# Inverse square law

## Homework Statement

The intensity (I) of sunlight (the received power per unit area) drops with distance (d) from the sun according to the inverse square law - i.e I2/I1 is proportional to (d1/d2)^2

What is the total power received at Earth (above the atmosphere) per unit of surface area?

P.S. The wavelength of peak emission from the surface = 4.99655x10^-7 m
AND the total radiated power per unit of surface area of the sun = 3.901x10^26

## Homework Equations

Inverse square law

## The Attempt at a Solution

Work out distance from earth to sun and square it? Im really not sure how this formula works, I havent used it before.

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Ok, do you have the Solar Radius?

No it wasnt given

You need to look up Solar Radius and Earth-Sun distance for this one.

Simon Bridge
Homework Helper
Can you think of some resource where you can look the solar radius up?
Do you know the distance from the Sun to the Earth?

total radiated power per unit of surface area of the sun = 3.901x10^26
... you left off the units: W/m^2 perhaps?

If this is I1 is the intensity at the radius of the Sun Rs, then what is the intensity at the radius of the Earth Re?

So the equation would look like this:

I2/3.901x10^26 is proportional to (Rs/Re)^2 ??

So the equation would look like this:

I2/3.901x10^26 is proportional to (Rs/Re)^2 ??
Yes.

And can you treat the "proportional to" sign as an equal sign and just rearrange the equation?

Proportionality generally is not the same as equals but when you are dealing with ratios as in this case, any constants cancel out. So yes.

1 person
SteamKing
Staff Emeritus
Homework Helper
Can you think of some resource where you can look the solar radius up?
Do you know the distance from the Sun to the Earth?
the total radiated power per unit of surface area of the sun = 3.901x10^26
... you left off the units: W/m^2 perhaps?

If this is I1 is the intensity at the radius of the Sun Rs, then what is the intensity at the radius of the Earth Re?
The figure 3.901*10^26 watts represents the total power emitted from the surface of the sun, not per sq.m. Anything*10^26 is a truly astounding number.

Oh so what units have I got it in? I just did the SA (4pi*r^2) multiplied by σT^4 to get the power

Simon Bridge
Homework Helper
You did SA (4pi*r^2) without knowing the radius of the Sun?
What did you find the SA of? Maybe it would help if you detailed your initial calculations?

Note:

If ##a \propto b## then ##a=kb## where k is the (unknown, in this case) constant of proportionality.
This is how you turn a proportional symbol into an equals sign.

then ##a_1 \propto b_1 \implies a_1=kb_1## (1) and ##a_2 \propto b_2\implies a_2=kb_2## (2)
... by the same logic. Notice the constant of proportionality is the same because it's the same relation.

then $$\frac{a_1=kb_1}{a_2=kb_2}\implies\frac{a_1}{a_2}=\frac{b_1}{b_2}$$
... because the constants divide out.

It's a very handy approach for getting rid of the bits of equations you don't know.

I used 6.955x10^8 as the r but im not sure where I got that from now?
The question also gives that "the surface area of the sun has an effective temperature of approx. 5800K"
and for the calculation to find the wavelength of peak emission i did:
(2.898x10^-3)/5800 = 4.9965x10^-7

Simon Bridge
Homework Helper
I used 6.955x10^8 as the r but im not sure where I got that from now?
It is good practice to always include the units with your numbers ... they are meaningless without.