Comparison of two data sets with corresponding mathematical expressions for analysis.

Let's analyze both Data Set 5 and Data Set 6, determine the mathematical relationships between the variables, and then write the corresponding mathematical expressions.

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🔹 Data Set 5

| R (a.u.) | T (yr) |
|---------|--------|
| 0.38 | 0.24 |
| 0.72 | 0.62 |
| 1.00 | 1.00 |
| 1.52 | 1.88 |
| 5.19 | 11.9 |
| 9.53 | 29.5 |
| 19.1 | 84.1 |
| 30.0 | 165 |
| 39.4 | 249 |

We are to find a relationship between R (distance in astronomical units) and T (orbital period in years).

This resembles Kepler’s Third Law, which states:

> $ T^2 \propto R^3 $

Or more precisely:
> $ T^2 = k \cdot R^3 $

Let’s test this by calculating $ T^2 $ and $ R^3 $ for a few values.

#### Try: $ T^2 / R^3 $
For each row, compute $ \frac{T^2}{R^3} $:

1. $ R = 0.38, T = 0.24 $
- $ T^2 = 0.0576 $
- $ R^3 = 0.054872 $
- $ T^2/R^3 ≈ 1.053 $

2. $ R = 0.72, T = 0.62 $
- $ T^2 = 0.3844 $
- $ R^3 = 0.373248 $
- $ T^2/R^3 ≈ 1.029 $

3. $ R = 1.00, T = 1.00 $
- $ T^2 = 1 $
- $ R^3 = 1 $
- $ T^2/R^3 = 1 $

4. $ R = 1.52, T = 1.88 $
- $ T^2 = 3.5344 $
- $ R^3 = 3.511808 $
- $ T^2/R^3 ≈ 1.006 $

5. $ R = 5.19, T = 11.9 $
- $ T^2 = 141.61 $
- $ R^3 ≈ 140.1 $
- $ T^2/R^3 ≈ 1.01 $

All values of $ T^2/R^3 $ are approximately 1, with small variations due to rounding.

✔ So, the relationship is:

> $ T^2 = R^3 $

Or:

> $ T = R^{3/2} $

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✔ Mathematical Expression #5

$$
\boxed{T = R^{3/2}}
$$

This is Kepler's Third Law for objects orbiting the Sun (in AU and years). It shows that the square of the orbital period is proportional to the cube of the semi-major axis.

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🔹 Data Set 6

| r (m) | F_g (N) |
|-------|--------|
| 0.2 | 425 |
| 0.5 | 68.3 |
| 1.0 | 16.5 |
| 2.0 | 4.26 |
| 5.0 | 0.67 |
| 10.0 | 0.18 |
| 20.0 | 0.042 |

We are to find the relationship between r (distance in meters) and F_g (gravitational force in Newtons).

This looks like inverse square law, such as Newton’s Law of Gravitation:

> $ F \propto \frac{1}{r^2} $

Let’s test this by computing $ F_g \cdot r^2 $ — if it's constant, then $ F_g \propto 1/r^2 $

Compute $ F_g \cdot r^2 $ for each row:

1. $ r = 0.2 $, $ F = 425 $
- $ r^2 = 0.04 $
- $ F \cdot r^2 = 425 × 0.04 = 17 $

2. $ r = 0.5 $, $ F = 68.3 $
- $ r^2 = 0.25 $
- $ F \cdot r^2 = 68.3 × 0.25 = 17.075 $

3. $ r = 1.0 $, $ F = 16.5 $
- $ r^2 = 1 $
- $ F \cdot r^2 = 16.5 $

4. $ r = 2.0 $, $ F = 4.26 $
- $ r^2 = 4 $
- $ F \cdot r^2 = 4.26 × 4 = 17.04 $

5. $ r = 5.0 $, $ F = 0.67 $
- $ r^2 = 25 $
- $ F \cdot r^2 = 0.67 × 25 = 16.75 $

6. $ r = 10.0 $, $ F = 0.18 $
- $ r^2 = 100 $
- $ F \cdot r^2 = 0.18 × 100 = 18 $

7. $ r = 20.0 $, $ F = 0.042 $
- $ r^2 = 400 $
- $ F \cdot r^2 = 0.042 × 400 = 16.8 $

Values are all around 17, with minor variation due to rounding or measurement error.

So, we can say:

> $ F_g \cdot r^2 \approx 17 $

Thus,

> $ F_g = \frac{17}{r^2} $

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✔ Mathematical Expression #6

$$
\boxed{F_g = \frac{17}{r^2}}
$$

This matches the inverse-square law for gravitational force, where the constant 17 depends on the masses involved and the gravitational constant.

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📌 Final Answers:

#### Mathematical Expression #5:
$$
\boxed{T = R^{3/2}}
$$

#### Mathematical Expression #6:
$$
\boxed{F_g = \frac{17}{r^2}}
$$

These are the best-fit mathematical models based on the data provided.

Let me know if you'd like graphs or further analysis!