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 I year on Mercury?

The correct option is 88 earth days.

To find the length of a year on Mercury, we can use 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 average orbital radius ($R$) of its orbit around the Sun:

$T^2\propto R^3$

Let:

• $T_M$ be the orbital period (year length) of Mars = 687 Earth days
• $T_{\mathrm{Me}}$ be the orbital period (year length) of Mercury
• $R_M$ be the orbital radius of Mars
• $R_{\mathrm{Me}}$ be the orbital radius of Mercury

According to the problem, the orbital radius of Mars is about 4 times the orbital radius of Mercury:

$R_M=4R_{\mathrm{Me}}\Rightarrow \frac{R_M}{R_{\mathrm{Me}}}=4$

Using the ratio form of Kepler's Third Law for Mars and Mercury:

${\left(\frac{T_M}{T_{\mathrm{Me}}}\right)}^2={\left(\frac{R_M}{R_{\mathrm{Me}}}\right)}^3$

Substitute the given ratio of the radii into the equation:

${\left(\frac{687}{T_{\mathrm{Me}}}\right)}^2=4^3$

${\left(\frac{687}{T_{\mathrm{Me}}}\right)}^2=64$

Taking the square root on both sides:

$\frac{687}{T_{\mathrm{Me}}}=\sqrt{64}$

$\frac{687}{T_{\mathrm{Me}}}=8$

Now, solve for the orbital period of Mercury ($T_{\mathrm{Me}}$):

$T_{\mathrm{Me}}=\frac{687}{8}\approx 85.9\text{Earth days}$

Since the problem states that the orbital radius of Mars is about 4 times that of Mercury, this calculated value is an approximation. Among the given options, the value closest to 85.9 Earth days is 88 Earth days.