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


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

$I f w e o b s e r v e w h e n l i g h t s o u r c e$ $i l l u m i n a t e s t h e s u r f a c e o f a n o b j e c t$ $l e t s s a y i t s a s p h e r e (f o o t b a l l), t h e a r e a$ $v i s i b l e z o n e d e c r e a s e s a s t h e l i g h t s o u r c e$ $c o m e s n e a r t h e s u r f a c e . S o c a l c u l a t e r a t e o f c h a n g e$ $v i s i b l e a r e a w h e n l i g h t s o u r c e i s c o m i n g$ $t o w a r d s t h e o b j e c t o r m o v i n g a w a y$ $f r o m t h e o b j e c t . A n d i n t h e c a s e o f e a r t h$ $f i n d t h e e q u a t i o n i n f u n c t i o n o f t i m e$ $I f t h e l i g h t o b j e c t i s c o m i n g f r o m s p a c e$ $t o w a r d s e a r t h, e x p r e s s t h a t e x p r e s s i o n$ $i n t h e f u n c t i o n o f t i m e$
Commented by mr W last updated on 04/Aug/24

$w h a t d o y o u m e a n w i t h ‘ ‘ {t i m e}^{″} i n$ $‘ ‘ a s t h e f u n c t i o n o f {t i m e}^{″} ?$
Commented by BHOOPENDRA last updated on 04/Aug/24

$y e s s o r r y t h a t w a s t y p i n g m i s t a k e$ $D e f i n e S a s a f u n c t i o n o f t i m e$
Commented by mr W last updated on 04/Aug/24

$w h a t d o y o u m e a n w i t h ‘ ‘ {t i m e}^{″} ?$
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

$I h a v e a t t a c h e d m y a p p r o c h s i r c a n y o u$ $t e l l m e w h e r e i h a v e m i s t a k e n ?$
	Answered by mr W last updated on 04/Aug/24

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

	$O S = l$ $\frac{d l}{d t} = - v$ $\cos θ = \frac{R}{l}$ $A = 2 π R^{2} (1 - \cos θ)$ $A = 2 π R^{2} (1 - \frac{R}{l})$ $l e t λ = \frac{R}{l}$ $\frac{d λ}{d t} = - \frac{R}{l^{2}} \times \frac{d l}{d t} = \frac{R v}{l^{2}} = \frac{λ^{2} v}{R}$ $A = 2 π R^{2} (1 - λ)$ $\frac{d A}{d t} = - 2 π R^{2} \times \frac{d λ}{d t}$ $\frac{d A}{d t} = - 2 π R^{2} \times (\frac{λ^{2} v}{R})$ $\Rightarrow \frac{d A}{d t} = - \frac{2 π R^{3} v}{l^{2}}$ $a = \frac{d v}{d t} = - v \frac{d v}{d l} = \frac{G M}{l^{2}}$ $- v d v = \frac{G M d l}{l^{2}}$ $\int_{0}^{v} v d v = - G M \int_{l_{0}}^{l} \frac{d l}{l^{2}}$ $\frac{v^{2}}{2} = G M (\frac{1}{l} - \frac{1}{l_{0}})$ $v = \sqrt{2 G M (\frac{1}{l} - \frac{1}{l_{0}})}$ $\Rightarrow \frac{d A}{d t} = - \frac{2 π R^{3}}{l^{2}} \sqrt{2 G M (\frac{1}{l} - \frac{1}{l_{0}})}$
	Commented by BHOOPENDRA last updated on 04/Aug/24

	$\Rightarrow \frac{d A}{d t} = - \frac{2 π R^{3} v}{l^{2}} N i c e c a n y o u t e l l m e$ $r e s t o f t h e t h i n g s a r e c o r r e c t ?$
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