Question

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? A. 2^1/3. B. 2^1/2. C. 2^2/3. D. 2^3/2

Answer 1

The given equation for the relationship between a planet's orbital period, T and the planet's mean distance from the sun, A is T^2 = A^3. Let the orbital period of planet X be T(X) and that of planet Y = T(Y) and let the mean distance of planet X from the sun be A(X) and that of planet Y = A(Y), then A(Y) = 2A(X) [T(Y)]^2 = [A(Y)]^3 = [2A(X)]^3 But [T(X)]^2 = [A(X)]^3 Thus [T(Y)]^2 = 2^3[T(X)]^2 [T(Y)]^2 / [T(X)]^2 = 2^3 T(Y) / T(X) = 2^3/2 Therefore, the orbital period increased by a factor of 2^3/2

Answer 2

Answer:

The orbital increased by factor $2^{\frac{3}{2}}$.Hence, correctander is option D.

Step-by-step explanation:

Given;$T^2=A^3$

Distance of the planet X from the sun = d

Time-period of the planet X:

$T^2=d^3$..(1)

Distance of the planet Y from the sun = d'= 2d

$T'^2=d'^2=(2d)^3=8d^3$..(2)

On dividing (1)and (2).

$\frac{T'}{T}=\frac{8d^3}{d^3}=8=2\sqrt{2}=2^1\times 2^{\frac{1}{2}}=2^{\frac{3}{2}}$

The orbital increased by factor $2^{\frac{3}{2}}$.Hence, correctander is option D.