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Question Number 209352 by Spillover last updated on 07/Jul/24

Answered by mr W last updated on 08/Jul/24

$${L}={mr}^{\mathrm{2}} \omega={constant} \\$$$$\omega_{{max}} =\frac{{L}}{{mr}_{{min}} ^{\mathrm{2}} }=\frac{{L}}{{m}\left({a}−{c}\right)^{\mathrm{2}} } \\$$$$\omega_{{min}} =\frac{{L}}{{mr}_{{max}} ^{\mathrm{2}} }=\frac{{L}}{{m}\left({a}+{c}\right)^{\mathrm{2}} } \\$$$${n}=\frac{\omega_{{max}} }{\omega_{{min}} }=\left(\frac{{a}+{c}}{{a}−{c}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{1}+{e}}{\mathrm{1}−{e}}\right)^{\mathrm{2}} \\$$$$\Rightarrow\frac{\mathrm{1}+{e}}{\mathrm{1}−{e}}=\sqrt{{n}} \\$$$$\Rightarrow{e}=\frac{\sqrt{{n}}−\mathrm{1}}{\:\sqrt{{n}}+\mathrm{1}}\:\checkmark \\$$

Answered by Spillover last updated on 12/Jul/24

$${from}\:{conservation}\:{of}\:{angular}\:{momentum} \\$$$${L}={mr}^{\mathrm{2}} \theta\:\:\:\:\:\: \\$$$$\:\theta_{{max}} =\frac{{L}}{{mr}_{{max}} ^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\theta_{{min}} =\frac{{L}}{{mr}_{{min}} ^{\mathrm{2}} }\:\: \\$$$${where}\:\:\:\:\:\:\:{r}_{{min}} ={a}\left(\mathrm{1}−{e}\right)\:\:\:\:\:\:\:\:\:\:{r}_{{max}} ={a}\left(\mathrm{1}+{e}\right) \\$$$${ratio}\:{of}\:{angular}\:{velocity}\:\left({n}\right) \\$$$${n}=\frac{\:\theta_{{max}} =\frac{{L}}{{mr}_{{max}} ^{\mathrm{2}} }\:\:\:}{\:\theta_{{min}} =\frac{{L}}{{mr}_{{min}} ^{\mathrm{2}} }\:\:}=\frac{\theta_{{max}} }{\theta_{{min}} }=\frac{{r}_{{max}} ^{\mathrm{2}} }{{r}_{{min}} ^{\mathrm{2}} } \\$$$${n}=\left(\frac{\mathrm{1}+{e}}{\mathrm{1}−{e}}\right)^{\mathrm{2}} \:\:\:\:\:\:\:\sqrt{{n}}=\:\:\:\:\frac{\mathrm{1}+{e}}{\mathrm{1}−{e}}\:\:\:\:\:\:\:\: \\$$$$\sqrt{{n}}\:\left(\mathrm{1}−{e}\right)=\mathrm{1}+{e} \\$$$${e}=\frac{\sqrt{{n}}−\mathrm{1}}{\:\sqrt{{n}}+\mathrm{1}} \\$$