Question

The equation 7^2=a^2 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 k times the mean distance from the sun as planet X, by what factor is the orbital period increased?

Answer 1

Answer:

The period of Y increases by a factor of $k^ {3/2}$ with respect to the period of X

Step-by-step explanation:

The equation $T ^ 2 = a ^ 3$ shows the relationship between the orbital period of a planet, T, and the average distance from the planet to the sun, A, in astronomical units, AU. If planet Y is k times the average distance from the sun as planet X, at what factor does the orbital period increase?  

For the planet Y:  

$T_y ^ 2 = a_y ^ 3$

For planet X:  

$T_x ^ 2 = a_x ^ 3$

To know the factor of aumeto we compared $T_x$ with $T_y$

We know that the distance "a" from planet Y is k times larger than the distance from planet X to the sun. So:  

$a_y ^ 3 = (a_xk) ^ 3$

So

$\frac{T_y ^ 2}{T_x ^ 2}=\frac{a_y ^ 3}{a_x^ 3}\\\\\frac{T_y ^ 2}{T_x ^ 2}=\frac{(a_xk)^3}{a_x ^ 3}\\\\\frac{T_y^ 2}{T_x^ 2}=\frac{k ^ {3}a_{x}^ 3}{a_{x}^ 3}\\\\\frac{T_{y}^ 2}{T_{x}^ 2}=k ^ 3\\\\T_{y}^ 2 = T_{x}^{2}k^{3}\\\\T_{y} =k^{\frac{3}{2}}T_x$

Then, the period of Y increases by a factor of $k^ {3/2}$ with respect to the period of X