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luckis11
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H. Young (page 319) says g=GM/R^2. This means g(R)=GM/R^2. So, g(t)=?
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luckis11 said:g(t)=?
Couldn't have said it betterDelta² said:g(t) depends on the problem. Assuming a particle that its trajectory is described by position vector r(t), then [tex]g(t)=\frac{GM}{\abs{r(t)}^2}[/tex]
Delta² said:Yes well hm, he probably is interested in the case of free fall where the differential equation is [tex]\frac{d^2r}{dt^2}=\frac{GM}{r^2}[/tex]. Hmm my math abandons me here, what is the solution to that ODE? Boundary conditions [tex]r(0)=c, r^{'}(0)=0 [/tex]
luckis11 said:R(t) is not known either. There is a problem with which symbolisms are the less wrong...anyway.
g(t)=GM/(R(t))^2=GM/(A-Δx)^2<=>f''(t)=GM/(A-f(t))^2
where A is the initial R where there was immobility.
You can correspond the problem with a weird spring that pulls a mass and it is supposed that g(R)=GM/R^2. Here you may discover the possibility that the logic of H.Young is wrong, but let’s find the solution according to that supposition first.
When I put e.g. f’’(t)=2/((5-f(t))^2 at
http://www.wolframalpha.com/input/?i=f’’(t)=2/((5-f(t))^2
I get a complication which is indefinite even when replacing c1=c2=0
Look, this is a silly request. The problem is a differential equation, and the energies are constants of motion which make the equation easier to solve. If you don't allow the use of energies then I don't think that you can get a closed-form solution and the best you can do is a numerical differentiation. If that is all you want then just plug in the equaqtion Delta² gave into your favorite math package, otherwise energy is going to be involved.luckis11 said:now I say PLEASE PLEASE do not mix or mention the Energies
Not all differential equations which can be obtained by Newton's equations have analytical solutions. Of those which have solutions, not all will have solutions without using some simplification technique, such as using constants of motion like energy. Elementary calculus with no simplifications does not get you very far.luckis11 said:Actually, the silly is that Energies are needed. Otherwise is like you are saying Newton would need the Energies to support his universal attraction. And that he could not solve it with elementary calculus. Potential Energy was invented afterwards, and I think kinetic energy too. You can start another thread with Energies, otherwise I will stop reading your posts if you mention Energies again.
Found the threads with the math. Credit to Arildno for figuring how to solve one of the integrals:rcgldr said:There are posts with solutions for the time it took for two objects at rest to collide due to gravity, one for fixed sized objects, the other for point (zero size). I'll see if I can find those old threads.
luckis11 said:I can't find the solution at that thread "for point masses". For point masses where r is the distance between the two masses
I don't recall, the newer thread was a continuation of the older thread, and there were prior threads. Someone just wanted to see the math worked out starting with basic gravity equations for acceleration versus distance between objects.luckis11 said:The solution to problem should be ready as the one of the "Two body problem". At least it seems so as I see at the wikipedia article on it. Why didn't you get the equations of that at the first place?
luckis11 said:Well, the math worked out are supposed to be explained at the article http://en.wikipedia.org/wiki/Kepler_orbit.
But already the equation m1r''1=... does not say how it is derived.
The formula for acceleration of gravity is g(t) = 9.8 m/s², where g(t) represents the acceleration due to gravity and is dependent on time (t).
The acceleration of gravity is calculated by dividing the force of gravity acting on an object by its mass. This results in the value of 9.8 m/s² for objects near the surface of the Earth.
Yes, the acceleration of gravity can change depending on the location and mass of the object. For example, the acceleration of gravity on the moon is 1.62 m/s², which is significantly lower than on Earth.
The acceleration of gravity causes falling objects to increase in speed as they fall towards the Earth. This is due to the constant force of gravity acting on the object.
Yes, the acceleration of gravity can be negative if the object is moving in the opposite direction of the gravitational force. For example, when an object is thrown upwards, the acceleration of gravity will be negative as it moves away from the Earth's surface.