The period of revolution of the planet $$A$$ round the sun is 8 times that of $$B.$$ The distance of $$A$$ from the sun is how many times greater than that of $$B$$ from the sun?
A.
5
B.
4
C.
3
D.
2
Answer :
4
Solution :
According to Kepler's third law $${T^2} \propto {r^3}$$
where, $$T =$$ Time period of revolution
$$r =$$ Semi major axis
$$\eqalign{
& \therefore \frac{{T_A^2}}{{T_B^2}} = \frac{{r_A^3}}{{r_B^3}} \cr
& \therefore \frac{{{r_A}}}{{{r_B}}} = {\left( {\frac{{{T_A}}}{{{T_B}}}} \right)^{\frac{2}{3}}} = {\left( 8 \right)^{\frac{2}{3}}} = {2^{3 \times \frac{2}{3}}} = 4 \cr
& {\text{or}}\,\,{r_A} = 4{r_B} \cr} $$
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