Calculate The Surface Temperature of Mercury

AI Thread Summary
To estimate the surface temperature of Mercury, the discussion emphasizes using the inverse square law of infrared radiation and the distance from the Sun. The distance from Mercury to the Sun is calculated as approximately 5.635 x 10^7 m, based on the Earth-Sun distance. The ratio method for temperature comparison suggests using temperatures in Kelvin for accuracy, as the average surface temperature of Mercury is debated, with estimates ranging from 97 degrees Celsius to around 1650 degrees Celsius. The lack of an atmosphere on Mercury means it loses heat rapidly at night, complicating temperature calculations. Overall, the discussion highlights the challenges in accurately determining Mercury's surface temperature due to its unique conditions.
TFM
Messages
1,016
Reaction score
0

Homework Statement



Estimate the surface temperature of Mercury (the planet). Mercury is distant 0.38 ORE from the Sun, where the Earth-Sun distance, ORE is 1.49.108m. The radii of the Sun, Earth and Mercury are 696.103, 6.38.103 and 2.42.103 km respectively.

Homework Equations



Heat (IR Radiation) follows inverse square Law

The Attempt at a Solution



See I am confused with why the Sun Earth and Mercury's radius are on here. I know that Infra Red radiation follows the inverse square law, so I would ave thought you would have used this to find the temperature on Mars, however, the radius are given, so I am slightly confused...is there more to this question?

Thanks in Advance,

TFM
 
Last edited:
Physics news on Phys.org
TFM,

It's asking you to find the surface temperature. When it references Mercury-Sun distance it's probably referencing the distance between their COMs.
 
Okay so the Distance from the Earth to the Sun is

1.49*10^8 m

So we take out the sun/Earth radius, giving:

1.483*10^8 m

the distance from Sun to the Mercury is 0.38x this, so:

5.635*10^7m

The Surface Temperature of Earth is 14 degrees C

So now Do we have to use ratios to find the temperature of Mercury using:

T \propto \frac{1}{d^2}

ie

\frac{T_{Earth}}{T_{Mercury}} \propto \frac{\frac{1}{d_{Earth}^2}}{\frac{1}{d_{Mercury}^2}}


?
 
Okay, I have tried out my ratio methiod, and it gives me a average surface temperature of 97 Degrees Celsius. Does this sound right?

Looking on the web, I have found a maximum Daytime temp of about 400, minimum of -200, which gives an average of 100 degrees C, so does seem to agree there...

Does this look okay?

TFM
 
Does the above look okay?
 
I've been wondering lately how to calculate Mercury's surface temperature, so I'm glad you asked this.

Your logic sounds good. However, when taking the ratios of the temperatures, they need to be in degrees Kelvin, which is Celsius plus 273 degrees, the absolute zero point in Celsius. If you do this and apply the inverse square law, you'll get an average temperature of 1650 deg Celsius, which is too high.

This approach would probably work better for gases rather than rock. Also, Mercury has no atmosphere to speak of, so most of whatever heat it receives during its day is radiated away at night, more so than it would be on the earth.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top