The mean radius of the earth's orbit roun | vedclass

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Question

$\frac{T_{1}^{2}}{q_{1}^{3}}=\frac{T_{2}^{2}}{r_{2}^{3}}$

$\frac{1}{\left(1 \cdot 5 	imes 10^{11}ight)^{3}}=\frac{T_{2}^{2}}{\left(6 	imes 10^{

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Nearly $\frac{1}{4}$ year

Vedclass · Answer

$\frac{T_{1}^{2}}{q_{1}^{3}}=\frac{T_{2}^{2}}{r_{2}^{3}}$

$\frac{1}{\left(1 \cdot 5 	imes 10^{11}ight)^{3}}=\frac{T_{2}^{2}}{\left(6 	imes 10^{10}ight)^{3}}$

$T_{2}=\sqrt{\left(\frac{6 	imes 10^{10}}{1.5 	imes 10^{11}}ight)^{3}}$

$=\frac{\sqrt{\left(\frac{4}{10}ight)^{3}}}{\sqrt{\frac{64}{1000}}}$

$=\frac{8}{10 \sqrt{10}}$

$=\frac{0.8}{3.14}$

$\simeq 0.25$

The mean radius of the earth's orbit round the sun is $1.5 \times 10^{11}.$ The mean radius of the orbit of mercury round the sun is $6 \times10^{10}\,m.$ The mercury will rotate around the sun in

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