Answer to Question #101001 in Astronomy | Astrophysics for kbeshnkoff

Question #101001
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
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Expert's answer
2020-01-06T10:16:23-0500

Kepler's 3rd Law states:

T12T22=a13a23T2=T1a23a13,\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:

T2=1.00(57.9106149.6106)30.24yearsT_2 = 1.00 \cdot \sqrt{\left( \frac{57.9 \cdot 10^6}{149.6 \cdot 10^6}\right)^3} \approx 0.24 \, \text{years}


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