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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
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 change visible area when light source is coming towards the object or moving away from the object.And in the case of earth find the equation in function of time If the light object is coming from space towards earth, express that expression in the function of time
$${If}\:{we}\:{observe}\:{when}\:{light}\:{source}\: \\ $$$${illuminates}\:{the}\:{surface}\:{of}\:{an}\:{object} \\ $$$${lets}\:{say}\:{its}\:{a}\:{sphere}\left({football}\right)\:,{the}\:{area} \\ $$$${visible}\:{zone}\:{decreases}\:{as}\:{the}\:{light}\:{source} \\ $$$${comes}\:{near}\:{the}\:{surface}.{So}\:\:{calculate}\:{rate}\:{of}\:{change} \\ $$$${visible}\:{area}\:{when}\:{light}\:{source}\:{is}\:{coming} \\ $$$${towards}\:{the}\:{object}\:{or}\:{moving}\:{away}\: \\ $$$${from}\:{the}\:{object}.{And}\:{in}\:{the}\:{case}\:{of}\:{earth} \\ $$$${find}\:{the}\:{equation}\:{in}\:{function}\:{of}\:{time} \\ $$$${If}\:{the}\:{light}\:{object}\:{is}\:{coming}\:{from}\:{space} \\ $$$${towards}\:{earth},\:{express}\:{that}\:{expression} \\ $$$${in}\:{the}\:{function}\:{of}\:{time} \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 04/Aug/24
what do you mean with “time” in “as the function of time”?
$${what}\:{do}\:{you}\:{mean}\:{with}\:“{time}''\:{in} \\ $$$$“{as}\:{the}\:{function}\:{of}\:{time}''? \\ $$
Commented by BHOOPENDRA last updated on 04/Aug/24
yes sorry that was typing mistake Define S as a function of time
$${yes}\:{sorry}\:{that}\:{was}\:{typing}\:{mistake} \\ $$$${Define}\:{S}\:{as}\:{a}\:{function}\:{of}\:{time} \\ $$
Commented by mr W last updated on 04/Aug/24
what do you mean with “time”?
$${what}\:{do}\:{you}\:{mean}\:{with}\:“{time}''? \\ $$
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
I have attached my approch sir can you tell me where i have mistaken?
$${I}\:{have}\:{attached}\:{my}\:{approch}\:{sir}\:{can}\:{you} \\ $$$${tell}\:{me}\:{where}\:{i}\:{have}\:{mistaken}? \\ $$
Answered by mr W last updated on 04/Aug/24
Commented by mr W last updated on 04/Aug/24
OS=l (dl/dt)=−v cos θ=(R/l) A=2πR^2 (1−cos θ) A=2πR^2 (1−(R/l)) let λ=(R/l) (dλ/dt)=−(R/l^2 )×(dl/dt)=((Rv)/l^2 )=((λ^2 v)/R) A=2πR^2 (1−λ) (dA/dt)=−2πR^2 ×(dλ/dt) (dA/dt)=−2πR^2 ×(((λ^2 v)/R)) ⇒(dA/dt)=−((2πR^3 v)/l^2 ) a=(dv/dt)=−v(dv/dl)=((GM)/l^2 ) −vdv=((GMdl)/l^2 ) ∫_0 ^v vdv=−GM∫_l_0 ^l (dl/l^2 ) (v^2 /2)=GM((1/l)−(1/l_0 )) v=(√(2GM((1/l)−(1/l_0 )))) ⇒(dA/dt)=−((2πR^3 )/l^2 )(√(2GM((1/l)−(1/l_0 ))))
$${OS}={l} \\ $$$$\frac{{dl}}{{dt}}=−{v} \\ $$$$\mathrm{cos}\:\theta=\frac{{R}}{{l}} \\ $$$${A}=\mathrm{2}\pi{R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$${A}=\mathrm{2}\pi{R}^{\mathrm{2}} \left(\mathrm{1}−\frac{{R}}{{l}}\right) \\ $$$${let}\:\lambda=\frac{{R}}{{l}} \\ $$$$\frac{{d}\lambda}{{dt}}=−\frac{{R}}{{l}^{\mathrm{2}} }×\frac{{dl}}{{dt}}=\frac{{Rv}}{{l}^{\mathrm{2}} }=\frac{\lambda^{\mathrm{2}} {v}}{{R}} \\ $$$${A}=\mathrm{2}\pi{R}^{\mathrm{2}} \left(\mathrm{1}−\lambda\right) \\ $$$$\frac{{dA}}{{dt}}=−\mathrm{2}\pi{R}^{\mathrm{2}} ×\frac{{d}\lambda}{{dt}} \\ $$$$\frac{{dA}}{{dt}}=−\mathrm{2}\pi{R}^{\mathrm{2}} ×\left(\frac{\lambda^{\mathrm{2}} {v}}{{R}}\right) \\ $$$$\Rightarrow\frac{{dA}}{{dt}}=−\frac{\mathrm{2}\pi{R}^{\mathrm{3}} {v}}{{l}^{\mathrm{2}} } \\ $$$${a}=\frac{{dv}}{{dt}}=−{v}\frac{{dv}}{{dl}}=\frac{{GM}}{{l}^{\mathrm{2}} } \\ $$$$−{vdv}=\frac{{GMdl}}{{l}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{{v}} {vdv}=−{GM}\int_{{l}_{\mathrm{0}} } ^{{l}} \frac{{dl}}{{l}^{\mathrm{2}} } \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}={GM}\left(\frac{\mathrm{1}}{{l}}−\frac{\mathrm{1}}{{l}_{\mathrm{0}} }\right) \\ $$$${v}=\sqrt{\mathrm{2}{GM}\left(\frac{\mathrm{1}}{{l}}−\frac{\mathrm{1}}{{l}_{\mathrm{0}} }\right)} \\ $$$$\Rightarrow\frac{{dA}}{{dt}}=−\frac{\mathrm{2}\pi{R}^{\mathrm{3}} }{{l}^{\mathrm{2}} }\sqrt{\mathrm{2}{GM}\left(\frac{\mathrm{1}}{{l}}−\frac{\mathrm{1}}{{l}_{\mathrm{0}} }\right)} \\ $$
Commented by BHOOPENDRA last updated on 04/Aug/24
⇒(dA/dt)=−((2πR^3 v)/l^2 ) Nice can you tell me rest of the things are correct?
$$\Rightarrow\frac{{dA}}{{dt}}=−\frac{\mathrm{2}\pi{R}^{\mathrm{3}} {v}}{{l}^{\mathrm{2}} }\:\:{Nice}\:{can}\:{you}\:{tell}\:{me}\: \\ $$$${rest}\:{of}\:\:{the}\:{things}\:{are}\:{correct}? \\ $$

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