Question
Derive an expression for work done against gravity.
✓
Answer
Potential energy of the body on the surface of the earth $=\frac{-\text{GMm}}{\text{R}}$
Potential energy of the body at a heighth from the surface of the earth $=-\frac{\text{GMm}}{(\text{R}+\text{h})}$
Work done $=\Big(-\frac{\text{GMm}}{\text{R}+\text{h}}\Big)-\Big(-\frac{\text{GMm}}{\text{R}}\Big)$
$=\frac{\text{GMm}}{\text{R}}-\frac{\text{GMm}}{\text{R}+\text{h}}$
$=\text{GMm}\Big(\frac{1}{\text{R}}-\frac{1}{\text{R}+\text{h}}\Big)$
$=\frac{\text{GMm}}{\text{R}}-\frac{\text{GMm}}{\text{R}+\text{h}}$
$=\text{GMm}\Big(\frac{1}{\text{R}}-\frac{1}{\text{R}+\text{h}}\Big)$
$=\frac{\text{GMmh}}{\text{R}(\text{R}+\text{h})}=\frac{\text{MgR}^2\text{h}}{\text{R}(\text{R}+\text{h})}$ $\Big[\because\text{g}=\frac{\text{GM}}{\text{R}^2}\Big]$
$=\frac{(\text{Mgh})}{(\text{R}+\text{h})}=\frac{\text{Mgh}}{1+\frac{\text{h}}{\text{R}}}$
Potential energy of the body at a heighth from the surface of the earth $=-\frac{\text{GMm}}{(\text{R}+\text{h})}$
Work done $=\Big(-\frac{\text{GMm}}{\text{R}+\text{h}}\Big)-\Big(-\frac{\text{GMm}}{\text{R}}\Big)$
$=\frac{\text{GMm}}{\text{R}}-\frac{\text{GMm}}{\text{R}+\text{h}}$
$=\text{GMm}\Big(\frac{1}{\text{R}}-\frac{1}{\text{R}+\text{h}}\Big)$
$=\frac{\text{GMm}}{\text{R}}-\frac{\text{GMm}}{\text{R}+\text{h}}$
$=\text{GMm}\Big(\frac{1}{\text{R}}-\frac{1}{\text{R}+\text{h}}\Big)$
$=\frac{\text{GMmh}}{\text{R}(\text{R}+\text{h})}=\frac{\text{MgR}^2\text{h}}{\text{R}(\text{R}+\text{h})}$ $\Big[\because\text{g}=\frac{\text{GM}}{\text{R}^2}\Big]$
$=\frac{(\text{Mgh})}{(\text{R}+\text{h})}=\frac{\text{Mgh}}{1+\frac{\text{h}}{\text{R}}}$
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