If-a-planet-is-suddenly-stopped-in-its-orbit-supposed-to-be-circular-then-it-would-fall-into-the-sun-in-a-time-T-4-2-where-T-is-the-time-period-of-revolution-Prove-this-

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Question Number 27445 by Tinkutara last updated on 07/Jan/18

$$Ifaplanetissuddenlystoppedinits$$$$orbit,supposedtobecircular,thenit$$$$wouldfallintothesuninatime\frac{T}{4\sqrt{2}},$$$$whereTisthetimeperiodof$$$$revolution.Provethis.$$

Answered by mrW1 last updated on 07/Jan/18

$$Gravitationforcebetweenplanet$$$$andsun:$$$$F=\frac{GMm}{L^2}$$$$withM=massofsun$$$$m=massofplanet$$$$L=distancebetweenthem$$$$F=\frac{{mv}^2}{L}=\frac{GMm}{L^2}$$$$\Rightarrow v=\sqrt{\frac{GM}{L}}=velocityofplanetinorbit$$$$T=\frac{2\pi L}{v}=2\pi L\sqrt{\frac{L}{GM}}$$$$$$$$freefallofplanetintosunfrom$$$$heightx=Lto0:$$$$F\left(x\right)=\frac{GMm}{x^2}=ma$$$$\Rightarrow a=\frac{GM}{x^2}=-v\frac{dv}{dx}$$$$\Rightarrow vdv=-GM\frac{dx}{x^2}$$$$\Rightarrow {\int }_0^{v}vdv=-GM{\int }_L^x\frac{dx}{x^2}$$$$\Rightarrow \frac{1}{2}{v}^2=GM(\frac{1}{x}-\frac{1}{L})=\frac{GM(L-x)}{Lx}$$$$\Rightarrow v=\sqrt{\frac{2GM}{L}}\sqrt{\frac{L-x}{x}}=-\frac{dx}{dt}$$$$\Rightarrow -\sqrt{\frac{2GM}{L}}dt=\sqrt{\frac{x}{L-x}}dx$$$$\Rightarrow -\sqrt{\frac{2GM}{L}}{\int }_0^{T_1}dt={\int }_L^0\sqrt{\frac{x}{L-x}}dx$$$$\Rightarrow -\sqrt{\frac{2GM}{L}}{T}_1={[-\sqrt{x(L-x)}-L{\tan }^{-1}\sqrt{\frac{L-x}{x}}]}_L^0=-\frac{L\pi }{2}$$$$\Rightarrow T_1=\frac{\pi L}{2}\times \sqrt{\frac{L}{2GM}}=\frac{1}{4\sqrt{2}}\times 2\pi L\sqrt{\frac{L}{GM}}$$$$\Rightarrow T_1=\frac{T}{4\sqrt{2}}$$

Commented by Tinkutara last updated on 07/Jan/18

How do you solved the integral? As in question 27400 it is giving another answer.

Commented by mrW1 last updated on 07/Jan/18

$$I={\int }_L^0\sqrt{\frac{x}{L-x}}dx$$$$=L{\int }_L^0\sqrt{\frac{\frac{x}{L}}{1-\frac{x}{L}}}d\frac{x}{L}$$$$=L{\int }_1^0\sqrt{\frac{\lambda }{1-\lambda }}d\lambda$$$$with\lambda =\frac{x}{L}$$$$$$$$let\frac{1}{t}=\sqrt{\frac{\lambda }{1-\lambda }}$$$$\Rightarrow 1-\lambda =t^2\lambda$$$$\Rightarrow \lambda =\frac{1}{1+t^2}$$$$d\lambda =-\frac{2t}{{(1+t^2)}^2}$$$$\int \sqrt{\frac{\lambda }{1-\lambda }}d\lambda =-2\int \frac{1}{{(1+t^2)}^2}dt$$$$=-2[\frac{t}{2(1+t^2)}+\frac{1}{2}{\tan }^{-1}t]$$$$=-\frac{t}{1+t^2}-{\tan }^{-1}t$$$$=-\sqrt{\lambda (1-\lambda )}-{\tan }^{-1}\sqrt{\frac{1-\lambda }{\lambda }}$$$$=-\sqrt{\frac{x}{L}(1-\frac{x}{L})}-{\tan }^{-1}\sqrt{\frac{1-\frac{x}{L}}{\frac{x}{L}}}$$$$=-\frac{1}{L}\sqrt{x(L-x)}-{\tan }^{-1}\sqrt{\frac{L-x}{x}}$$$$$$$$I={\int }_L^0\sqrt{\frac{x}{L-x}}dx$$$$={[-\sqrt{x(L-x)}-L{\tan }^{-1}\sqrt{\frac{L-x}{x}}]}_L^0$$$$=-\frac{L\pi }{2}$$

Commented by Tinkutara last updated on 07/Jan/18

Commented by Tinkutara last updated on 07/Jan/18

This was book's solution. Where is mistake here then? The integral comes out negative but here it shows positive.

Commented by mrW1 last updated on 07/Jan/18

$$Tobeexact,the3rdand4thlineinbookisnotcorrect.$$$$in3rdlineitshouldbe{(-\frac{dx}{dt})}^{2}as$$$$replacementfor{v}^2,since$$$$withthedefinitionforx-axiswe$$$$havev=-\frac{dx}{dt},ononesideof4thline$$$$a“{-}^{″}signismissing.$$$$$$$$sincef\left(x\right)=\sqrt{\frac{x}{r-x}}ispositive,$$$${\int }_r^{0}f\left(x\right)dxisalwaysnegative.$$$$$$$$Ithinkthebook{doesn}^{\prime }tworkso$$$$cleanly.$$

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