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

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)^{(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)^{(3)/(2)}=k^{(3)/(2)}\cdot (A)^{(3)/(2)}` ....[2]

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

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

Simplify:

`(T_(Y))/(T_(X)) =k^{(3)/(2)}`

or

`T_(Y) =k^(3)/(2) T_X`

Therefore, the orbital period is increased by factor
`k^(3)/(2)`

Answer 2

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.