The Sun appears 400,000 times brighter than the full Moon in Earth's sky. How far from the Sun (in astronomical units) would you have to go for the Sun to appear only as bright as the full Moon appears in Earth's nighttime sky? How does the distance you would have to travel compare to the semimajor axis of Neptune's orbit?

The inverse square law is an important concept in physics and astronomy. It states that the brightness or intensity of light (or any other quantity that spreads out evenly in all directions) decreases with the square of the distance from the source. Mathematically, it can be represented as: \( B \propto \frac{1}{d^2} \), where \( B \) is the brightness and \( d \) is the distance.

For example, if you double the distance from a light source, the brightness decreases by a factor of four (\( 2^2 = 4 \)). This is because the light is spreading out over a larger area.

In the context of our exercise, we use the inverse square law to find the distance from the Sun at which its brightness would equal that of the full Moon on Earth. Since the Sun initially appears 400,000 times brighter, we use the following equation:

\[ \frac{B_{\text{Sun}}}{B_{\text{desired}}} = 400,000 = \left( \frac{d_{\text{desired}}}{d_{\text{Earth}}} \right)^2 \].

The desired distance can be isolated and solved using this principle.

Astronomical units (AU) are a standard unit of measurement in astronomy. One AU is defined as the average distance between the Earth and the Sun, approximately 93 million miles or about 150 million kilometers.

This unit makes it easier to compare distances within our solar system without dealing with very large numbers. For instance, Earth's orbit is 1 AU in size.

In our exercise, we need to express the distance where the Sun's brightness is reduced by a factor of 400,000 in AU. Using the mathematical solution from earlier, we found that this distance is approximately 632 AU.

The semimajor axis is the longest radius of an elliptical orbit. It is half the longest diameter of the ellipse. For planets in our solar system, this measurement defines the size of their orbits around the Sun.

Understanding the semimajor axis is crucial because it helps us compare different planetary orbits.

In our exercise, we compare the calculated distance (632 AU) to the semimajor axis of Neptune's orbit, which is about 30 AU. This helps us understand how far we would need to travel to reduce the Sun's brightness to the level of the full Moon.

Neptune is the eighth and farthest planet from the Sun in our solar system. Its orbit has a semimajor axis of approximately 30 AU. This means the average distance from Neptune to the Sun is 30 times the distance between the Earth and the Sun.

In our exercise, we calculated that the distance needed to make the Sun's brightness appear as a full Moon is 632 AU.
This is roughly 21 times the distance of Neptune's orbit.

Having Neptune's orbit as a reference helps us grasp the vast distances involved in our solar system.

Brightness comparison allows us to understand how the intensity of light changes with distance. In optics and astronomy, we often compare the brightness of different objects to make sense of their relative distances and luminance.

In this exercise, we know the Sun appears 400,000 times brighter than the full Moon when viewed from Earth. This drastic difference guides us in finding how far we would need to go from the Sun for it to appear only as bright as the full Moon.

Using the inverse square law, we discovered that this distance is approximately 632 AU.
Relating this to the known distances within our solar system, specifically Neptune's orbit, provides a concrete frame of reference to understand how faint the Sun would need to be for such a comparison.