- #1

haaj86

- 17

- 0

[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.