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The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A, in astronomical units, AU. If planet Y is twice the mean distance from the sun as planet X, by what factor is the orbital period increased?

Verified Answer :

maddieboss123

From a series of derivations in the equation, I got C. 2^2/3 though I'm not very sure with that. Hope that helps.

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SociometricStar

Answer:

\(2\sqrt2\) or 2.828

Step-by-step explanation:

The given equation is

\(T^2=A^3\)

Let the mean distance from the sun of planet X is A then

mean distance from the sun of planet Y is 2A.

Therefore, we have

For planet X-

\(T_x^2=A^3....(i)\)

For planet Y-

\(T_y^2=(2A)^3\\\\T_y^2=8A^3....(ii)\)

Divide equation (i) and (ii)

\(\frac{T_x^2}{T_y^2}=\frac{A^3}{8A^3}\\\\\frac{T_x^2}{T_y^2}=\frac{1}{8}\\\\\frac{T_x}{T_y}=\frac{1}{2\sqrt2}\\\\T_y=2\sqrt2T_x\)

Therefore, we can conclude that orbital period is increase by a factor of \(2\sqrt2\) or 2.828

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