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Grade 12Mechanics

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 the
centre 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 a
maximum altitude equal to the thrice the radius of the planet from its surface, is equal to

Utkarsh mishra

8 Years agoGrade 12

2 Answers

Eshan

8 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

8 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=

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