- #1
haaj86
- 17
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Hi, I am trying to solve the following problem (but note you can skip the details and look at the last equation and help me with the integral): a small planet (mass m) is found at rest relative to the Sun (mass M) with a separation distance [tex]x_{0}[/tex]. Consider now the small planet move in response to the gravitational force of the Sun (the sun is assumed so massive such that it doesn't move) what is the position function of the small planet as a function of time t? So,
[tex]m\frac{dv}{dt}=-\frac{GMm}{x^{2}}[/tex]
[tex]\int^{v}_{0}{mvdv}=-\int^{x}_{x_{0}}{\frac{GMm}{x^{2}}}[/tex]
[tex]\frac{1}{2}mv^{2}=GMm(\frac{1}{x}-\frac{1}{x_{0}})[/tex]
Therefore,
[tex]\frac{dx}{dt}= \pm[2GM(\frac{1}{x}-\frac{1}{x_{0}})]^{\frac{1}{2}}[/tex]
We take the minus sign because we know that the velocity is in the negative direction (the planet moves toward the sun), hence:
[tex]t=-\frac{1}{\sqrt{2GM}}\int^{x}_{x_{0}}{(\frac{1}{x'}-\frac{1}{x_{0}})^{-\frac{1}{2}}}dx'[/tex]
So far I haven't found this integral solved in any textbook, please if the calculation is long and you can't type it then tell me where to find it.
[tex]m\frac{dv}{dt}=-\frac{GMm}{x^{2}}[/tex]
[tex]\int^{v}_{0}{mvdv}=-\int^{x}_{x_{0}}{\frac{GMm}{x^{2}}}[/tex]
[tex]\frac{1}{2}mv^{2}=GMm(\frac{1}{x}-\frac{1}{x_{0}})[/tex]
Therefore,
[tex]\frac{dx}{dt}= \pm[2GM(\frac{1}{x}-\frac{1}{x_{0}})]^{\frac{1}{2}}[/tex]
We take the minus sign because we know that the velocity is in the negative direction (the planet moves toward the sun), hence:
[tex]t=-\frac{1}{\sqrt{2GM}}\int^{x}_{x_{0}}{(\frac{1}{x'}-\frac{1}{x_{0}})^{-\frac{1}{2}}}dx'[/tex]
So far I haven't found this integral solved in any textbook, please if the calculation is long and you can't type it then tell me where to find it.