5. now lets make period measurements for different orbits of the solar system. then, calculate the relation…

Explanation:

Step1: Recall orbital - period and semi - major axis data

For Mercury: The period $T_{Mercury}\approx0.241$ years and semi - major axis $a_{Mercury}\approx0.387$ AU. Then $\frac{T_{Mercury}}{a_{Mercury}}=\frac{0.241}{0.387}\approx0.623$ AU/years.

Step2: Calculate for Earth

For Earth: $T_{Earth} = 1$ year and $a_{Earth}=1$ AU. Then $\frac{T_{Earth}}{a_{Earth}}=\frac{1}{1}=1$ AU/years.

Step3: Calculate for Mars

For Mars: $T_{Mars}\approx1.88$ years and $a_{Mars}\approx1.524$ AU. Then $\frac{T_{Mars}}{a_{Mars}}=\frac{1.88}{1.524}\approx1.234$ AU/years.

Step4: Analyze patterns

a. The relation $\frac{T}{a}$ changes. The differences are: $\Delta_{1}=1 - 0.623 = 0.377$ (between Earth and Mercury) and $\Delta_{2}=1.234 - 1=0.234$ (between Mars and Earth). b. $\frac{T}{a}$ is bigger for Mars compared to Earth and Mercury. c. The graph of $T$ vs $a$ is a curve. As $a$ increases, $T$ increases, but not linearly.

Step5: Find constant - relation exponents

Kepler's third law states that $T^{2}=k\cdot a^{3}$ (where $k$ is a constant). If we rewrite it in terms of a constant $\frac{T^{n}}{a^{m}} = 1$, we find that when $n = 2$ and $m = 3$, $\frac{T^{2}}{a^{3}}$ is constant for all orbits in the solar - system. The graph of $T^{2}$ vs $a^{3}$ is a straight line passing through the origin.

Answer:

Planet	Period, T (years)	Semi - major axis, a (AU)	T/a (AU/years)
Mercury	0.241	0.387	0.623
Earth	1	1	1
Mars	1.88	1.524	1.234
a. The relation $\frac{T}{a}$ changes. Differences are 0.377 (Earth - Mercury) and 0.234 (Mars - Earth).
b. It is bigger for Mars.
c. The graph of $T$ vs $a$ is a curve.
For the constant relation, period power $n = 2$, semi - major axis power $m = 3$, and the graph of $T^{2}$ vs $a^{3}$ is a straight line.