Physics

$$v^2-u^2 \neq 2as$$

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When a particle is thrown up vertically from the surface of the earth (radius R), we have v2-u2=2(-g)h (h << R)  if the effects due to the atmosphere of the earth are neglected (symbols mean the usual things). However, if h << R condition is not applicable, prove that \displaystyle {{v}^{2}}-{{u}^{2}}=2(-g)h\frac{1}{\left( 1+\frac{h}{R} \right)}.

Solution

\displaystyle \frac{1}{2}m{{u}^{2}}-\frac{GMm}{R}=\frac{1}{2}m{{v}^{2}}-\frac{GMm}{R+h} [Conservation of mechanical energy]

\displaystyle \Rightarrow {{v}^{2}}-{{u}^{2}}=2GM\left( \frac{1}{R+h}-\frac{1}{R} \right)

\displaystyle \because GM=g{{R}^{2}}

\displaystyle {{v}^{2}}-{{u}^{2}}=2g{{R}^{2}}\left( \frac{1}{R+h}-\frac{1}{R} \right)

\displaystyle {{v}^{2}}-{{u}^{2}}=2gR\frac{-h}{(R+h)}

\displaystyle \Rightarrow {{v}^{2}}-{{u}^{2}}=2(-g)h\frac{1}{\left( 1+\frac{h}{R} \right)}

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