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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
