Where will a body weigh more, \({2}{k}{m}\) above the surface of earth or \({2}{k}{m}\) below the surface of earth ?
Answer
At height \({h}{\left(={2}{k}{m}\right)}\) above the surface of earth.
\({{g}_{{{h}}}=}{g{{\left({1}-\frac{{{2}{h}}}{{{R}}}\right)}}}={g{{\left({1}-\frac{{{2}\times{2}}}{{{R}}}\right)}}}={g{{\left({1}-\frac{{{4}}}{{{R}}}\right)}}}\)
and below the surface of earth at depth
\({d}{\left(={2}{k}{m}\right)}\)
\({{g}_{{{d}}}=}{g{{\left({1}-\frac{{{d}}}{{{R}}}\right)}}}={g{{\left({1}-\frac{{{2}}}{{{R}}}\right)}}}\)
From above, it is clear that
\({{g}_{{{d}}}\gt}{{g}_{{{h}}}}\) or \({m}{{g}_{{{d}}}\gt}{m}{{g}_{{{h}}}}\)
Hence body well weigh more at a depth \({2}{k}{m}\) below the surface of earth than \({2}{k}{m}\) above the surface of earth.
\({{g}_{{{h}}}=}{g{{\left({1}-\frac{{{2}{h}}}{{{R}}}\right)}}}={g{{\left({1}-\frac{{{2}\times{2}}}{{{R}}}\right)}}}={g{{\left({1}-\frac{{{4}}}{{{R}}}\right)}}}\)
and below the surface of earth at depth
\({d}{\left(={2}{k}{m}\right)}\)
\({{g}_{{{d}}}=}{g{{\left({1}-\frac{{{d}}}{{{R}}}\right)}}}={g{{\left({1}-\frac{{{2}}}{{{R}}}\right)}}}\)
From above, it is clear that
\({{g}_{{{d}}}\gt}{{g}_{{{h}}}}\) or \({m}{{g}_{{{d}}}\gt}{m}{{g}_{{{h}}}}\)
Hence body well weigh more at a depth \({2}{k}{m}\) below the surface of earth than \({2}{k}{m}\) above the surface of earth.
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