# The distance of two planets from sun are nearly 1014 and 1012 meters. Assuming that they move in circular orbits, their periodic times will be in the ratio Option 1) 100 Option 2) 10 Option 3) 1000 Option 4) $100\sqrt{10}$

As we learnt in

Kepler's 3rd law -

$T^{2}\: \alpha\: a^{3}$

From fig.

$AB=AF+FB$

$2a=r_{1}+r_{2}$

$\therefore\; a=\frac{r_{1}+r_{1}}{2}$

$a=$ semi major Axis

$r_{1}=$ Perigee

- wherein

Known as law of periods

$r_{2}=$ apogee

$T^{2}\: \alpha \: \left ( \frac{r_{1}+r_{2}}{2} \right )^{3}$

${r_{1}+r_{2}= 2a$

$T^{2}\propto R^{3}\Rightarrow \left ( \frac{T_{2}}{T_{1}} \right )^{2}=\left ( \frac{R_{2}}{R_{1}} \right )^{3}=\left ( \frac{10^{12}}{10^{14}} \right )^{3}=10^{-6}$

$\therefore T_{2}=10^{-3}T_{1}$    or     $\frac{T_{1}}{T_{2}}=1000$

Option 1)

100

This is incorrect option

Option 2)

10

This is incorrect option

Option 3)

1000

This is correct option

Option 4)

$100\sqrt{10}$

This is incorrect option

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