Assume the planetary orbits to be circular and T be the time period of revolution of the earth around the sun of mass M. If the Jupiter’s period of revolution around the sun is K times that of the earth, the velocity of Jupiter in heliocentric reference frame is ${[\frac{2 πGM}{{TK}^{1 / y}}]}^{1 / x}$ .The product of x and y is

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answer is 3.

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Detailed Solution

${[\frac{T_{J}}{T}]}^{2} = {[\frac{R_{J}}{R}]}^{3} = K^{2}$

$\Rightarrow R_{J} = K^{2 / 3} R \dots \dots \dots . . (1)$

For earth:

$\frac{{GMm}_{e}}{R^{2}} = m_{e} R \frac{4 π^{2}}{T^{2}}$
$\Rightarrow T^{2} = \frac{4 π^{2} R^{3}}{GM}$

$\Rightarrow R = {[\frac{T^{2} GM}{4 π^{2}}]}^{1 / 3}$ ............(2)

For Jupiter:
$\frac{{GMm}_{J}}{R_{J}^{2}} = \frac{m_{J} \cdot V^{2}}{R_{J}}$

$\Rightarrow V = \sqrt{\frac{GM}{R_{J}}} = \sqrt{\frac{GM}{K^{2 / 3} R}}$

$∴ V = {[\frac{2 πGM}{TK}]}^{1 / 3}$

$\Rightarrow xy = 3$

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