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 k times the mean distance from the sun as planet X, by what factor is the orbital period increased? <<

Answer 1

If planet Y is k times the mean distance from the sun than planet X, the right side of the equation becomes (kA)^3. which is k^3 times the left side, T^2. To equate both sides of the equation, multiply T by k^3/2 so that the left side becomes ((k^3/2) x T)^2 which simplifies into (k^3) x (T^2). Therefore, the answer is k^3/2. 

Answer 2

Answer:

Given 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

then:

For planet X:

Orbital period is:

$T_{X} = (A)^{\frac{3}{2}}$            .....[1]

As per the statement:

If planet Y is k times the mean distance from the sun as planet X.

⇒ A planet Y= kA mean distance from the sun as planet X.

then orbital period of Planet Y is:

$T_{Y} = (kA)^{\frac{3}{2}}=k^{\frac{3}{2}}\cdot (A)^{\frac{3}{2}}$    ....[2]

Divide equation [2] by [1] we have;

$\frac{T_{Y}}{T_{X}} = \frac{k^{\frac{3}{2}}\cdot (A)^{\frac{3}{2}}}{A^{\frac{3}{2}}}$

Simplify:

$\frac{T_{Y}}{T_{X}} =k^{\frac{3}{2}}$

or

$T_{Y} =k^\frac{3}{2} T_X$

Therefore, the orbital period is increased by factor $k^\frac{3}{2}$