Answer to Question #101001 in Astronomy | Astrophysics for kbeshnkoff

Question

the mean distance of the earth from the sun is 149.6*10^6 km and the mean distance of mercury from the sun is 57.9*10^6km thm the period of earth's revolution is 1 year use keepler's law of planetary motion to determine the period of mercury's revolution

Answer 1

Kepler's 3rd Law states:

$\frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3} \quad \Rightarrow T_2= T_1 \sqrt{\frac{a_2^3}{a_1^3}},$

where index "1" relates to the Earth whereas index "2" describes Mercury.

Substituting the numerical values, we obtain:

$T_2 = 1.00 \cdot \sqrt{\left( \frac{57.9 \cdot 10^6}{149.6 \cdot 10^6}\right)^3} \approx 0.24 \, \text{years}$