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A small ball of mass $‘m’$ is released at a height $‘R’$ above the earth surface, as shown in the figure above. If the maximum depth of the ball to which it goes is $R/2$ inside the earth through a narrow grove before coming to rest momentarily. The grove, contain an ideal spring of spring constant $K$ and natural length $R,$ find the value of $K$ if $R$ is radius of earth and $M$ mass of earth
$\frac{{3GMm}}{{{R^3}}}$
$\frac{{6GMm}}{{{R^3}}}$
$\frac{{9GMm}}{{{R^3}}}$
$\frac{{7GMm}}{{{R^3}}}$
Solution
By energy conservation $K_{i}+U_{i}=K_{f}+U_{f}$
$0-\frac{G M m}{2 R}=\frac{1}{2} K\left(\frac{R}{2}\right)^{2} \frac{11 G M m}{8 R}$
$\frac{G M m}{R}\left[\frac{11}{8}-\frac{1}{2}\right]=\frac{1}{2} K\left(\frac{R^{2}}{4}\right)$
$\frac{7 G M m}{8 R}=\frac{K R^{2}}{8}$
$K=\frac{7 G M m}{R^{3}}$
