- #1
physics baws
- 14
- 0
Hey,
I was trying to prove to myself an expression for the total energy, and I got stuck
Here is the picture (it's transparent, so I won't embed it here).
The problem I seem to be having is when I observe these two points where the speed is in its highest and its lowest. Since in these points is the force perpendicular to the velocity vector, I conclude that the acceleration in these points must only have its centripetal component, and therefore the centripetal force and the gravity force must be the same
[itex]\frac{mv_1 ^2}{r_1} = \frac{GMm}{r_1 ^2 },[/itex]
and similarly
[itex]\frac{mv_2 ^2}{r_2} = \frac{GMm}{r_2 ^2 }.[/itex]
The total energy must remain constant
[itex]E = K + U = \frac{mv_1 ^2}{2} - \frac{GMm}{r_1} = \frac{mv_2 ^2}{2} - \frac{GMm}{r_2},[/itex]
and using the results above, I get
[itex]E = - \frac{GMm}{2r_1 } = - \frac{GMm}{2r_2 },[/itex] which is obviously wrong.
I can't seem to see where my reasoning went wrong? Maybe someone could help me? Any help is appreciated!
Thanks.
I was trying to prove to myself an expression for the total energy, and I got stuck
Here is the picture (it's transparent, so I won't embed it here).
The problem I seem to be having is when I observe these two points where the speed is in its highest and its lowest. Since in these points is the force perpendicular to the velocity vector, I conclude that the acceleration in these points must only have its centripetal component, and therefore the centripetal force and the gravity force must be the same
[itex]\frac{mv_1 ^2}{r_1} = \frac{GMm}{r_1 ^2 },[/itex]
and similarly
[itex]\frac{mv_2 ^2}{r_2} = \frac{GMm}{r_2 ^2 }.[/itex]
The total energy must remain constant
[itex]E = K + U = \frac{mv_1 ^2}{2} - \frac{GMm}{r_1} = \frac{mv_2 ^2}{2} - \frac{GMm}{r_2},[/itex]
and using the results above, I get
[itex]E = - \frac{GMm}{2r_1 } = - \frac{GMm}{2r_2 },[/itex] which is obviously wrong.
I can't seem to see where my reasoning went wrong? Maybe someone could help me? Any help is appreciated!
Thanks.