Question

Numerical problems:

The mean distance of Earth from the Sun is 149.6 × 10⁶ km and the mean distance of mercury from the sun is 57.9× 10⁶km. The period of Earth’s revolution is 1 year. What is the period of Mercury’s revolution?

Answer 1

Mean distance of Earth from Sun, r1 = 149.6 × 10 ⁶
2. Mean distance of Mercury from the sun, r2 = 57.9 × 10⁶
3. The period of Earth’s revolution, T1 = 1 year
4. The period of Mercury’s revolution, T2 =?
According to Kepler's Third Law

`"T"_1^2/"r"_1^3 = "T"_2^2/"r"_2^3` ⇒ `"T"_2^2` = `("T"_1^2 × "r"_2^3)/ "r"_1^3`

`"T"_2^2` = `((1)^2 × (57.9 × 10^6)^3)/ (149.6 × 10^6)^3` = `(1,94,105 × 10^18)/(33,48,071 ×10^18` YEAR = 0. 0579

`"T"_2` = `√0.0579`

= 0.2407 year.

`"T"_2` = 0.2407 year = 87.85 days.