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Question Number 64452 by Tanmay chaudhury last updated on 18/Jul/19

Commented by Tanmay chaudhury last updated on 18/Jul/19

$t h a n k y o u . . .$
Commented by Tanmay chaudhury last updated on 18/Jul/19

$u n s o l v e d p r o b l e m . . .$
Commented by ajfour last updated on 18/Jul/19

$i w i l l s o l v e i t b y e v e n i n g, T a n m a y S i r . .$
Commented by mr W last updated on 18/Jul/19

$c o u l d i t b e 12 ?$
Commented by Tanmay chaudhury last updated on 18/Jul/19

$s i r s o m e o n e s h a r e d m e t h e p r o b l e m$ $t h e a n s w e r o f \frac{b - a}{3} = 5 a s p e r s e n d e r c o m m e n t$
	Answered by ajfour last updated on 19/Jul/19

	Commented by ajfour last updated on 19/Jul/19
	$potential within a solid ball of mass M radius R. V(r)=((GMr^3 )/(R^3 (r)))+∫_r ^( R) ((Gρ(4πz^2 )dz)/z) = ((GMr^2 )/R^3 )+((GM(4π))/(((4/3)πR^3 )))(((R^2 −r^2 )/2)) V(r)= ((GM)/R^3 )(((3R^2 −r^2 )/2)) __________________________ gravitational potential energy b/w full ball of radius R , density ρ and the small ball of radius R/2, density −ρ. −∫dU_1 =∫_0 ^( R) ρAVdr =∫_0 ^( R) ρ[2π(r^2 −(r^3 /R))((GM)/R^3 )(((3R^2 −r^2 )/2))dr =((πρGM)/R^3 )∫_0 ^( R) {3R^2 r^2 −r^4 −3Rr^3 +(r^5 /R)}dr = ((3GM^2 )/(4R))(1−(1/5)−(3/4)+(1/6)) U_1 =−((3GM^2 )/(4R))(((13)/(60))) =−((13GM^2 )/(80R)) And gravitational potential energy of small sphere with small sphere be U_2 . U_2 =∫_0 ^( R/2) [V_(R/2) (r)](4πρr^2 )dr U_2 = ∫((G((M/8)))/(((R/2))^3 ))[((3((R/2))^2 −r^2 )/2)](4πρr^2 )dr =((4πGM)/(2R^3 ))(((3M)/(4πR^3 )))∫_0 ^( R/2) (((3R^2 r^2 )/4)−r^4 )dr =((3GM^2 )/(2R))((1/(32))−(1/(5×32))) U_2 =((3GM^2 )/(80R)) W_(reauired) = −(U_1 +U_2 ) = −((GM^2 )/R)(−((13)/(80))+(3/(80))) =((GM^2 )/(8R))$
	$p o t e n t i a l w i t h i n a s o l i d b a l l o f m a s s M$ $r a d i u s R .$ $V (r) = \frac{{G M r}^{3}}{R^{3} (r)} + \int_{r}^{R} \frac{G ρ (4 π z^{2}) d z}{z}$ $= \frac{{G M r}^{2}}{R^{3}} + \frac{G M (4 π)}{(\frac{4}{3} π R^{3})} (\frac{R^{2} - r^{2}}{2})$ $V (r) = \frac{G M}{R^{3}} (\frac{3 R^{2} - r^{2}}{2})$ $__________________________$ $g r a v i t a t i o n a l p o t e n t i a l e n e r g y$ $b / w f u l l b a l l o f r a d i u s R,$ $d e n s i t y ρ a n d t h e s m a l l b a l l o f$ $r a d i u s R / 2, d e n s i t y - ρ .$ $- \int {d U}_{1} = \int_{0}^{R} ρ A V d r$ $= \int_{0}^{R} ρ [2 π (r^{2} - \frac{r^{3}}{R}) \frac{G M}{R^{3}} (\frac{3 R^{2} - r^{2}}{2}) d r$ $= \frac{π ρ G M}{R^{3}} \int_{0}^{R} {3 R^{2} r^{2} - r^{4} - 3 {R r}^{3} + \frac{r^{5}}{R}} d r$ $= \frac{3 {G M}^{2}}{4 R} (1 - \frac{1}{5} - \frac{3}{4} + \frac{1}{6})$ $U_{1} = - \frac{3 {G M}^{2}}{4 R} (\frac{13}{60}) = - \frac{13 {G M}^{2}}{80 R}$ $A n d g r a v i t a t i o n a l p o t e n t i a l e n e r g y$ $o f s m a l l s p h e r e w i t h s m a l l s p h e r e$ $b e U_{2} .$ $U_{2} = \int_{0}^{R / 2} [V_{R / 2} (r)] (4 π ρ r^{2}) d r$ $U_{2} = \int \frac{G (\frac{M}{8})}{{(\frac{R}{2})}^{3}} [\frac{3 {(\frac{R}{2})}^{2} - r^{2}}{2}] (4 π ρ r^{2}) d r$ $= \frac{4 π G M}{2 R^{3}} (\frac{3 M}{4 π R^{3}}) \int_{0}^{R / 2} (\frac{3 R^{2} r^{2}}{4} - r^{4}) d r$ $= \frac{3 {G M}^{2}}{2 R} (\frac{1}{32} - \frac{1}{5 \times 32})$ $U_{2} = \frac{3 {G M}^{2}}{80 R}$ $W_{r e a u i r e d} = - (U_{1} + U_{2})$ $= - \frac{{G M}^{2}}{R} (- \frac{13}{80} + \frac{3}{80})$ $= \frac{{G M}^{2}}{8 R}$
	Commented by Tanmay chaudhury last updated on 19/Jul/19

	$t h a n k y o u s i r . .$
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