What am I supposed to do to solve this?

Question

The intensity of the sunlight that reaches Earth's upper atmosphere is approximately 1400 W/m2.

What is the intensity of sunlight incident on Mercury, which is 5.7e10 m from the Sun?
Answer in W/m^2

Answer 1

Intensity varies as the square of the distance.
That should help

Answer 2

I = P / A
I * A = P

P is a constant, it's the power of the radiation generated by the Sun.

The intensity of sunlight is the power divided by area. The area is that of a sphere with r = (distance from Earth to Sun). Look up the equation and distance...prob in the back of your physics book.
Compute P.

Now P is a constant, but A gets smaller as you get closer to the sun. Find A2, the area of a sphere with r = 5.7e10 (given in the problem). Substitute in the P value you computed above. I = P/A.

The solution looks like this:
I(1) = P(1) / A(1)

where (1) means a subscript. Let's say (1) is a location at the Earth, and (2) is a location at Mercury.

I(2) = P(2) / A(2)

I(2) is the intensity of the light on Mercury, which is what we are to give as the answer.

A(1) = 4*pi*r*r
where r = 149,000,000 km

A(1) = 2.79 e23 sq m

A(2) = 4*pi*r*r
where r= 5.7 e10 meters

A(2) = 4.08 e 22 sq m

P(1) = P(2) = a constant

I(2) = I(1) * A(1) / A(2)

I(2) = 1400 W/m^2 * 2.79 e23 sq m / 4.08 e 22 sq m
I(2) = 9566 W/m^2


Figures and lookups should be checked.

Answer 3

light leaving the sun forms a sphere. there is one sphere that includes mercury. after some time a bigger sphere coincides with Earth. We need the distance of earth from Sun. I assume that's in your book or chapter from whence this question came.

the formula for surface area of sphere would be nice.
sa = 4 pi r^2 (looked it up)

the guy before me, good formula I=P/A
so for earth, 1400 W/m2 = P/A
well for A lets look at 1 m2 (2 as superscript).

1400=P/(4* pi* De^2) (De=distance sun to earth)

same P for mercury
I = P/(4 pi Dm^2)

setting P's equal to each other leaving I for mercury we get
1400 (4piDe^2) = I (4piDm^2)
leaving
I=1400(De^2)/(Dm^2)

and it seems to make sense.
if we use De=2 Dm=1
we get the I equals 4 times as much intensity for mercury as the intensity of earth.