If-we-observe-when-light-source-illuminates-the-surface-of-an-object-lets-say-its-a-sphere-football-the-area-visible-zone-decreases-as-the-light-source-comes-near-the-surface-So-calculate-rate-of-

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Question Number 210265 by BHOOPENDRA last updated on 04/Aug/24

$$Ifweobservewhenlightsource$$$$illuminatesthesurfaceofanobject$$$$letssayitsasphere\left(football\right),thearea$$$$visiblezonedecreasesasthelightsource$$$$comesnearthesurface.Socalculaterateofchange$$$$visibleareawhenlightsourceiscoming$$$$towardstheobjectormovingaway$$$$fromtheobject.Andinthecaseofearth$$$$findtheequationinfunctionoftime$$$$Ifthelightobjectiscomingfromspace$$$$towardsearth,expressthatexpression$$$$inthefunctionoftime$$$$$$$$$$

Commented by mr W last updated on 04/Aug/24

$$whatdoyoumeanwith“{time}^{″}in$$$$“asthefunctionof{time}^{″}?$$

Commented by BHOOPENDRA last updated on 04/Aug/24

$$yessorrythatwastypingmistake$$$$DefineSasafunctionoftime$$

Commented by mr W last updated on 04/Aug/24

$$whatdoyoumeanwith“{time}^{″}?$$

Commented by BHOOPENDRA last updated on 04/Aug/24

Commented by BHOOPENDRA last updated on 04/Aug/24

Commented by BHOOPENDRA last updated on 04/Aug/24

Commented by BHOOPENDRA last updated on 04/Aug/24

Commented by BHOOPENDRA last updated on 04/Aug/24

Commented by BHOOPENDRA last updated on 04/Aug/24

Commented by BHOOPENDRA last updated on 04/Aug/24

Commented by BHOOPENDRA last updated on 04/Aug/24

$$Ihaveattachedmyapprochsircanyou$$$$tellmewhereihavemistaken?$$

Answered by mr W last updated on 04/Aug/24

Commented by mr W last updated on 04/Aug/24

$$OS=l$$$$\frac{dl}{dt}=-v$$$$\cos \theta =\frac{R}{l}$$$$A=2\pi R^2(1-\cos \theta )$$$$A=2\pi R^2(1-\frac{R}{l})$$$$let\lambda =\frac{R}{l}$$$$\frac{d\lambda }{dt}=-\frac{R}{l^2}\times \frac{dl}{dt}=\frac{Rv}{l^2}=\frac{{\lambda }^{2}v}{R}$$$$A=2\pi R^2(1-\lambda )$$$$\frac{dA}{dt}=-2\pi R^2\times \frac{d\lambda }{dt}$$$$\frac{dA}{dt}=-2\pi R^2\times \left(\frac{{\lambda }^{2}v}{R}\right)$$$$\Rightarrow \frac{dA}{dt}=-\frac{2\pi R^{3}v}{l^2}$$$$a=\frac{dv}{dt}=-v\frac{dv}{dl}=\frac{GM}{l^2}$$$$-vdv=\frac{GMdl}{l^2}$$$${\int }_0^{v}vdv=-GM{\int }_{l_0}^l\frac{dl}{l^2}$$$$\frac{v^2}{2}=GM(\frac{1}{l}-\frac{1}{l_0})$$$$v=\sqrt{2GM(\frac{1}{l}-\frac{1}{l_0})}$$$$\Rightarrow \frac{dA}{dt}=-\frac{2\pi R^3}{l^2}\sqrt{2GM(\frac{1}{l}-\frac{1}{l_0})}$$

Commented by BHOOPENDRA last updated on 04/Aug/24

$$\Rightarrow \frac{dA}{dt}=-\frac{2\pi R^{3}v}{l^2}Nicecanyoutellme$$$$restofthethingsarecorrect?$$

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