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We are given the brightness formula:

\[
b = \frac{L}{4\pi d^2}
\]

where:
- \( b \) is the brightness,
- \( L = 3.9 \times 10^{26} \) watts is the luminosity of the sun,
- \( d \) is the distance from the sun (we're interested in Jupiter's distance, \( 7.79 \times 10^{11} \) meters).

We need to find the derivative of \( b \) with respect to \( d \), which is \( \frac{db}{dd} \).

### Step 1: Differentiate the brightness equation with respect to \( d \).

The equation for \( b \) is:

\[
b = \frac{L}{4\pi d^2}
\]

Differentiating both sides with respect to \( d \), we apply the power rule to \( d^{-2} \):

\[
\frac{db}{dd} = \frac{d}{dd}\left(\frac{L}{4\pi d^2}\right) = \frac{-2L}{4\pi d^3} = \frac{-L}{2\pi d^3}
\]

### Step 2: Plug in the values for \( L \) and \( d \).

Using \( L = 3.9 \times 10^{26} \) watts and \( d = 7.79 \times 10^{11} \) meters, we substitute into the derivative expression:

\[
\frac{db}{dd} = \frac{-(3.9 \times 10^{26})}{2\pi (7.79 \times 10^{11})^3}
\]

I'll calculate this next.The derivative of the brightness \( b \) with respect to the distance \( d \) at Jupiter's distance from the Sun is approximately:

\[
\frac{db}{dd} \approx -1.31 \times 10^{-10} \, \text{watts per square meter per meter}
\]

Would you like more details or explanations? Here are five related questions:

1. How does the brightness change as the distance increases?
2. What is the physical interpretation of the derivative of brightness with respect to distance?
3. How would the brightness change if you were at half the distance from the sun?
4. How does the inverse-square law relate to other fields, like gravity or radiation intensity?
5. What would the brightness at Earth’s distance (1.496 × 10¹¹ m) be?

**Tip**: The inverse-square law applies in many areas of physics, not just light but also sound and gravity!

The brightness b of the sun (in watts per square meter) at a distance d meters from the sun is expressed as an inverse-square law in the form b = L / (4πd^2), where L is the luminosity of the sun and equals 3.9 × 10^26 watts. What is the derivative of b with respect to d at the Jupiter's distance from the sun (7.79 × 10^11 m)?

Find the derivative of the brightness of the sun with respect to distance at Jupiter's distance (7.79 × 10^11 m) from the sun using the inverse-square law. Learn how to apply differentiation and scientific notation to solve this advanced calculus problem.

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