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A planet of radius R has an acceleration due to gravity of gs on its surface. A deep smooth tunnel is dug on this planet, radially inward, to reach a point P located at a distance of R/2 from thecentre of the planet. Assume that the planet has uniform density. The kinetic energy required to be given to a small body of mass m, projected radially outward from P, so that it gains amaximum altitude equal to the thrice the radius of the planet from its surface, is equal to

Utkarsh mishra , 6 Years ago

Grade 12

2 Answers

Eshan

Last Activity: 6 Years ago

Dear student,

Gravitational potential energy of mass at point P=

$-\dfrac{GMm}{R}(\dfrac{3}{2}-\dfrac{r^2}{2R^2}) =-\dfrac{GMm}{R}(\dfrac{3}{2}-\dfrac{1}{8})=-\dfrac{11}{8}\dfrac{GMm}{R}$

Gravitational potential energy of mass at the given altitude= $-\dfrac{GMm}{4R}$

Hence the gain in gravitational potential energy is compensated by the loss in kinetic energy.

$\implies \dfrac{1}{2}mv^2=\dfrac{11}{8}\dfrac{GMm}{R}-\dfrac{GMm}{4R}=\dfrac{9}{8}\dfrac{GMm}{R}$
$\implies v=\sqrt{\dfrac{9GM}{4R}}$

Gitanjali Rout

Last Activity: 6 Years ago

$\implies v=\sqrt{\dfrac{9GM}{4R}}$ $\implies \dfrac{1}{2}mv^2=\dfrac{11}{8}\dfrac{GMm}{R}-\dfrac{GMm}{4R}=\dfrac{9}{8}\dfrac{GMm}{R}$ Hence the gain in gravitational potential energy is compensated by the loss in kinetic energy. $-\dfrac{GMm}{4R}$ Gravitational potential energy of mass at the given altitude= $-\dfrac{GMm}{R}(\dfrac{3}{2}-\dfrac{r^2}{2R^2}) =-\dfrac{GMm}{R}(\dfrac{3}{2}-\dfrac{1}{8})=-\dfrac{11}{8}\dfrac{GMm}{R}$ Gravitational potential energy of mass at point P=