The Sun Is 150 Million Kilometers From The Earth, And The Earth's Equatorial Radius Is 6378 Km. How Much Would You Expect The Sun's Intensity To Drop, Based Only On This Additional Distance?

To calculate how much the Sun's intensity would drop based on the additional distance, we need to understand how the intensity of sunlight decreases with distance.

The intensity of sunlight follows the inverse square law, which states that the intensity is inversely proportional to the square of the distance. In other words, as the distance from a point source (like the Sun) increases, the intensity decreases exponentially.

The formula to calculate the intensity (I) is:

I = P / (4πr²)

where P is the power (or energy) output of the Sun and r is the distance from the Sun.

In this case, we are given that the distance from the Earth to the Sun is 150 million kilometers (or 150,000,000 km).

Let's assume the power output of the Sun remains constant. Therefore, to find the expected drop in intensity, we can compare the intensity at the current distance (r1) to the intensity at the additional distance (r2).

The drop in intensity can be calculated as:

Drop in intensity = (I2 - I1) / I1

Now, let's plug in the values:

r1 = 150 million kilometers
r2 = 150 million kilometers + 6378 kilometers

We need to convert the distance from kilometers to meters for the equation. So:

r1 = 150,000,000 km * 1000 m/km = 150,000,000,000 meters
r2 = (150,000,000 km + 6378 km) * 1000 m/km = 150,006,378,000 meters

With these values, we can calculate:

I1 = P / (4πr₁²)
I2 = P / (4πr₂²)

Finally, we can calculate the drop in intensity:

Drop in intensity = (I2 - I1) / I1

Keep in mind that we need the power output of the Sun to perform this calculation, which we haven't been provided.