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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?

Answer

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
\({F}=\frac{{{G}{m}_{{{1}}}{m}_{{{2}}}}}{{{r}^{{{2}}}}}=\frac{{{3}{g}}}{{{4}\pi{R}\rho}}\times\frac{{{m}_{{{1}}}{m}_{{{2}}}}}{{{r}^{{{2}}}}}\)
\(=\frac{{{3}\times{9.8}\times{48.15}\times{4.815}\times{10}^{{{2}}}}}{{{4}\times{3.14}\times{\left({6.37}\times{10}^{{{6}}}\right)}\times{5.53}\times{10}^{{{3}}}\times{\left({1}\right)}^{{{2}}}}}\)
\(={15.4}\times{10}^{{-{11}}}{N}\)
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