Problem 56 - University Physics with Modern Physics

Question

What is the radiation pressure due to Betelgeuse (which has a luminosity, or power output, 10,000 times that of the Sun) at a distance equal to that of Uranus's orbit from it?

Step-by-Step Solution

Answer

Answer: The radiation pressure due to Betelgeuse at a distance equal to Uranus's orbit is approximately 7.42 x 10^(-9) N/m^2.

Step 1: Constants

Write down the constants: Luminosity of the Sun (L_sun) = 3.8 x 10^26 W Luminosity of Betelgeuse (L_B) = 10,000 x L_sun Distance of Uranus's orbit from the Sun (d) = 2.9 x 10^12 m Speed of light (c) = 3.0 x 10^8 m/s

Step 2: Calculate the luminosity of Betelgeuse

Calculate the luminosity of Betelgeuse by multiplying the given factor by the solar luminosity: L_B = 10,000 × L_sun = 10,000 × (3.8 × 10^26 W) = 3.8 × 10^30 W

Step 3: Calculate the radiation pressure

Use the formula for radiation pressure and substitute the values: P = L_B / (4 * π * d^2 * c) Substitute the values: P = (3.8 × 10^30 W) / (4 * π * (2.9 x 10^12 m)^2 * 3.0 x 10^8 m/s)

Step 4: Solve for P

Calculate the radiation pressure: P ≈ 7.42 × 10^(-9) N/m^2 So the radiation pressure due to Betelgeuse at a distance equal to that of Uranus's orbit from it is approximately 7.42 x 10^(-9) N/m^2.

Key Concepts

Luminosity of Stars

When we talk about the luminosity of stars, we're referring to the amount of energy a star emits per second. It is effectively the star's power output, measured in units of watts (W). This property is a crucial indicator of many other stellar characteristics, including mass and temperature. For instance, our Sun has a luminosity of approximately 3.8 x 10^26 W, which may seem formidable, but is moderate compared to some other stars.

Now, consider Betelgeuse, a red supergiant star, whose luminosity is a staggering 10,000 times that of the Sun. To calculate Betelgeuse's luminosity, we multiply this factor by the Sun's luminosity to yield a luminous near 3.8 x 10^30 W. This immense energy output is central to understanding the impacts Betelgeuse can have, including radiation pressure exerted at a distance, as seen in our exercise.

Distance and Radiation Relationship

The relationship between distance and the radiation received from a star is a fundamental principle of astrophysics known as the inverse-square law. This law states that the intensity of an effect (such as gravity, light, sound, or in our case, radiation pressure) is inversely proportional to the square of the distance from the source of that effect.

In simpler terms, if you double the distance from the star, the radiation pressure you experience is not just halved; it is reduced to a quarter of its original value. This radical decrement is due to the dispersion of radiation over a larger area as you move away from the source. This concept is why the potentially devastating radiation from Betelgeuse, at a distance equivalent to Uranus's orbit, is lessened to a pressure that's much easier to comprehend (and far less hazardous!).

Radiation Pressure Formula

The formula for radiation pressure is given by: \( P = \frac{L}{4 \pi d^2 c} \) where \( P \) is the radiation pressure, \( L \) is the luminosity of the star, \( d \) is the distance from the star, and \( c \) is the speed of light. Here, the formula encapsulates the essence of both the inverse-square law and the direct proportionality of pressure to the star's luminosity, mitigated by the constant that is the speed of light.

To solve for \( P \) in the context of our exercise involving Betelgeuse and its staggering luminosity, we plugged in the appropriate values and determined the radiation pressure on a hypothetical object located at the distance of Uranus's orbit. The calculation demonstrates a direct application of the formula and provides an understanding of the immense power of stellar objects, even at vast distances.