Question

now lets make period measurements for different orbits of the solar system. then, calculate the relation t/a. for this, remember to use the target orbit panel to recreate real world orbits. target orbit. mars period, t (years) semi - major axis, a (au) t/a (au/years) mercury earth mars 6. do you see any patterns of period vs semi - major axis? a. does the relation t/a change? how much? b. for which planets is it bigger? c. draw the shape of the graph:

Explanation:

Step1: Recall known values

The period $T$ and semi - major axis $a$ values for Mercury, Earth and Mars are:
Mercury: $T_{Mercury}=0.2408$ years, $a_{Mercury} = 0.3871$ AU.
Earth: $T_{Earth}=1$ year, $a_{Earth}=1$ AU.
Mars: $T_{Mars}=1.8808$ years, $a_{Mars}=1.5237$ AU.

Step2: Calculate $T/a$ for Mercury

$T_{Mercury}/a_{Mercury}=\frac{0.2408}{0.3871}\approx0.622$ AU/years

Step3: Calculate $T/a$ for Earth

$T_{Earth}/a_{Earth}=\frac{1}{1} = 1$ AU/years

Step4: Calculate $T/a$ for Mars

$T_{Mars}/a_{Mars}=\frac{1.8808}{1.5237}\approx1.234$ AU/years

Step5: Analyze the pattern of $T/a$

The relation $T/a$ changes. The difference between the $T/a$ of Earth and Mercury is $1 - 0.622=0.378$ AU/years. The difference between the $T/a$ of Mars and Earth is $1.234 - 1 = 0.234$ AU/years.
The value of $T/a$ is bigger for Mars compared to Earth and Mercury.

Step6: Consider the graph shape

According to Kepler's third law, $T^{2}\propto a^{3}$, or $T = k\sqrt{a^{3}}$ where $k$ is a constant. So the graph of $T$ vs $a$ is a non - linear curve, and the graph of $T/a$ vs $a$ is also a non - linear curve. It will show an increasing trend as $a$ increases.

Planet	Period, T (years)	Semi - major axis, a (AU)	T/a (AU/years)
Earth	1	1	1
Mars	1.8808	1.5237	1.234

Answer:

a. Yes, the relation $T/a$ changes. The differences in $T/a$ values between planets are as calculated above.
b. The value of $T/a$ is bigger for Mars.
c. The graph of $T/a$ vs $a$ is a non - linear increasing curve.