The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?

The correct option is 88 earth days.

To find the length of 1 year on Mercury, we can apply Kepler's Third Law of Planetary Motion. Kepler's Third Law states that the square of the orbital period ($T$) of a planet is directly proportional to the cube of the semi-major axis of its orbit ($R$):

$T^2\propto R^3$

This relationship can be written as a ratio for two different planets, Mars ($M$) and Mercury ($m$):

${\left(\frac{T_m}{T_M}\right)}^2={\left(\frac{R_m}{R_M}\right)}^3$

From the problem statement, we are given:
- The orbital radius of Mars is about 4 times that of Mercury: $R_M=4R_m$, which means $\frac{R_m}{R_M}=\frac{1}{4}$
- The orbital period of Mars is: $T_M=687\text{days}$

Substituting these values into Kepler's Third Law relation:

${\left(\frac{T_m}{687}\right)}^2={\left(\frac{1}{4}\right)}^3$

${\left(\frac{T_m}{687}\right)}^2=\frac{1}{64}$

Taking the square root of both sides gives:

$\frac{T_m}{687}=\sqrt{\frac{1}{64}}$

$\frac{T_m}{687}=\frac{1}{8}$

Solving for $T_m$:

$T_m=\frac{687}{8}\approx 85.875\text{Earth days}$

Comparing this value to the given choices, the value is closest to 88 Earth days, which matches option D.