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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?

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44 times

4 times

40 times

14 times

answer is B.

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Detailed Solution

The square of time period(T) is directly proportional to cubic square of radius(R) of orbit in which the satellite is revolving i.e.

T^{2} \propto R^{3}

⸫ (\frac{T_{A}^{2}}{T_{B}^{2}}) = (\frac{R_{A}^{3}}{R_{B}^{3}})

w h ere, T_{A}^{2} = square of time period by satellite A

T_{B}^{2} = square of time period 0 f satellite B

R_{A}^{3} = square cube of radius of orbit of satellite A,

R_{B}^{3} = square cube of radius of orbit of satellite B

Given, T_{A} = 8 T_{B}

⸫ (\frac{T_{A}^{2}}{T_{B}^{2}}) = (\frac{R_{A}^{3}}{R_{B}^{3}})

(\frac{{8 T}_{A}}{T_{B}}) = {(\frac{R_{A}}{R_{B}})}^{3 / 2}

8 = {(\frac{R_{A}}{R_{B}})}^{\frac{3}{2}}

R_{A} = 8^{\frac{3}{2}} R_{B}

R_{A} = 4 R_{B}

Therefore, the distance of satellite A is 4times greater than satellite B from sun.

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