Question: 31. 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?
(1) 88 earth days
(2) 225 earth days
(3) 172 earth days
(4) 124 earth days
Answer: Option (1)
Explanation:
For planets revolving around the Sun, Kepler’s third law is applicable.
According to Kepler’s third law,
the square of the time period of revolution T is proportional
to the cube of the radius of the orbit r:
T^{2} \propto r^{3}This can be written for two planets as:
\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}Here, let the subscript M denote Mars and
m denote Mercury.
It is given that:
r_{M} = 4 r_{m} T_{M} = 687 \text{ days}Using Kepler’s third law:
\frac{T_{M}^{2}}{T_{m}^{2}} = \left(\frac{r_{M}}{r_{m}}\right)^{3}Substitute the given values:
\frac{687^{2}}{T_{m}^{2}} = 4^{3} = 64So,
T_{m}^{2} = \frac{687^{2}}{64}Taking square root on both sides:
T_{m} = \frac{687}{8}Calculating this value:
T_{m} \approx 85.9 \text{ days}This is approximately equal to 88 Earth days.
Hence, the length of one year on Mercury is about 88 Earth days,
which corresponds to Option (1).