Mars has an average surface temperature of about 200K. Pluto has an average surface temperature of about 40K. Which planet emits more energy per square meter of surface area per second? By a factor of how much?

Question

Mars has an average surface temperature of about 200K.  Pluto has an average surface temperature of about 40K.  Which planet emits more energy per square meter of surface area per second? By a factor of how much?

Leah Angelini · Accepted Answer

Mars emits #625# times more energy per unit of surface area than Pluto does.

We already know that Mars will emit more energy than Pluto because it is obviously the case that a hotter object will emit more black body radiation; the only thing left to determine is how much more. In order to solve this problem, the energy of the black body radiation that both planets emit must be calculated. This energy is expressed as a function of temperature and emission frequency: #E(nu,T) = (2pi^2 nu)/c (h nu)/(e^((hnu)/(kT)) - 1)# Integrating over frequency gives the total power per unit area as a function of temperature: #int_0^infty E(nu,T) = (pi^2c(kT)^4)/(60(barhc)^3)# (note that the above equation uses #barh#, the reduced Planck's constant, rather than #h#. It is difficult to read in Socratic's notation) Solving for the ratio between the two, then, the result is incredibly simple. If #T_p# is Pluto's temperature and #T_m# is Mars' temperature then the factor #a# can be calculated with: #(pi^2c(kT_m)^4)/(60(barhc)^3)=a(pi^2c(kT_p)^4)/(60(barhc)^3)# #cancel(pi^2ck^4)/cancel(60(barhc)^3) T_m^4=acancel(pi^2ck^4)/cancel(60(barhc)^3) T_p^4# #(T_m/T_p)^4 = a = (200/40)^4 =5^4 = 625# times as much

Henry Asbill · Answer

undefined The Stefan-Boltzmann law can be used to compare the energy emitted per square meter of surface area per second: E = σ * T^4
Where: T = temperature in Kelvin
E = energy emitted per square meter per second
σ = Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4)
T_mars = 200K
E_mars = 5.67 x 10^-8 * (200^4)
For Pluto: T_pluto = 40K
E_pluto = 5.67 x 10^-8 * (40^4)
E_mars / E_pluto = (200^4) / (40^4)
E_mars / E_pluto = (160,000,000) / (2,560,000)
E_mars / E_pluto = 62.5
Therefore, Mars emits approximately 62.5 times more energy per square meter of surface area per second than Pluto.

Everett Abrams · Answer

undefined The rate at which a planet emits energy per square meter of surface area per second can be calculated using the Stefan-Boltzmann law, which states that the power radiated per unit area of a black body is directly proportional to the fourth power of the body's absolute temperature.

Using the formula:

$$ E = \sigma \cdot T^4 $$

where:
$ E $ = energy emitted per square meter per second,
$ \sigma $ = Stefan-Boltzmann constant ($ 5.67 \times 10^{-8} \, \text{W m}^{-2} \, \text{K}^{-4} $),
$ T $ = absolute temperature in Kelvin.

For Mars:
$ T_{\text{Mars}} = 200 \, \text{K} $

For Pluto:
$ T_{\text{Pluto}} = 40 \, \text{K} $

Substituting the temperatures into the formula:

For Mars:
$$ E_{\text{Mars}} = 5.67 \times 10^{-8} \times (200)^4 $$

For Pluto:
$$ E_{\text{Pluto}} = 5.67 \times 10^{-8} \times (40)^4 $$

Calculating the values:

For Mars:
$$ E_{\text{Mars}} \approx 1960 \, \text{W/m}^2 $$

For Pluto:
$$ E_{\text{Pluto}} \approx 0.0005 \, \text{W/m}^2 $$

Comparing the energy emitted per square meter per second:

$$ \frac{E_{\text{Mars}}}{E_{\text{Pluto}}} \approx \frac{1960}{0.0005} \approx 3,920,000 $$

Therefore, Mars emits approximately $3,920,000$ times more energy per square meter of surface area per second than Pluto.