7.Gravitation
normal

The distance of neptune and saturn from the sun is nearly $10^{13}$ and $10^{12}$ meter respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio

A

$10$

B

$100$

C

$10\sqrt {10} $

D

$1000$

Solution

${T^2} \propto {R^3}\,\left( {According\,to\,kepler's\,law} \right)$

$T_1^2 \propto {\left( {{{10}^{13}}} \right)^3}\,and\,T_2^2 \propto {\left( {{{10}^{12}}} \right)^3}$

$\therefore \frac{{T_1^2}}{{T_2^2}} = {\left( {10} \right)^3}\,\,or\,\,\frac{{{T_1}}}{{{T_2}}} = 10\sqrt {10} $

Standard 11
Physics

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