Calculate Solar Energy Incident on Earth at 1.5x10^11 m from Sun

[student123]

Hi, so I have been asked to find an expression that governs the rate of solar energy incident upon the Earth, if the Earth lies at a distance, D from the Sun of 1.5x10^11 m, and the Earth has a radius, RE. I also know the Sun has a power output of about 4x10^26 watts.

At this point I have considered looking at P = W/T, I think the main issue is I have no clue what the rate of solar energy incident upon the Earth is supposed to look like written out in equation form. I know the question is asking the rate at which the Earth absorbs energy from the sun, but I'm not sure what to do or where to go from here. Please help me! Thank you!

 

[CWatters]

The answer will have units "Watts per square meter".

Imagine a glass sphere of radius D surrounding the sun. All the light from the sun must pass through that sphere. It has area = ??. The Earth blocks what fraction of that area?

 

[haruspex]

The answer will have units "Watts per square meter".

My reading is that it's the total power incident on the Earth that's required, so that will be in Watts.

I have considered looking at P = W/T

Not useful here since there is no mention of an amount of work or a period of time. Just follow CWatters' hint.

 

[CWatters]

My reading is that it's the total power incident on the Earth that's required, so that will be in Watts.

I've had another read of the question and think haruspex is right. They probably want the total power in watts rather than the power per square meter.

 

[HalfLostAndSearching]

Hi there,

To calculate the solar energy incident on Earth at a distance of 1.5x10^11 m from the Sun, we can use the inverse square law, which states that the intensity of radiation decreases with the square of the distance from the source. This means that the further away an object is from the Sun, the less solar energy it receives.

To calculate the rate of solar energy incident on Earth, we can use the formula:

P = (L/4πD^2)A

Where P is the power (in watts), L is the luminosity of the Sun (4x10^26 watts), D is the distance from the Sun (1.5x10^11 m), and A is the surface area of Earth (4πRE^2).

Plugging in the given values, we get:

P = (4x10^26/4π(1.5x10^11)^2)(4π(6.37x10^6)^2)

P = 1.74x10^17 watts

This is the rate at which solar energy is incident on Earth at a distance of 1.5x10^11 m from the Sun. Keep in mind that this is the total amount of solar energy that reaches Earth's atmosphere, and not all of it is absorbed by the Earth.

I hope this helps! Let me know if you have any further questions.