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- Thread starter PrakashPhy
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Simon Bridge

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For the sake of the thought experiment, we can imagine a universe with only two masses in it.Suppose a case where two masses are essentially far away form all other bodies so that we can assume the two masses are not affected by any other interactions going on beside them

In theory, not everything is possible, that is why we have theories. However, we can imagine that the objects have particular positions at time t=0 whenand they do not have the property of gravitation until a time( they are perfectly isolated{in theory everything is possible}).

If suddenly at t=0 the gravitational field starts how do we find the position of the two masses as a function of time due to gravitational interaction ( and nothing else)??

You mean apart from finding their mutual acceleration by Newton's gravitation?

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olivermsun

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You simply apply Newton's laws,F= mawhere the force is given by Newton's law of gravitational. Break vectors into components and integrate the acceleration..

You don't even need to decompose vectors, since there are only two masses in this universe and I assume gravity attracts them "toward" each other.

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Matterwave

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One can use conservation of energy to get the velocity at any point in the trajectory, but I forget if this greatly simplifies the solution for the full equation of motion.

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Simon Bridge

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multiply both sides by [itex]\dot{x}[/itex]

[tex]\dot{x}\ddot{x} = \frac{K\dot{x}}{x^2}[/tex]

Now you can integrate both sides, rearrange to get:

[tex]\dot{x} = \sqrt{ \frac{2K}{x} +C }[/tex]

which you can do. It's not pretty.

[tex]t = \frac{\dot{x}}{C} - \frac{K\ln|2K + 2C\dot{x}|}{C^{3/2}} + D [/tex]

More generally:

http://www.berkeleyscience.com/pm.htm

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The resulting differential equation d^2x/dt^2 = K/x^2 (where K is some constant) is not trivial to solve though since it's a non-linear ODE. Unless I'm forgetting something basic.

I meant setting up the problem is straight-forward. Solving the differential equation is not always doable analytically. When not, you an always solve differential equations numerically, either by using per-packaged software or writing your own code.

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Simon Bridge

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The nice quadratic ballistic equations we learn in school only work under special conditions ... linear gravity and no other forces.

Actual accurate satellite orbits are determined numerically and cannot be expressed analytically like this.

OTOH: the 2-body problem described was worked out quite a while ago.

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Can't the two body problem be reduced to a similar one body problem? Taking the Lagrangian of the system, assuming the origin to be the COM ([itex]r_1m_1 = -r_2m_2[/itex]) and defining [itex]r = r_1 - r_2[/itex]you have

[tex] L = \frac{1}{2}m_1\dot{r_1}^2 + \frac{1}{2}m_2\dot{r_2}^2 - U(r)[/tex]

We can express the position vectors as [itex] r_1 = \frac{m_2r}{m_1+m_2} [/itex] and [itex] r_2 = \frac{m_1r}{m_1+m_2} [/itex], then define [itex]m = \frac{m_1m_2}{m_1+m_2} [/itex] which then reduces the Lagrangian to

[tex] L = \frac{1}{2}m\dot{r}^2 - U(r) [/tex]

Where

[tex]U(r) = -\frac{m_1m_2 G}{r} [/tex]

I guess this would be more of a question than an answer though? :)

[tex] L = \frac{1}{2}m_1\dot{r_1}^2 + \frac{1}{2}m_2\dot{r_2}^2 - U(r)[/tex]

We can express the position vectors as [itex] r_1 = \frac{m_2r}{m_1+m_2} [/itex] and [itex] r_2 = \frac{m_1r}{m_1+m_2} [/itex], then define [itex]m = \frac{m_1m_2}{m_1+m_2} [/itex] which then reduces the Lagrangian to

[tex] L = \frac{1}{2}m\dot{r}^2 - U(r) [/tex]

Where

[tex]U(r) = -\frac{m_1m_2 G}{r} [/tex]

I guess this would be more of a question than an answer though? :)

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