$m$ દળ ધરાવતો એક ઉપગ્રહ, પૃથ્વી (ત્રિજયા $R$) ની આસપાસ સપાટીથી $h$ ઊંચાઇએ પરિક્રમણ કરે છે. જો પૃથ્વીની સપાટી પર ગુરુત્વપ્રવેગ $g_0$ હોય, તો ઉપગ્રહની કુલઊર્જા $g_0$ ના પદમાં કઇ હશે?

Question

A$\frac{{2m{g_0}{R^2}}}{{R + h}}$
B$-$$\;\frac{{2m{g_0}{R^2}}}{{R + h}}$
C$\;\frac{{{mg_0}{R^2}}}{{2\left( {R + h} \right)}}$
D$-$$\;\frac{{{mg_0}{R^2}}}{{2\left( {R + h} \right)}}$

NEET 2016, Diffcult

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Solution

d
Total energy of satellite at height $h$ from the earth surface,

$E=PE+KE$

$ = - \frac{{GMm}}{{\left( {R + h} \right)}} + \frac{1}{2}m{v^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

Also, $\frac{{m{v^2}}}{{\left( {R + h} \right)}} = \frac{{GMm}}{{\left( {R + {h^2}} \right)}}$

or, ${v^2} = \frac{{GM}}{{R + h}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

From eqns, $(i)$ and $(ii)$,

$E = - \frac{{GMm}}{{\left( {R + h} \right)}} + \frac{1}{2}\frac{{GMm}}{{\left( {R + h} \right)}} = - \frac{1}{2}\frac{{GMm}}{{\left( {R + h} \right)}}$

$ = - \frac{1}{2}\frac{{GM}}{{{R^2}}} \times \frac{{m{R^2}}}{{\left( {R + h} \right)}}$

$ = - \frac{{m{g_0}{R^2}}}{{2\left( {R + h} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\left( {{g_0} = \frac{{GM}}{{{R^2}}}} \right)$

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