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Question Number 138896 by ajfour last updated on 19/Apr/21

	Answered by ajfour last updated on 20/Apr/21
	$Let A be origin and x axis vertically down towards center of cube. −dg_x =(((Gdm)x)/((x^2 +y^2 +z^2 )^(3/2) )) −dg_x =(((ρG)xdxdydz)/((x^2 +y^2 +z^2 )^(3/2) )) x∈[0,2a] ; y∈[−a,a] ; z∈[−a,a] −(g_x /(ρG))=∫dz∫dy{(1/( (√(y^2 +z^2 ))))−(1/( (√(4a^2 +y^2 +z^2 ))))} =∫_(−a) ^( a) {ln (((a+(√(a^2 +z^2 )))/(−a+(√(a^2 +z^2 )))))−ln( ((a+(√(5a^2 +z^2 )))/(−a+(√(5a^2 +z^2 )))))}dz ρ=(M/(8a^3 )) & g_0 =−((GM)/R^2 )=−((ρG(8a^3 ))/((((6a^3 )/π))^(2/3) )) as (4/3)πR^3 =8a^3 ⇒ g_0 =−((ρG(8a))/(((6/π))^(2/3) )) ⇒ −ρG=(g_0 /(8a))((6/π))^(2/3) −ρG=((3g_0 )/(4πR)) as a=R((π/6))^(1/3) ⇒ ((g_x (4πR))/(3g_0 ))=∫_(−a) ^( a) {ln (((a+(√(a^2 +z^2 )))/(−a+(√(a^2 +z^2 )))))−ln( ((a+(√(5a^2 +z^2 )))/(−a+(√(5a^2 +z^2 )))))}dz let (z/a)=t ⇒ dz=adt for z=a, t=1 & z=−a, t=−1 so (g_x /g_0 )=((3a)/(4πR))∫_(−1) ^( 1) {ln (((1+(√(1+t^2 )))/(−1+(√(1+t^2 )))))−ln (((1+(√(5+t^2 )))/(−1+(√(5+t^2 )))))}dt (g_x /g_0 )=(1/8)((6/π))^(2/3) ( I_0 )$
	$L e t A b e o r i g i n a n d x a x i s$ $v e r t i c a l l y d o w n t o w a r d s c e n t e r$ $o f c u b e .$ $- {d g}_{x} = \frac{(G d m) x}{{(x^{2} + y^{2} + z^{2})}^{3 / 2}}$ $- {d g}_{x} = \frac{(ρ G) x d x d y d z}{{(x^{2} + y^{2} + z^{2})}^{3 / 2}}$ $x \in [0, 2 a]; y \in [- a, a]; z \in [- a, a]$ $- \frac{g_{x}}{ρ G} = \int d z \int d y {\frac{1}{\sqrt{y^{2} + z^{2}}} - \frac{1}{\sqrt{4 a^{2} + y^{2} + z^{2}}}}$ $= \int_{- a}^{a} {\ln (\frac{a + \sqrt{a^{2} + z^{2}}}{- a + \sqrt{a^{2} + z^{2}}}) - \ln (\frac{a + \sqrt{5 a^{2} + z^{2}}}{- a + \sqrt{5 a^{2} + z^{2}}})} d z$ $ρ = \frac{M}{8 a^{3}} & g_{0} = - \frac{G M}{R^{2}} = - \frac{ρ G (8 a^{3})}{{(\frac{6 a^{3}}{π})}^{2 / 3}}$ $a s \frac{4}{3} π R^{3} = 8 a^{3} \Rightarrow$ $g_{0} = - \frac{ρ G (8 a)}{{(\frac{6}{π})}^{2 / 3}} \Rightarrow - ρ G = \frac{g_{0}}{8 a} {(\frac{6}{π})}^{2 / 3}$ $- ρ G = \frac{3 g_{0}}{4 π R} a s a = R {(\frac{π}{6})}^{1 / 3}$ $\Rightarrow \frac{g_{x} (4 π R)}{3 g_{0}} = \int_{- a}^{a} {\ln (\frac{a + \sqrt{a^{2} + z^{2}}}{- a + \sqrt{a^{2} + z^{2}}}) - \ln (\frac{a + \sqrt{5 a^{2} + z^{2}}}{- a + \sqrt{5 a^{2} + z^{2}}})} d z$ $l e t \frac{z}{a} = t \Rightarrow d z = a d t$ $f o r z = a, t = 1 & z = - a, t = - 1$ $s o$ $\frac{g_{x}}{g_{0}} = \frac{3 a}{4 π R} \int_{- 1}^{1} {\ln (\frac{1 + \sqrt{1 + t^{2}}}{- 1 + \sqrt{1 + t^{2}}}) - \ln (\frac{1 + \sqrt{5 + t^{2}}}{- 1 + \sqrt{5 + t^{2}}})} d t$ $\frac{g_{x}}{g_{0}} = \frac{1}{8} {(\frac{6}{π})}^{2 / 3} (I_{0})$
	Commented by mr W last updated on 20/Apr/21
	$I_0 =∫_(−1) ^( 1) {ln (((1+(√(1+t^2 )))/(−1+(√(1+t^2 )))))−ln (((1+(√(5+t^2 )))/(−1+(√(5+t^2 )))))}dt ≈5.193793 g_x =ρGI_0 a g_0 =((GM)/R^2 )=(G/R^2 )×((ρ4πR^3 )/3)=((4ρGπR)/3) 8a^3 =((4πR^3 )/3) ⇒(a/R)=((π/6))^(1/3) (g_x /g_0 )=((3I_0 a)/(4πR))=(I_0 /8)×((6a)/(πR))=(I_0 /8)((6/π))^(2/3) ≈0.999376$
	$I_{0} = \int_{- 1}^{1} {\ln (\frac{1 + \sqrt{1 + t^{2}}}{- 1 + \sqrt{1 + t^{2}}}) - \ln (\frac{1 + \sqrt{5 + t^{2}}}{- 1 + \sqrt{5 + t^{2}}})} d t$ $\approx 5.193793$ $g_{x} = ρ {G I}_{0} a$ $g_{0} = \frac{G M}{R^{2}} = \frac{G}{R^{2}} \times \frac{ρ 4 π R^{3}}{3} = \frac{4 ρ G π R}{3}$ $8 a^{3} = \frac{4 π R^{3}}{3} \Rightarrow \frac{a}{R} = {(\frac{π}{6})}^{\frac{1}{3}}$ $\frac{g_{x}}{g_{0}} = \frac{3 I_{0} a}{4 π R} = \frac{I_{0}}{8} \times \frac{6 a}{π R} = \frac{I_{0}}{8} {(\frac{6}{π})}^{\frac{2}{3}} \approx 0.999376$
	Commented by ajfour last updated on 20/Apr/21

