Question

Power of sunlight on Earth The Sun emits about 3.9 * 1026 J of electromagnetic radiation each second. (a) Estimate the power that each square meter of the Sun’s surface radiates. (b) Estimate the power that 1 m2 of Earth’s surface receives. (c) What assumptions did you make in part (b)? The distance from Earth to the Sun is about 1.5 * 1011 m and the diameter of the Sun is about 1.4 * 109 m

Answer 1

The power per square meter reaching Earth's surface from the Sun can be estimated by considering the power output of the Sun and the distance between the Sun and Earth. The maximum power that reaches Earth's surface is approximately 1.30 kW/m². The assumptions made in this estimation include the perfect absorption of sunlight by the Earth's surface and the average radius of the Earth.

Step-by-step explanation:

The power per square meter reaching Earth's surface from the Sun can be estimated by considering the power output of the Sun and the distance between the Sun and Earth. The power output of the Sun is given as 4.00 × 10^26 W. To estimate the power reaching Earth's surface, we need to take into account that part of the Sun's radiation is absorbed and reflected by the atmosphere. The maximum power that reaches Earth's surface is approximately 1.30 kW/m².

To estimate the power received by 1 m² of Earth's surface, we can use the fact that the area of the Earth facing the Sun is πR², where R is the radius of the Earth. Assuming the Earth is a perfect sphere, we can take the average radius of the Earth as Re ≈ 6,371 km. Therefore, the power received by 1 m² of Earth's surface can be calculated as (1.30 kW/m²) × (πR²) / (4πRe²), where Re is the radius of the Earth.

In part (b), the assumptions made include the perfect absorption of sunlight by the Earth's surface, the average radius of the Earth, and the assumption that the Earth is a perfect sphere.

Answer 2

a)6.34 x
`10^(7)`W/m²

b)1.37 x
`10^{3` W/m²

c) see explanation.

Step-by-step explanation:

a)The relation of intensity'I' of the radiation and area 'A' is given by:

I= P/A

where P= power of sunlight i.e 3.9 x
`10^(26)` J

and the area of the sun is given by,

A= 4π
`R_(sun)` => 4π
`((1.4*10^(9) )/(2) )^(2)`

A=6.15 x
`10^(18)`m²

`I_(sun)` = 3.9 x
`10^(26)` / 6.15 x
`10^(18)` => 6.34 x
`10^(7)`W/m²

b) First determine the are of the sphere in order to determine intensity at the surface of the virtual space

A= 4π
`R`

Now R= 1.5 x
`10^{11`m

A= 4π x 1.5 x
`10^{11` =>2.83 x
`10^{23` m²

The power that each square meter of Earths surface receives

`I_{earth` = 3.9 x
`10^(26)`/2.83 x
`10^{23` =>1.37 x
`10^{3` W/m²

c) in part (b), by assuming the shape of the wavefront of the light emitted by the Sun is a spherical shape so each point has the same distance from the source i.e sun on the wavefront.