How Does Sunlight Intensity Translate to Force on Earth?

[sgoeke]

Homework Statement

The intensity of light from the Sun is 1.3 kW/m^2 at the distance of the Earth. The Earth’s radius is 6.4 × 10^6 m. If all the sunlight that hits the Earth were absorbed, what would be the magnitude of the resulting force on the Earth?

Homework Equations

F=IA/c

The Attempt at a Solution

I converted KW/m^2 to W/m^2 and got 1300 W/m^2. For area I used 4pi*r^2 and plugged those values into the equation but I can't seem to get the correct answer. Any help?

 

[alphysicist]

Hi sgoeke,

Homework Statement

The intensity of light from the Sun is 1.3 kW/m^2 at the distance of the Earth. The Earth’s radius is 6.4 × 10^6 m. If all the sunlight that hits the Earth were absorbed, what would be the magnitude of the resulting force on the Earth?

Homework Equations

F=IA/c

The Attempt at a Solution

I converted KW/m^2 to W/m^2 and got 1300 W/m^2. For area I used 4pi*r^2

This formula gives the surface area of a sphere. Do you see why this is the wrong area to use for this problem?

and plugged those values into the equation but I can't seem to get the correct answer. Any help?

 

[sgoeke]

would i need to find the volume of the earth?

 

[alphysicist]

would i need to find the volume of the earth?

No, the formula needs an area; the question is which area to use. It's not the surface area of the Earth (after all, the "back" of the Earth is not getting sunlight at all). And it is not just half of the total surface area. What type of area is important here? Remember that the idea is to find how much of the sunlight the Earth is "catching", in a sense.

 

[rwisz]

I would approach this problem by first clarifying the units used. The intensity of light is typically measured in units of watts per square meter (W/m^2), while force is measured in units of newtons (N). It is important to use consistent units in order to properly calculate the resulting force.

Using the given values, the intensity of light at the distance of the Earth can be calculated as 1.3 kW/m^2 or 1,300 W/m^2. The Earth's radius can be converted to meters as 6.4 x 10^6 m.

Next, we can use the equation F=IA/c to calculate the resulting force on the Earth. Here, I represents the intensity of light, A represents the area of the Earth that is exposed to the sunlight, and c represents the speed of light. Plugging in the values, we get:

F = (1,300 W/m^2) x (4π x (6.4 x 10^6 m)^2) / (2.998 x 10^8 m/s) = 1.7 x 10^17 N

This is a very large force, but it is important to note that not all of the sunlight that hits the Earth is absorbed. Some is reflected or scattered back into space, so the actual force on the Earth would be less than this calculated value.

In summary, the magnitude of the resulting force on the Earth from the intensity of light is approximately 1.7 x 10^17 N.