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Vector calc derivation |
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| May12-09, 03:27 PM | #1 |
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Vector calc derivation
Hi,
This is probably really simple but I can't figure out how they go from eq. 2 to eq. 3 here: http://img179.imageshack.us/my.php?image=vcalc.jpg First term I see, however the - mu / r I don't. Shouldn't this be + (1/2) mu / r? Asked my professor but he said "just do the derivations". :( |
| May12-09, 03:44 PM | #2 |
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Your professor was right. Lets take the second term of the second equation first.
[tex] \frac{1}{2} \frac{\mu}{r^3}\frac{d (\vec{r} \cdot \vec{r})}{dt}=\frac{1}{2} \frac{\mu}{r^3}\frac{d r^2}{dt}=\frac{1}{2}\frac{\mu}{r^3} 2 r \frac{dr}{dt}=\frac{\mu}{r^2}\frac{dr}{dt} [/tex] Now the second term of the third equation. [tex] \frac{d}{dt}\frac{-\mu}{r}=\frac{-1}{r^2}\frac{d-\mu r}{dt}=\frac{\mu}{r^2}\frac{dr}{dt} [/tex] |
| May12-09, 03:45 PM | #3 |
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[tex] \frac{1}{2} \frac{\mu}{r^3} \frac{d}{dt}r\cdot r = \frac{1}{2} \frac{\mu}{r^3} \frac{d}{dt}r^2 = \frac{1}{2} \frac{\mu}{r^3} 2r \frac{dr}{dt} = \frac{\mu}{r^2}\frac{dr}{dt} = -\mu \frac{dr^{-1}}{dt} = \frac{d}{dt} \frac{-\mu}{r} [/tex]
EDIT: Cyosis beat me to it! |
| May12-09, 03:56 PM | #4 |
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Vector calc derivation
Shame on me. Thank you.
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