Imagine that a hole is drilled from one point on the earth surface to the other side through the diameter of the earth and small of mass ^′ m^′ released into the hole. show that the hole of mass excute simple harmonic motion and find its period of revolution


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Question Number 177240 by peter frank last updated on 02/Oct/22

$Imagine that a hole is drilled$ $from one point on the earth$ $surface to the other side through the$ $diameter of the earth and small$ $of mass^{^{'}} m^{^{'}} released into the hole .$ $show that the hole of mass excute$ $simple harmonic motion and$ $find its period of revolution$
Commented by JDamian last updated on 02/Oct/22
What is "the hole of mass"?
Commented by mr W last updated on 02/Oct/22

$h e m e a n s t h e ‘ ‘ m a s s i n t h e {h o l e}^{″},$ $i t h i n k .$
	Answered by mr W last updated on 02/Oct/22

	Commented by mr W last updated on 02/Oct/22

	$a s s u m e t h a t t h e e a r t h i s a n u n i f o r m$ $s o l i d s p h e r e w i t h d e n s i t y ρ a n d$ $r a d i u s R .$ $w h e n t h e s m a l l m a s s m i n t h e h o l e$ $i s a t a d i s t a n c e x t o t h e c e n t e r o f$ $t h e e a r t h, i t i s a t t r a c t e d b y t h e$ $g r a v i t a t i o n a l f o r c e F o f t h e e a r t h .$ $t h e g r a v i t a t i o n a l f o r c e f r o m t h e p a r t$ $o f t h e e a r t h i n s i d e t h e r a d i u s x$ $(m a s s M_{1}) i s$ $F_{1} = \frac{{G m M}_{1}}{x^{2}}$ $w i t h M_{1} = \frac{4 π x^{3} ρ}{3}$ $\Rightarrow F_{1} = \frac{G m}{x^{2}} \times \frac{4 π x^{3} ρ}{3} = \frac{4 π G m ρ x}{3}$ $t h e g r a v i t a t i o n a l f o r c e f r o m t h e p a r t$ $o f t h e e a r t h o u t s i d e t h e r a d i u s x$ $(m a s s M_{2}) i s F_{2} = 0 .$ $F = F_{1} + F_{2} = \frac{4 π G m ρ x}{3}$ $t h e a c c e l e r a t i o n o f t h e s m a l l m a s s i s$ $a = \frac{d^{2} x}{{d t}^{2}} .$ $w e h a v e$ $m a = - F = - \frac{4 π G m ρ x}{3}$ $w i t h k = \frac{4 π G ρ}{3} = c o n s t a n t$ $\Rightarrow \frac{d^{2} x}{{d t}^{2}} + k x = 0$ $t h i s i s t h e e q u a t i o n o f a s i m p l e$ $h a r m o n i c m o t i o n .$ $ω = \sqrt{k} = \sqrt{\frac{4 π G ρ}{3}} .$ $g = \frac{G M}{R^{2}} = \frac{4 π G ρ R}{3} = k R$ $\Rightarrow k = \frac{g}{R}$ $\Rightarrow ω = \sqrt{\frac{g}{R}}$ $t h e p e r i o d i s T = \frac{2 π}{ω} = 2 π \sqrt{\frac{R}{g}}$ $w i t h R = 6381 k m, g = 9.81 m / s^{2}$ $w e g e t$ $T = 2 π \sqrt{\frac{6381000}{9.81}} = 5067 s \approx 1 h 24 m$
	Commented by peter frank last updated on 02/Oct/22

	$thank you . I understand very well$
	Commented by mr W last updated on 03/Oct/22

	$i t m e a n s w h e n y o u c o u l d t a k e t h i s$ $s h o r t c u t, y o u w o u l d b e a b l e t o t r a v e l$ $f r o m t h e n o r t h p o l e t o t h e s o u t h p o l e$ $i n o n l y 42 m i n u t e s!$
	Commented by peter frank last updated on 03/Oct/22

	$thank you$
	Commented by Tawa11 last updated on 03/Oct/22

	$Great sir$
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