cannot define one formula. http://en.m.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion Key note on center of mass.
DimaMazin Posted May 30, 2015 Author Posted May 30, 2015 [latex]m_0[/latex] is just the sum of rest masses of Earth and Moon. Mass of the Earth is [latex]5.97*10^{24}kg[/latex] and mass of the Moon is [latex]0.073*10^{24}kg[/latex] so the [latex]m_0 = 6.043*10^{24}kg[/latex]. For further info Google Keplers laws. Unfortunately we cannot define one formula. http://en.m.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion Key note on center of mass. Is Moon kinetic energy relatively of the Earth a part of total mass?
Strange Posted May 30, 2015 Posted May 30, 2015 (edited) Is Moon kinetic energy relatively of the Earth a part of total mass? Momentum is defined in terms of rest mass. So, no. Also, you are measuring things from your frame of reference, so if you want to consider kinetic energy, you would have to consider the velocity of both the Earth and the Moon relative to you. The velocity of the Moon relative to Earth is only relevant if you are measuring things from an Earth-centred frame of reference. In which case the momentum would be based on the (rest) mass of the Moon only. (Which is why the question in post #3 is relevant.) Edited May 30, 2015 by Strange
DimaMazin Posted May 30, 2015 Author Posted May 30, 2015 (edited) Momentum is defined in terms of rest mass. So, no. Also, you are measuring things from your frame of reference, so if you want to consider kinetic energy, you would have to consider the velocity of both the Earth and the Moon relative to you. The velocity of the Moon relative to Earth is only relevant if you are measuring things from an Earth-centred frame of reference. In which case the momentum would be based on the (rest) mass of the Moon only. (Which is why the question in post #3 is relevant.) I don't understand why we can't consider system of Earth&Moon like system of warmed gas in cylinder. There a sum of all molecules masses is less than total mass. Edited May 30, 2015 by DimaMazin
Strange Posted May 30, 2015 Posted May 30, 2015 I don't understand why we can't consider system of Earth&Moon like system of warmed gas in cylinder. There a sum of all molecules masses is less than total mass. Well, that's a different question, isn't it? If you calculate the "relativistic mass" of the Earth-Moon system then it will be greater than the rest mass of the two bodies. This is why the gamma appears in the equation for momentum when you take relativity into account: it accounts for the energy of the bodies as relativistic mass.
Robittybob1 Posted May 30, 2015 Posted May 30, 2015 I don't understand why we can't consider system of Earth&Moon like system of warmed gas in cylinder. There a sum of all molecules masses is less than total mass. Is that a generally known fact or is it a subject of recent research? Who did this experiment?
Janus Posted May 30, 2015 Posted May 30, 2015 I don't understand why we can't consider system of Earth&Moon like system of warmed gas in cylinder. There a sum of all molecules masses is less than total mass. You can, but with some additional complications. You just can't consider the kinetic energy of the Moon, you have to consider its orbital energy with respect to the Earth. ( the sun of its kinetic and gravitational potential energy) . you also have to consider the rotational kinetic energy of both the Moon and Earth. Overall this should increase the relativistic mass of the system by ~ 1/1,000,000,000 of the rest mass of both bodies. 1
Strange Posted May 30, 2015 Posted May 30, 2015 Is that a generally known fact or is it a subject of recent research? Who did this experiment? It is a consequence of the equivalence of mass and energy. If you increase the energy of a system, then you increase its mass. In fact, this is where nearly all the mass of atoms comes from: the binding energy of the nucleons. 1
Robittybob1 Posted May 30, 2015 Posted May 30, 2015 It is a consequence of the equivalence of mass and energy. If you increase the energy of a system, then you increase its mass. In fact, this is where nearly all the mass of atoms comes from: the binding energy of the nucleons. So that mass added from this energy adds to the gravitational and inertial mass too does it? If it does that is interesting.
DimaMazin Posted May 31, 2015 Author Posted May 31, 2015 Well, that's a different question, isn't it? If you calculate the "relativistic mass" of the Earth-Moon system then it will be greater than the rest mass of the two bodies. This is why the gamma appears in the equation for momentum when you take relativity into account: it accounts for the energy of the bodies as relativistic mass. Again you confuse us. gamma appears only for relativistic speed in calculation of momentum. Mass is Internal Kinetic Energy / c2. It is a consequence of the equivalence of mass and energy. If you increase the energy of a system, then you increase its mass. In fact, this is where nearly all the mass of atoms comes from: the binding energy of the nucleons. Now correct.
pavelcherepan Posted May 31, 2015 Posted May 31, 2015 Again you confuse us. It seems more like you're trying to confuse yourself, really... gamma appears only for relativistic speed in calculation of momentum. You can use gamma at any speed you want. It's just that at low speeds gamma would essentially be 1. Mass is Internal Kinetic Energy / c2 Only when an object is at rest. In most FoRs that you might choose either Moon, Earth or both would be in motion. So it doesn't apply to your question.
DimaMazin Posted May 31, 2015 Author Posted May 31, 2015 (edited) It seems more like you're trying to confuse yourself, really... You can use gamma at any speed you want. It's just that at low speeds gamma would essentially be 1. Only when an object is at rest. In most FoRs that you might choose either Moon, Earth or both would be in motion. So it doesn't apply to your question. I don't understand what you try to say. Here only answer of Janus is completely exact. He use relativistic mass, but his relativistic mass doesn't confirm relativistic mass of Strange in that post. Gamma for calculus of momentum of the system isn't the same gammas for calculus of relativistic masses inside the system. Edited May 31, 2015 by DimaMazin
pavelcherepan Posted May 31, 2015 Posted May 31, 2015 (edited) I don't understand what you try to say. Same here. Gamma for calculus of momentum of the system isn't the same gammas for calculus of relativistic masses inside the system. Gamma in all formulas relating to relativity is always the same: [latex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/latex] EDIT: Forgot square root in the formula Edited May 31, 2015 by pavelcherepan
Recommended Posts
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 accountSign in
Already have an account? Sign in here.
Sign In Now