Question

Highway 6563 in New Mexico has signs along a roughly \(30 \mathrm{~km}\) stretch of road corresponding to the solar system scale from the Sun to Neptune. On this scale, how large would Earth, Sun, and Jupiter be, in diameter? Express in convenient units appropriate to the scale.

Short answer

Answer: On the scale of Highway 6563 in New Mexico, Earth would have a diameter of around 0.08509 meters, the Sun would have a diameter of about 9.279 meters, and Jupiter would have a diameter of approximately 0.933 meters.

Step-by-step solution

Find the scale factor

To find the scale factor, we first need to know the actual distance from the Sun to Neptune. The average distance between the Sun and Neptune is approximately 4.5 billion kilometers (4.5 x 10^9 km). Given the model's 30 km distance representation, we can find the scale factor using the formula: Scale factor = Model distance / Actual distance Let's calculate the scale factor: Scale factor = 30 km / (4.5 x 10^9 km) ≈ 6.67 x 10^(-9) This means that 1 km in the model represents about 6.67 x 10^(-9) times the actual distance in the solar system.

Calculate the scaled diameter of Earth

To calculate the scaled diameter of Earth, we need to know Earth's actual diameter, which is approximately 12,742 km. Using the scale factor found previously, we can find the scaled diameter: Scaled diameter of Earth = Actual diameter of Earth × Scale factor Scaled diameter of Earth ≈ 12,742 km × 6.67 x 10^(-9) ≈ 8.509 x 10^(-5) km To express this in a more convenient unit, we can convert it to meters: Scaled diameter of Earth ≈ 8.509 x 10^(-5) km × (1000 m/km) ≈ 0.08509 m

Calculate the scaled diameter of the Sun

To calculate the scaled diameter of the Sun, we need to know its actual diameter, which is approximately 1,391,000 km. Using the scale factor found previously, we can find the scaled diameter: Scaled diameter of the Sun = Actual diameter of the Sun × Scale factor Scaled diameter of the Sun ≈ 1,391,000 km × 6.67 x 10^(-9) ≈ 0.009279 km To express this in a more convenient unit, we can convert it to meters: Scaled diameter of the Sun ≈ 0.009279 km × (1000 m/km) ≈ 9.279 m

Calculate the scaled diameter of Jupiter

To calculate the scaled diameter of Jupiter, we need to know its actual diameter, which is approximately 139,822 km. Using the scale factor found previously, we can find the scaled diameter: Scaled diameter of Jupiter = Actual diameter of Jupiter × Scale factor Scaled diameter of Jupiter ≈ 139,822 km × 6.67 x 10^(-9) ≈ 0.000933 km To express this in a more convenient unit, we can convert it to meters: Scaled diameter of Jupiter ≈ 0.000933 km × (1000 m/km) ≈ 0.933 m In conclusion, on the scale of Highway 6563 in New Mexico, Earth would have a diameter of around 0.08509 meters, the Sun would have a diameter of about 9.279 meters, and Jupiter would have a diameter of approximately 0.933 meters.

Key concepts

Scale Factor Calculation

Understanding the scale factor is essential when creating a scale model of the solar system, like the one found along Highway 6563 in New Mexico. The scale factor represents the ratio between the actual size of objects or distances in space and their corresponding size or distance in a model.

The calculation process involves dividing the model distance by the actual distance. In the exercise example, for the solar system model stretching 30 km, representing the distance to Neptune, we used the actual distance to Neptune, which is about 4.5 billion kilometers, to determine the scale factor. The result, approximately 6.67 x 10^-9, indicates how many kilometers in the solar system are equivalent to one kilometer in the model.

With the scale factor, we can accurately depict astronomical sizes in a way that's tangible for students. It allows us to represent the vast distances and sizes in the solar system on a comprehensible human scale without losing the proportional relationships among celestial bodies.

Astronomical Unit Conversion

Astronomical units are a convenient way to express the massive distances in space. One astronomical unit (AU) is roughly the distance from the Earth to the Sun, about 150 million kilometers.

To make astronomical sizes more approachable, conversion to AUs can be used, especially when addressing interplanetary distances within our solar system. However, in our example involving the scale model, we worked with kilometers and meters for greater accuracy and easier understanding when relating to the scale factor.

In real-world applications, converting planetary diameters to kilometers before applying the scale factor ensures precision. But when working with enormous distances between celestial objects, the AU serves as a manageable unit, bridging the daunting gap between human-scale distances and the expansive reaches of space.

Planetary Diameters

The diameters of planets are key to appreciating their relative sizes and imagining them on a smaller scale, like the model along Highway 6563. Planetary diameters vary widely, with Earth's diameter being 12,742 km, Jupiter's massive girth at 139,822 km, and the Sun's colossal figure at 1,391,000 km.

In scale models, we use these actual diameters in conjunction with the scale factor to obtain a scaled-down representation that maintains proportionality. This allows us to visualize the significant size differences between planets, such as the relatively small Earth compared to the giant gas planet Jupiter, and the sheer size of the Sun.

It's important to note that these diameters are normally average values, as planets are not perfect spheres and their measurements can vary slightly at different points. Nonetheless, for educational representations and scale models, average diameters suffice to illustrate the sheer diversity in our solar system's design.