Question

Estimate the average power output of the sun, given that about 1350 w/m2 reaches the upper atmosphere of the earth.

Answer 1

the average power output of the sun is 3.8 × $10^{26}$ W given that about 1350 w/m2 reaches the upper atmosphere of the earth.

As we know intensity falling over a surface is equal to Energy falling per unit area per unit volume.

I = $\frac{Energy}{area*time}$

Since Power is equal to energy/time

I = power/area

or

Power = Intensity × area

Now, Distance from Sun to Earth , r = 149.6 ×$10^{9}$  m

So, Surface area of the sphere of radius is:

A = 4 π $r^{2}$

= 4 × 3.14 × (149.6 × $10^{9}$)²

=  2.81 × $10^{23}$m²

Thus Average Power output of the sun will be:

P = 1350 × 2.81 × $10^{23}$

= 3.7935 × $10^{26}$

P = 3.8 ×    $10^{26}$  W

So the average power output of the sun is 3.8 × $10^{26}$ W.

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Answer 2

Answer:

Average power output of sun will be 3.8x10^26 W

Explanation:

As

Power= energy/time                             eq(1)                                                                      

And,      Intensity= energy/area x time      eq(2)                                                              

                                                 From eq(1) & eq(2), we get

Intensity= Power/Area

So Power =intensity x area

And area is 4 π r^2                i.e= surface area of sphere  and radius of earth=r = 149.6 × 10^9 m

 A= 4x3.14x(149.6x10^9)2

A= 2.81x10^23  m2  

And Power will be

P=1350x2.81x10^23

So we will get

P=3.8x10^26 W

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