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

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

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

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