Question-138896

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

Answered by ajfour last updated on 20/Apr/21

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} = \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/