Question

The mean distance of Earth from the Sun is $149.6\times {10}^6$ km and the mean distance of Mercury from the Sun is $57.9\times {10}^6$ km. The period of Earth’s revolutions is 1 year, what is the period of Mercury’s revolution?

• A. Infinite
• B. 0.24 year on earth
• C. 3.2 year on earth
• D. 2.1 year on earth

Answer 1

The correct option is B 0.24 year on earth

$\frac{T_1^2}{r_1^3}=\frac{T_2^2}{r_2^3}$

$1=$ Earth; $2=$ Mercury

$\frac{1^2}{{(149.6\times {10}^6)}^3}=\frac{T_2^2}{{(57.9\times {10}^6)}^3}$

$\Rightarrow \frac{1}{3348071.936\times {10}^{18}}=\frac{T_2^2}{194104.539\times {10}^{18}}\Rightarrow T_2^2=0.057975\Rightarrow T_2=\sqrt{0.057975}$

$T_2=0.24$ year on Earth

$1$ year on Earth $=365$ days