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Betelgeuse is 100,000 times more luminous than our sun, which means that it releases an estimated 3.846 x 1031 W of luminous light. If an exoplanet existed with the same mass (5.972 x 10^24 kg) and twice the radius of earth (1.27 x 10^7 m) and half the distance (7.5 x 10^10 m) from Betelgeuse what would the intensity of the light at the surface of that "earth" look like? a) What is the intensity of Betelgeuse at the "earth’s" surface?

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ajeigbeibraheem

Answer:

5.4  × 10⁸ W/m²

Explanation:

Given that:

The Power (P) of Betelgeuse is estimated to release 3.846 × 10³¹ W

the mass of the exoplanet = 5.972 × 10²⁴ kg

radius of the earth = 1.27 × 10⁷ m

half the distance (i.e radius r ) = 7.5  × 10¹⁰ m

a) What is the intensity of Betelgeuse at the "earth’s" surface?

The Intensity of  Betelgeuse  can be determined by using the formula:

[tex]Intensity \ I = \frac{P}{4 \pi r^2}[/tex]

[tex]I = \frac{3.846*10^{31}}{4 \pi (7.5*10^{10})^2}[/tex]

I = 544097698.8 W/m²

I = 5.4  × 10⁸ W/m²

mavila18

Answer:

1.97*10^14 W/m^2

Explanation:

To find the intensity of the light emitted by Betelgeuse you taken into account that the light from Betelgeuse expands spherically in space.

You use the following formula:

[tex]I=\frac{P}{A}=\frac{P}{4\pi r^2}[/tex]     (1)

I: intensity of light

r: distance from Betelgeuse to the exoplanet = (1/2)*7.5*10^10m

P: power = 100,000*3.486*10^31 W

By replacing the values of the parameters in (1) you obtain:[tex]I=\frac{100000(3.486*10^{31}W}{4\pi (0.5*7.5*10^{10}m)^2}=1.97*10^{14}\frac{W}{m^2}[/tex]

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