	$I f i t i s c o r r e c t, {i t}^{'} s r a t h e r s t r a n g e . .$ $\Rightarrow I_{0} = 8 {(\frac{g_{c u b e}}{g_{s p h e r e}})}_{A} {(\frac{π}{6})}^{2 / 3}$
	Commented by mr W last updated on 20/Apr/21

	$i s e e t h i s r e s u l t a s u n d e r s t a n d a b l e .$ $t h e p o i n t A i s c l o s e r t o t h e c e n t e r o f$ $e a r t h, s o g_{x} s h o u l d b e l a r g e r t h a n g_{0}$ $a t t h e s u r f a c e o f s p h e r i c a l e a r t h . b u t$ $t h e m a s s o f t h e c u b e i s l e s s$ $c o n c e n t r a t e d a s t h e s p h e r e, s o g_{x}$ $s h o u l d b e s m a l l e r t h a n g_{0} . b o t h e f f e c t s$ $a b r o g a t e e a c h o t h e r, s o g_{x} i s a l m o s t$ $t h e s a m e a s g_{0} a t p o i n t A .$
	Commented by mr W last updated on 20/Apr/21

	Commented by mr W last updated on 20/Apr/21

	Commented by mr W last updated on 20/Apr/21
	https://possiblywrong.wordpress.com/2011/09/09/if-the-earth-were-a-cube/
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