Question

Consider a bright star in our night sky. Assume its distance from the Earth is 20.0 light-years (ly) and its power output is 4.00 3 1028 W, about 100 times that of the Sun. (a) Find the intensity of the starlight at the Earth. (b) Find the power of the starlight the Earth intercepts. One light-year is the distance traveled by

Answer 1

intensity is 8.9 x
`10^(-8)` W/m²

power is 11.3 x
`10^(6)` W

Step-by-step explanation:

Given data

distance = 20 light year

power output = 4.00 x 10^28

to find out

intensity of the starlight and the power of the starlight

solution

we know the intensity formula that is

intensity = power / area

so here power is 4.00 x 10^28 and area = 4π(20ly)²

intensity = 4.00 x 10^28 / (4π (20×9.46×10^15)²

intensity = 8.9 x
`10^(-8)`

so intensity is 8.9 x
`10^(-8)` W/m²

and

power is Intensity × Area

so power = 8.9 x
`10^(-8)` × (π (6.37 × 10^6)²

power = 11.3 x
`10^(6)`

so power is 11.3 x
`10^(6)` W