Joshua Foley

2022-07-05

How did he get the fraction with fraction power?
So we have a simple equation that is from Kepler.
${\left(\frac{{\overline{r}}_{1}}{{\overline{r}}_{2}}\right)}^{3}={\left(\frac{{T}_{1}}{{T}_{2}}\right)}^{2}$
In an explanation of a physics book, you can resolve for ${r}_{2}$ like this:
${r}_{2}={r}_{1}{\left(\frac{{T}_{1}}{{T}_{2}}\right)}^{2/3}$
And I found
${r}_{1}=\sqrt[3]{\frac{{T}_{1}^{2}}{{T}_{2}^{2}}{r}_{2}^{3}}$
First question, is my approach correct? My second and main question is how did he get the ${r}_{2}$ equation that I stated first. The physics book doesn't explain how to get from the main equation to ${r}_{2}={r}_{1}{\left(\frac{{T}_{1}}{{T}_{2}}\right)}^{2/3}$. Can someone explain me, please? (By the way, of course the equation for ${r}_{1}$ and ${r}_{2}$ should be different).
Thank you!

Karla Hull

Your equation looks good. Check this out:
${r}_{1}=\sqrt[3]{\frac{{T}_{1}^{2}}{{T}_{2}^{2}}{r}_{2}^{3}}=\sqrt[3]{{r}_{2}^{3}}\sqrt[3]{{\left(\frac{{T}_{1}}{{T}_{2}}\right)}^{2}}={r}_{2}\sqrt[3]{{\left(\frac{{T}_{1}}{{T}_{2}}\right)}^{2}}$
Now all you need to know is that ${x}^{\frac{a}{b}}$ is defined as $\sqrt[b]{{x}^{a}}$
Does that help?

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