hi, im having trouble understanding what happens to the negative sign when i try to derive keplers law.(adsbygoogle = window.adsbygoogle || []).push({});

v=d/t

where d is the distance around one orbit, so d=2rpi

and t is the time for one orbit in seconds which i will call T

so v=2rpi/T

now using Ep=-Gm1m2/r

Ep is gravitational potential energy or work,

so Ep=force x distance

where we expand the terms Ep = (mass x acceleration) x distance

further expansion i get (mass x distance /seconds^2) x distance

simplifying to get Ep= v^2 x mass

now equating v^2 x mass = -Gm1m2/r

dividing mass, v^2 = -Gm/r

rooting, v=sqrt(Gm/r) ----- (ignoring the negative)

(cant understand why the negative should dissapear)

equating v=2rpi/T and v=sqrt(Gm/r) to get:

2rpi/T=sqrt(Gm/r)

squaring both sides, 4(r^2)(pi^2)/T^2 = Gm/r

now multiplying both sides by r and dividing by 4pi^2

i end up with r^3/T^2 = Gm/4pi^2

valid method? im not sure.

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Deriving keplers law

Loading...

Similar Threads - Deriving keplers | Date |
---|---|

I How to derive full solar sprectrum from gamma rays | Sep 9, 2016 |

Possible to derive an approximation of G from a Saros ? | Oct 10, 2015 |

Derive the Kepler equation | Sep 12, 2015 |

Deriving Oort Constants From Kepler's Third Law | Apr 2, 2013 |

**Physics Forums - The Fusion of Science and Community**