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7.Gravitation
medium
If the distance of the earth from Sun is $1.5 \times 10^6\,km$. Then the distance of an imaginary planet from Sun, if its period of revolution is $2.83$ years is $.............\times 10^6\,km$
A
$0.6$
B
$6$
C
$3$
D
$0.3$
(JEE MAIN-2023)
Solution
$T ^2 \propto R ^3 \Rightarrow\left(\frac{ T _1}{ T _2}\right)^2=\left(\frac{ R _1}{ R _2}\right)^3$
$\Rightarrow\left(\frac{1}{2.83}\right)^2=\left(\frac{1.5 \times 10^6}{ R _2}\right)^3$
$\Rightarrow R _2=\left[(2.83)^2 \times\left(1.5 \times 10^6\right)^3\right]^{1 / 3}$
$=8^{1 / 3} \times 1.5 \times 10^6=3 \times 10^6\,km$
Standard 11
Physics
