Estimate the rms electric field in the sunlight that hits Mars...

Question

Estimate the rms electric field in the sunlight that hits Mars, knowing that the Earth receives about 1350 W/m² and that Mars is 1.52 times farther from the Sun (on average) than is the Earth.

Accepted Answer

Welcome back. Everyone in this problem, a theoretical physicist is trying to calculate the R MS electric field of sunlight that reaches Jupiter. Given that Jupiter is 5.2 times farther from the sun than earth. And the intensity of sunlight at earth is 1370 watts per square meter. What must her calculation be? A says it's 22 volts per meter. B 35 volts per meter, C 110 volts per meter and D 140 volts per meter. Now let's try to visualize what's going on here. OK. So let's say that we have the sun and then we have the earth, the path of the earth can rotate or revolving around the sun and then we have the path of Jupiter. OK. So let's just put that in here and let's just make some space for the earth right here. So let's put E for earth and then let's make some space for Jupiter. OK? It's a little bigger and then we can put J for Jupiter. OK. So we're looking at this situation and now we know that Jupiter is 5.2 times farther from the earth. So that means the radius of Jupiter's orbit is going to be equal to 5.2 multiplied by the radius of earth's orbit around the sun. OK. And on our diagram, we can have RJ and we can have RE OK. And no, we want to figure out the R MS electric field of sunlight that reaches Jupiter given information about the earth, that is the intensity of sunlight that the earth is 1370 watts per square meter. So how can we figure out that value? How can we find the RMS electric field of sunlight that reaches Jupiter? Well, we know that the sun, the sun sorry has the same power at all distances. OK. So in other words, its power at Jupiter is going to be equal to its power at the earth. If we consider spheres at various distances centered around, centered at the sun, the total power through each of them due to the sunlight would be the same. So since their powers are the same, we can equate their powers. And now we also know that their power generally power is equal to the magnitude of the pointing vector which is equal to the intensity multiplied by the surface area. A OK. The areas of the spheres at the planets center at the sun. So with this idea, then we could rewrite our formula as the pointing vector for Jupiter multiplied by the surface area of Jupiter is going to be equal to the pointing vector for the earth multiplied by the surface area of the earth. Now, what do we know about both of these? Well, we know that the surface area is going to be four pi R squared, OK, four pi R squared. And OK, we also know that our pointing vector, the magnitude of our pointing vector equals our average intensity which is going to be equal to the speed of light multiplied by the primitive of free space multiplied by the electric field, the RMS electric field squared. So if we apply that idea here to replace or pointing vector for um or pointing vector for Jupiter. With that expression, we should be able to solve for the R MS electric field of sunlight that reaches Jupiter because no, that tells us then that SJ which is gonna be C epsilon knot er MS J squared. OK. Multiplied by the surface area for Jupiter four pi RJ squared. OK is going to be equal two se multiplied by four pi R squared. Now here, if we, if we were to look at, if we were to look at our formula four pi is in command for both terms, so we can divide through OK. And know if we were to solve for MS J, then ERM SJ squared is going to be equal to se divided by C epsilon knot multiplied by re squared divided by RJ squared, which we could rewrite as re OK. Divided by RJ. I'll quit. Now, if we were to square root both sides of our equation, then we get our RMS electric field or Jupiter to be equal to the square root of S divided by C epsilon knot multiplied by re divided by RJ, which we know that RJ is 5.2 multiplied by re. Now we can go ahead and solve for our Armis electric field because we can substitute our values. Like we said, we know that the intensity of sunlight at earth is 1370 watts per square meter. And now we're dividing all of that, ok? By the speed of light three multiplied by 10 to the 8 m per second, multiplied by the permittivity of free space. 8.85 multiplied by 10 to the negative 12 square coulombs per Newton square meters multiplied by one divided by 5.2. Because here, notice that ie cancers. Now when we go ahead and try to solve, then we should get the R MS field for J, ok? For Jupiter to be equal to 137.13 volts per meter. And if we were to write at two significant figures, it's approximately 140 volts per meter. Thus, that tells us that the army electric field of sunlight that reaches Jupiter is 142 volts per meter. Therefore, D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Estimate the rms electric field in the sunlight that hits Mars, knowing that the Earth receives about 1350 W/m² and that Mars is 1.52 times farther from the Sun (on average) than is the Earth.

Key Concepts

Inverse Square Law

RMS Electric Field

Solar Constant

Watch next

Estimate the rms electric field in the sunlight that hits Mars, knowing that the Earth receives about 1350 W/m2 and that Mars is 1.52 times farther from the Sun (on average) than is the Earth.

Key Concepts

Inverse Square Law

RMS Electric Field

Solar Constant

Watch next

Estimate the rms electric field in the sunlight that hits Mars, knowing that the Earth receives about 1350 W/m² and that Mars is 1.52 times farther from the Sun (on average) than is the Earth.