Mercury orbits the sun at a distance of 0.4 AU. What is mercury’s orbital period in earth
years? Keplers’s Third Law : T^2=d^3. T: orbital period in earth year. D: distance to sun
in astronomical units. 1 AU=150,000,000 or 1.5x10^8 km.
Application of Kepler's third law gives us the following:
"T^2=d^3\u2192\nT=\\sqrt{ d^3}=0.25\\text{ years}."
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