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

$I f a p l a n e t i s s u d d e n l y s t o p p e d i n i t s$ $o r b i t, s u p p o s e d t o b e c i r c u l a r, t h e n i t$ $w o u l d f a l l i n t o t h e s u n i n a t i m e \frac{T}{4 \sqrt{2}},$ $w h e r e T i s t h e t i m e p e r i o d o f$ $r e v o l u t i o n . P r o v e t h i s .$
	Answered by mrW1 last updated on 07/Jan/18

	$G r a v i t a t i o n f o r c e b e t w e e n p l a n e t$ $a n d s u n :$ $F = \frac{G M m}{L^{2}}$ $w i t h M = m a s s o f s u n$ $m = m a s s o f p l a n e t$ $L = d i s t a n c e b e t w e e n t h e m$ $F = \frac{{m v}^{2}}{L} = \frac{G M m}{L^{2}}$ $\Rightarrow v = \sqrt{\frac{G M}{L}} = v e l o c i t y o f p l a n e t i n o r b i t$ $T = \frac{2 π L}{v} = 2 π L \sqrt{\frac{L}{G M}}$ $f r e e f a l l o f p l a n e t i n t o s u n f r o m$ $h e i g h t x = L t o 0 :$ $F (x) = \frac{G M m}{x^{2}} = m a$ $\Rightarrow a = \frac{G M}{x^{2}} = - v \frac{d v}{d x}$ $\Rightarrow v d v = - G M \frac{d x}{x^{2}}$ $\Rightarrow \int_{0}^{v} v d v = - G M \int_{L}^{x} \frac{d x}{x^{2}}$ $\Rightarrow \frac{1}{2} v^{2} = G M (\frac{1}{x} - \frac{1}{L}) = \frac{G M (L - x)}{L x}$ $\Rightarrow v = \sqrt{\frac{2 G M}{L}} \sqrt{\frac{L - x}{x}} = - \frac{d x}{d t}$ $\Rightarrow - \sqrt{\frac{2 G M}{L}} d t = \sqrt{\frac{x}{L - x}} d x$ $\Rightarrow - \sqrt{\frac{2 G M}{L}} \int_{0}^{T_{1}} d t = \int_{L}^{0} \sqrt{\frac{x}{L - x}} d x$ $\Rightarrow - \sqrt{\frac{2 G M}{L}} T_{1} = {[- \sqrt{x (L - x)} - L \tan^{- 1} \sqrt{\frac{L - x}{x}}]}_{L}^{0} = - \frac{L π}{2}$ $\Rightarrow T_{1} = \frac{π L}{2} \times \sqrt{\frac{L}{2 G M}} = \frac{1}{4 \sqrt{2}} \times 2 π L \sqrt{\frac{L}{G M}}$ $\Rightarrow T_{1} = \frac{T}{4 \sqrt{2}}$
	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

	$I = \int_{L}^{0} \sqrt{\frac{x}{L - x}} d x$ $= L \int_{L}^{0} \sqrt{\frac{\frac{x}{L}}{1 - \frac{x}{L}}} d \frac{x}{L}$ $= L \int_{1}^{0} \sqrt{\frac{λ}{1 - λ}} d λ$ $w i t h λ = \frac{x}{L}$ $l e t \frac{1}{t} = \sqrt{\frac{λ}{1 - λ}}$ $\Rightarrow 1 - λ = t^{2} λ$ $\Rightarrow λ = \frac{1}{1 + t^{2}}$ $d λ = - \frac{2 t}{{(1 + t^{2})}^{2}}$ $\int \sqrt{\frac{λ}{1 - λ}} d λ = - 2 \int \frac{1}{{(1 + t^{2})}^{2}} d t$ $= - 2 [\frac{t}{2 (1 + t^{2})} + \frac{1}{2} \tan^{- 1} t]$ $= - \frac{t}{1 + t^{2}} - \tan^{- 1} t$ $= - \sqrt{λ (1 - λ)} - \tan^{- 1} \sqrt{\frac{1 - λ}{λ}}$ $= - \sqrt{\frac{x}{L} (1 - \frac{x}{L})} - \tan^{- 1} \sqrt{\frac{1 - \frac{x}{L}}{\frac{x}{L}}}$ $= - \frac{1}{L} \sqrt{x (L - x)} - \tan^{- 1} \sqrt{\frac{L - x}{x}}$ $I = \int_{L}^{0} \sqrt{\frac{x}{L - x}} d x$ $= {[- \sqrt{x (L - x)} - L \tan^{- 1} \sqrt{\frac{L - x}{x}}]}_{L}^{0}$ $= - \frac{L π}{2}$
	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

	$T o b e e x a c t, t h e 3 r d a n d 4 t h l i n e i n b o o k i s n o t c o r r e c t .$ $i n 3 r d l i n e i t s h o u l d b e {(- \frac{d x}{d t})}^{2} a s$ $r e p l a c e m e n t f o r v^{2}, s i n c e$ $w i t h t h e d e f i n i t i o n f o r x - a x i s w e$ $h a v e v = - \frac{d x}{d t}, o n o n e s i d e o f 4 t h l i n e$ $a ‘ ‘ -^{″} s i g n i s m i s s i n g .$ $s i n c e f (x) = \sqrt{\frac{x}{r - x}} i s p o s i t i v e,$ $\int_{r}^{0} f (x) d x i s a l w a y s n e g a t i v e .$ $I t h i n k t h e b o o k {d o e s n}^{'} t w o r k s o$ $c l e a n l y .$
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