the equation ( t^{2}=a^{3} ) shows the relationship between a planets orbital period, ( t ), and the planets…

the equation ( t^{2}=a^{3} ) shows the relationship between a planets orbital period, ( t ), and the planets 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?\n( k^{\frac{1}{3}} )\n( k^{\frac{1}{2}} )\n( k^{\frac{2}{3}} )\n( k^{\frac{3}{2}} )

the equation ( t^{2}=a^{3} ) shows the relationship between a planets orbital period, ( t ), and the planets 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?\n( k^{\frac{1}{3}} )\n( k^{\frac{1}{2}} )\n( k^{\frac{2}{3}} )\n( k^{\frac{3}{2}} )

Answer

Explanation:

Step1: Find the orbital period of planet X

Let the mean - distance of planet X from the sun be (A_X). Using the formula (T^{2}=A^{3}), for planet X, we have (T_X^{2}=A_X^{3}), so (T_X = A_X^{\frac{3}{2}}).

Step2: Find the orbital period of planet Y

The mean - distance of planet Y from the sun is (A_Y=kA_X). Using the formula (T^{2}=A^{3}), for planet Y, we have (T_Y^{2}=A_Y^{3}=(kA_X)^{3}=k^{3}A_X^{3}). Then (T_Y=\sqrt{k^{3}A_X^{3}}=k^{\frac{3}{2}}A_X^{\frac{3}{2}}).

Step3: Find the factor by which the orbital period is increased

We want to find (\frac{T_Y}{T_X}). Substitute (T_X = A_X^{\frac{3}{2}}) and (T_Y=k^{\frac{3}{2}}A_X^{\frac{3}{2}}) into (\frac{T_Y}{T_X}). Then (\frac{T_Y}{T_X}=\frac{k^{\frac{3}{2}}A_X^{\frac{3}{2}}}{A_X^{\frac{3}{2}}}=k^{\frac{3}{2}}).

Answer:

(k^{\frac{3}{2}}) (the fourth option)