Question Number 211992 by ajfour last updated on 26/Sep/24
Commented by ajfour last updated on 26/Sep/24
$${Find}\:{r}\:{in}\:{terms}\:{of}\:{R}\:{and}\:{a}. \\ $$
Commented by ajfour last updated on 26/Sep/24
https://youtu.be/xI8mXZpX-qM?si=hInwvqv42gbBwtVx
Answered by mr W last updated on 27/Sep/24
![cos 2α=(a/R)=1−2 sin^2 α sin^2 α=((1−(a/R))/2), cos^2 α=((1+(a/R))/2) tan α=(r/(R+(√((R−r)^2 −r^2 )))) [(r/(R+(√(R(R−2r)))))]^2 =((1−(a/R))/(1+(a/R))) say k=(r/R), λ=(a/R) ⇒[(k/(1+(√(1−2k))))]^2 =((1−λ)/(1+λ))](https://www.tinkutara.com/question/Q212009.png)
$$\mathrm{cos}\:\mathrm{2}\alpha=\frac{{a}}{{R}}=\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\alpha \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\alpha=\frac{\mathrm{1}−\frac{{a}}{{R}}}{\mathrm{2}},\:\mathrm{cos}^{\mathrm{2}} \:\alpha=\frac{\mathrm{1}+\frac{{a}}{{R}}}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\alpha=\frac{{r}}{{R}+\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }} \\ $$$$\left[\frac{{r}}{{R}+\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}}\right]^{\mathrm{2}} =\frac{\mathrm{1}−\frac{{a}}{{R}}}{\mathrm{1}+\frac{{a}}{{R}}} \\ $$$${say}\:{k}=\frac{{r}}{{R}},\:\lambda=\frac{{a}}{{R}} \\ $$$$\Rightarrow\left[\frac{{k}}{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}{k}}}\right]^{\mathrm{2}} =\frac{\mathrm{1}−\lambda}{\mathrm{1}+\lambda} \\ $$
Commented by mr W last updated on 27/Sep/24
$${yes},\:{thanks}. \\ $$
Commented by ajfour last updated on 27/Sep/24
![⇒[(k/(1+(√(1−2k))))]^2 =((1−λ)/(1+λ))=m^2 =tan^2 α k=m+m(√(1−2k)) k^2 +m^2 −2km=m^2 −2m^2 k ⇒ (k/2)=(r/(2R))=m−m^2 r=2R{(√((1−λ)/(1+λ)))−(((1−λ)/(1+λ)))}](https://www.tinkutara.com/question/Q212021.png)
$$ \\ $$$$\Rightarrow\left[\frac{{k}}{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}{k}}}\right]^{\mathrm{2}} =\frac{\mathrm{1}−\lambda}{\mathrm{1}+\lambda}={m}^{\mathrm{2}} =\mathrm{tan}\:^{\mathrm{2}} \alpha \\ $$$${k}={m}+{m}\sqrt{\mathrm{1}−\mathrm{2}{k}} \\ $$$${k}^{\mathrm{2}} +\cancel{{m}^{\mathrm{2}} }−\mathrm{2}{km}=\cancel{{m}^{\mathrm{2}} }−\mathrm{2}{m}^{\mathrm{2}} {k} \\ $$$$\Rightarrow\:\:\frac{{k}}{\mathrm{2}}=\frac{{r}}{\mathrm{2}{R}}={m}−{m}^{\mathrm{2}} \\ $$$$\:\:\:{r}=\mathrm{2}{R}\left\{\sqrt{\frac{\mathrm{1}−\lambda}{\mathrm{1}+\lambda}}−\left(\frac{\mathrm{1}−\lambda}{\mathrm{1}+\lambda}\right)\right\} \\ $$