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$R$ ત્રિજયા અને $M$ દળ ધરાવતો ગોળાના કેન્દ્રથી ${r_1}$ અને ${r_2}$ અંતરે ગુરુત્વાકર્ષણ બળ ${F_1}$ અને ${F_2}$ હોય,તો
$\frac{{{F_1}}}{{{F_2}}} = \frac{{{r_1}}}{{{r_2}}}$ if ${r_1} < R$ , ${r_2} < R$
$\frac{{{F_1}}}{{{F_2}}} = \frac{{r_1^2}}{{r_2^2}}$ if ${r_1} < R$ and ${r_2} < R$
$\frac{{{F_1}}}{{{F_2}}} = \frac{{{r_1}}}{{{r_2}}}$ if ${r_1} > R$ and ${r_2} > R$
$\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{r_{1}^{2}}{r_{2}^{2}}$ if ${{r}_{1}} < R$ and ${r_2} < R$
Solution
$F \propto I \propto \frac{1}{{{r^2}}}$, $r > R$ [$I = \frac{{GM}}{{{r^2}}}$]
$\therefore \frac{{{F_1}}}{{{F_2}}} = \frac{{r_2^2}}{{r_1^2}}$ ,${r_1} > R$ , ${r_2} > R$
$F \propto I \propto \,r$ , $r < R$ [ $I = \frac{4}{3}\pi \rho Gr$]
$\therefore \frac{{{F_1}}}{{{F_2}}} = \frac{{{r_1}}}{{{r_2}}}$ ,${r_1} < R$ , ${r_2} < R$.
