tonyza2006 Posted September 3, 2009 Posted September 3, 2009 Hi there, I don't understand why the equal of Newton's law of gravitation F = Gm1m2/d(squared) is different for no spherical objects. In the image of my attachment there are one spherical object and a no spherical object. So why if there are not very distant the equal used is F = Gm1m2/d(d+L). How It's used (d).(d+L) on the contrary (d).(d), which is the original equal? Thanks Tony PS: Sorry about my english.
baxtonduglonn Posted September 4, 2009 Posted September 4, 2009 The distance d refers to the distance between the centres of gravity of the two objects, not from the centre of the spherical one to the edge of the other.
newphysics Posted September 15, 2009 Posted September 15, 2009 Read the enclosed file Speed of gravitation.doc
insane_alien Posted September 15, 2009 Posted September 15, 2009 stick to accepted science. the speed of propagation of gravity has been measured to be approximately c with good reason to suggest that it is exactly c. but that is besides the point, the question concerns newtonian theory and NOTHING else.
D H Posted September 15, 2009 Posted September 15, 2009 newphysics, You want us to open a Word document posted by someone with one whole post under behind their name? Do you think we're insane? Besides, this has nothing to do with the speed of gravity.
smoove Posted October 4, 2010 Posted October 4, 2010 Hello, I have a different question than tonyza2006, but concerning Newton's law of gravitation, I hope I was right to post it here then. I am trying to create a program simulating the solar system's behaviour, using Newton's first and second law: F = G m1 m2 / r squared F = m a I understand that these two equations are precise enough approximations to explain and calculate our actual observation of the solar system's behaviour. Note that I already managed to create a stable system containing the Sun, Mercure, Earth and Jupiter. Now let us assume that we have the following static situation: (SUN) -------------------- (MOON) ---- (EARTH) If I apply "a = F / m" with the correct values (with the help of Wikipedia), I obtain the following results (you are welcome to re-calculate them to check): Sun radius: 695 * 10^6 meters Earth radius: 6 * 10^6 meters Moon radius: 1.7 * 10^6 meters Distance Earth to sun: 150 * 10^9 meters Distance Moon to earth: 400 * 10^6 meters Sun mass: 2 * 10^30 kg Earth mass: 6 * 10^24 kg Moon mass: 7.34 * 10^22 kg Acceleration from the earth to the sun : 0.0058776287291389753714553965137712 Acceleration from the moon to the earth: 0.0024092077399680867365033264189728 Acceleration from the moon to the sun : 0.00596459693833060296825931239269 Meaning that the Moon is more attracted by the Sun than the Earth actually is (I am not even sure it is logical...); but more important, that the Moon is two times more attracted by the Sun than by the Earth. My program then basically makes the Moon rotate around the Sun without maintaining an orbit around the Earth. What am I missing to find the appropriate balance of that system? Remark: we first assumed that the system was static, but the problem is even worse if we imagine that the Moon actually managed to rotate once around the Earth. It will then have a none null speed and will eventually be directed straight to the Sun, making it even more difficult for the Earth to maintain its satelit's orbit. --------| | (SUN) --------------------- (EARTH) | (MOON) | <--------| Thank you for your listening, Sorry about the last scheme: it was supposed to show the Moon starting to rotate from one side of the Earth, then finally poping out on the other side.
matterdoc Posted December 4, 2010 Posted December 4, 2010 Smoove, Sun, earth and moon are free bodies, moving in space. No free body can orbit around another moving body. Simple mechanics should show this fact. We may observe orbital motion around a central body only if we consider the central body is static in space. That is, by using relative reference frame. In reality, sun, earth and moon are not static bodies. They move together in wavy paths about a median path around galactic center. A stable galaxy is static in space. These movements, when considered in any relative reference frame, appear to be orbits around the respective central body. When earth is considered static, both moon and sun appear to orbit around the earth. When sun is considered as static, both earth and moon appear to orbit around earth. Etc.
D H Posted December 4, 2010 Posted December 4, 2010 Sun, earth and moon are free bodies, moving in space. No free body can orbit around another moving body. Simple mechanics should show this fact. Utter nonsense. Meaning that the Moon is more attracted by the Sun than the Earth actually is (I am not even sure it is logical...); but more important, that the Moon is two times more attracted by the Sun than by the Earth. My program then basically makes the Moon rotate around the Sun without maintaining an orbit around the Earth. In a sense, the Moon does indeed orbit the Sun. One obvious sense is that the specific energy of the Moon in a Sun-centered frame, [math]v^2/2 - GM_{\mbox{sun}}/r[/math], is negative. Another sense is that suppose we could magically make the Earth vanish. The Moon's orbit about the Sun would remain more or less the same (more or less meaning within about 10%). On the other hand, that the Moon's acceleration toward the Earth is about 1/2 that of the Moon toward the Sun means that the nearby presence of the Earth is a bit more than a perturbative effect. Effects should be less than an order of magnitude smaller than the primary effect to be considered 'perturbative'. The presence of the Earth is considerably more than just a perturbation of the Moon's orbit about the Sun. So, does the Moon orbit the Earth? Certainly. First off, the specific energy of the Moon in an Earth-centered frame is once again negative. The Moon is gravitationally bound to the Earth. Just as before, if we could magically make the Sun vanish, the Moon's orbit about the Earth would remain more or less unchanged. From the perspective of an Earth-centered frame, the effect of the Sun's gravitational influence on the Moon's is the inertial frame acceleration of the Moon toward the Sun less the inertial frame acceleration of the Earth toward the Sun. This is greatest when the Earth is a perihelion, the Moon is at apogee, and the Moon is new. Even then, it is only about 1.4% of the Moon's gravitational acceleration toward the Earth. This is a small perturbative effect. Yet another way to look at it is in terms of spheres of influence. There are two competing definitions of a gravitational sphere of influence of a planet on some object, Laplace's sphere of influence and the Hill sphere. With either definition, the Moon's orbit about the Earth is well inside Earth's gravitational sphere of influence. The Moon orbits the Earth (and the Sun, and the Milky Way, and the local group, and ...).
zheng sheng ming Posted December 6, 2010 Posted December 6, 2010 I discovery the origin ofe gravitation
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