Question

The period of revolution of planet A around 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 ??????

Answer 1

Here you need to approach accordance to keplars 3rd law
That explain relation of time period to mean distance

Answer 2

The distance of A from  the Sun is 4 times greater than that of B  from the Sun?

Explanation:

From Kepler's law we know that if T is the time period and r is the distance of the planet from the Sun then

$T^2\propto r^3$

Terefore, for the planets A and B

$\frac{T_A^2}{T_B^2}=\frac{r_A^3}{r_B^3}$

Given that

$\frac{T_A}{T_B}=8$

$\implies 8^2=\frac{r_A^3}{r_B^3}$

$\implies \frac{r_A^3}{r_B^3}=64$

$\implies \frac{r_A}{r_B}=\sqrt[3]{64}$

$\implies \frac{r_A}{r_B}=4$

$\implies r_A=4r_B$

Thus, the distance of A from the Sun is 4 times greater than that of B

Hope this helps.

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