(a) Using the solar constant, estimate the rate at which the w...

Question

(a) Using the solar constant, estimate the rate at which the whole Earth receives energy from the Sun. (b) Assume the Earth radiates an equal amount back into space (that is, the Earth is in equilibrium). Then, assuming the Earth is a perfect emitter, (∈ = 1.0) estimate its average surface temperature. [Hint: Discuss why you use area A = πr²_E or A = 4πr²_E in each part.]

Accepted Answer

Welcome back. Everyone in this problem, astronomers are observing a planet of radius R equals 3.6 multiplied by 10 to the 6 m that is orbiting a star. They have determined that the power of the star reaching a unit area perpendicular to the light rays at the planet is 1500 watts. First, we want to calculate the rate at which the star's energy is delivered to the entire planet. And then if the planet had an emissivity of one and it radiated the same amount of energy in space, it received from the star at the same rate. What would be its surface temperature? In part A in part one, we're taking the area A to be pi R squared. And in part two, we're taking the area A to be four pi R squared A says that the rate at which the stars energy is delivered to the planet is 6.1 multiplied by 10 to the 16 watts. And its surface temperature for an emissivity of one would be 290. Kelvin B says there are 6.1 multiplied by 10 to the 16th watts and 400 Kelvin C says there are 2.4 multiplied by 10 to the 17th watts and 290 Kelvin. And these says they are 2.4 multiplied by 10 to the 17th watts and 400 Kelvin respectively. Now, let's start our problem by trying to calculate the power received. So let's call that I, ok. Now, what do we know that is going to help us to find that power? That rate? Let me put that in right here. All right, let me put it in blue, that rate at which the star's energy is delivered to the entire planet. Well, we're told that the power of the star reaching a unit area perpendicular to the light rays at the planet is 1500 watts. OK. So that's our power density. In other words, the power density, the power density due to the star for the planet, OK is equal to 1500 watts per square meter. Now, this power density or I radiance is calculated for an area that is perpendicular to the light rays, OK, to the light rays emitted from the star. So we can think about it this way. But no, that means we're going to consider the cross sectional area of the planet and that cross sectional area is a circle. So in other words, we have our light rays here going to our cross sectional area of a circle. OK? And that is going to be perpendicular to the light rays emitted from the star. Now with a power density of 1500 watts per square meter that just tells us how much watts per unit of area. Ok. So we could rewrite that as P divided by a equals 1500 watts per square meter, which then means the power that we want is going to be equal to 1500 watts per square meter multiplied by our cross sectional area, which is pi R squared. Now, we are, we were told that the radius of the planet is 3.6 multiplied by 10 to the sixth meters. So if we substitute that, we're gonna have 1500 watts per square meter multiplied by pi multiplied by our radius squared. So that's how we can find our po I know when we go ahead and multiply we get the power to be 6.107 multiplied by 10 to the 16th watts or to two significant figures, approximately 6.1 multiplied by 10 to the 16th watts. That's going to be the rate at which the star's energy is delivered to the entire planet. Next, let's try to solve part two of our problem. Now, in part two, like we said here, we're given some more conditions. We're told that the planet has an emissivity of one, it radiates the same amount of energy and space it receives from the star at the same rate. And we want to use that information to help us figure out the surface temperature. Now, what do we know that can help that, that can relate the information we have the power to the surface temperature, taking into account the emissivity and the radiation. Well, again, we can use the Stefan Boltzmann equation to help us recall that the Stefan Boltzmann equation tells us that the power is directly proportional to the temperature by the formula that says the power equals the emissivity multiplied by the Stefan Boltzmann con constant multiplied by the surface area multiplied by the fourth power of the temperature. Now here in this case, the entire planet would emit the radiation. And we know the emissivity is one. And we know that since it's the entire planet, the surface area is four pi R squared, which we know the radius. So we know all of the values here and we know the power from our previous uh part of the problem. So if we can substitute those known values, we will be able to solve for our surface temperature. So first, let's isolate T in our, in our formula and then substitute our known values. Now, if this is what the power equals. OK, then that means the surface temperature to the fourth is going to be equal to the power divided by the emissivity multiplied by the constant multiplied by the surface area, we can find the fourth root of both sides. OK. So T is going to be equal to the fourth root of our expression. And we know that our surface area is four pi R squared. OK. So we could rewrite this thing as P divided by emissivity multiplied by our constant multiplied by four pi R squared. And that's all raised to the fourth pore. Now, like we said, we know what the power is. OK. And we know all of our values so we can go ahead and solve for our surface temperature, which means it's going to be equal to P 6.107 multiplied by 10 to the 16th watts. OK. Divided by our emissivity of one. I always have trouble saying that word multiplied by our Stefan Boltzmann constant 5.67 multiplied by 10 to the negative eight watts per square meter per Kelvin to the fourth multiplied by our surface area. So that's four pi multiplied by a radius, 3.6 multiplied by 10 to 6 m squared. OK? And now we're gonna raise all of this to the fourth power. Now let's substitute this into a calculator and see what we get. Now, when you do that, you should get the temperature, OK? You should get the temperature to be equal to 285.09 Kelvin. And if we write that to two significant figures, it's going to be approximately 290 Kelvin. Thus, the surface temperature is 290 Kelvin, the power is 6.1 multiplied by 10 to the 16 watts which tells us then that a must be the correct answer. Thanks for watching everyone. I hope this video helped.

Key Concepts

Solar Constant

Equilibrium Temperature

Surface Area Calculations

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