Question #5af5e

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Cosmic Defect
Dec 3, 2017

#T^2\proptoa^3; \qquad \rightarrow (T_{"ast"}/T_{\oplus})^2 = (a_{"ast"}/a_{\oplus})^3#
#T_{"ast"} = T_{\oplus}(\frac{a_{"ast"}}{a_{\oplus}})^{3/2} = (1\quad Yr)(\frac{2.1\quad AU}{1.0\quadAU})^{3/2} = 3.04\quad Yrs#

Explanation:

Kepler's Third Law: The square of the orbital time period is proportional to the cube of the "planet's" mean distance from the "sun".

#T^2 = (\frac{4\pi^2}{GM_{\odot}}) a^3#

Earth: #\qquad T_{\oplus} = 1 \quad Yr; \qquad a_{\oplus} = 1\quad AU;#
Asteroid: #\qquad T_{"ast"} = ?;\qquad a_{"ast"} = 2.1\quad AU;#

#T^2\proptoa^3; \qquad \rightarrow (T_{"ast"}/T_{\oplus})^2 = (a_{"ast"}/a_{\oplus})^3#

#T_{"ast"} = T_{\oplus}(\frac{a_{"ast"}}{a_{\oplus}})^{3/2} = (1\quad Yr)(\frac{2.1\quad AU}{1.0\quadAU})^{3/2} = 3.04\quad Yrs#

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