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QuestionPhysicsClass 11

Two lead spheres of $${20}{c}{m}$$ and $${2}{c}{m}$$ diametre respectively are planet with centres $${100}{c}{m}$$ apart. Calculate the attraction between them, given the radius of the Earth as $${6.37}\times{10}^{{{8}}}{c}{m}$$ and its mean density as $${5.53}\times{10}^{{{3}}}{k}{g}{m}^{{-{3}}}$$. Speciffic gravity of lead $$={11.5}$$. If the lead spheres are replaced by bress sphere of the same radii, would the force of attraction be the same?

Here, $${r}_{{{1}}}={0.20}/{2}={0.1}{m}$$,
$${r}_{{{2}}}={0.02}/{2}={0.01}{m},{r}={1.0}{m}$$,
$$\rho={5.53}\times{10}^{{{3}}}{k}{g}{m}^{{-{3}}}$$,
$$\rho'={11.5}\times{10}^{{{3}}}{k}{g}/{m}^{{{3}}}$$
Mass of first lead sphere, $${m}_{{{1}}}=\frac{{{4}}}{{{3}}}\pi{{r}_{{{1}}}^{{{3}}}}\rho'$$
$$=\frac{{{4}}}{{{3}}}\times{3.14}\times{\left({0.1}\right)}^{{{3}}}\times{11.5}\times{10}^{{{3}}}$$
$$={4.815}\times{10}^{{-{2}}}{k}{g}$$.
We know that, $${g}=\frac{{{G}{M}}}{{{R}^{{{2}}}}}=\frac{{{G}}}{{{R}^{{{2}}}}}\times\frac{{{4}}}{{{3}}}\pi{R}^{{{3}}}\rho$$
or $${G}=\frac{{{3}{g}}}{{{4}\pi{R}\rho}}$$
Force of attraction between lead spheres will
$${F}=\frac{{{G}{m}_{{{1}}}{m}_{{{2}}}}}{{{r}^{{{2}}}}}=\frac{{{3}{g}}}{{{4}\pi{R}\rho}}\times\frac{{{m}_{{{1}}}{m}_{{{2}}}}}{{{r}^{{{2}}}}}$$
$$=\frac{{{3}\times{9.8}\times{48.15}\times{4.815}\times{10}^{{{2}}}}}{{{4}\times{3.14}\times{\left({6.37}\times{10}^{{{6}}}\right)}\times{5.53}\times{10}^{{{3}}}\times{\left({1}\right)}^{{{2}}}}}$$
$$={15.4}\times{10}^{{-{11}}}{N}$$