# Hw help, its been a while since last physics course

1. Jan 31, 2006

### justinbaker

The Sun emits energy at a rate of about 3.9 · 1026W. At Earth, this sunshine gives
an incident energy flux Ie of about 1.4kWm−2. In this problem, you’ll investigate
whether any other planets in our solar systemcould support the sort of water-based
life we find on Earth.
Consider a planet orbiting at distance d from the Sun (and let de be Earth’s distance).
The Sun’s energy flux at distance d is I = Ie(de/d)2, because energy flux decreases
as the inverse square of distance. Call the planet’s radius R, and suppose that
it absorbs a fraction α of the incident sunlight, reflecting the rest back into space. The
planet intercepts a disk of sunlight of area πR2, so it absorbs a total power of πR2αI.
The Sun has been shining for a long time, but Earth’s temperature is roughly
stable: The planet is in a steady state. For this to happen, the absorbed solar energy
must get reradiated back to space as fast as it arrives (see Figure 1.2). Because the rate
at which a body radiates heat depends on its temperature, we can find the expected
mean temperature of the planet, using the formula

In this formula, σ denotes the number 5.7·10−8Wm−2 K−4 (the “Stefan–Boltzmann
constant”). The formula gives the rate of energy loss per unit area of the radiating
body (here, the Earth). You needn’t understand the derivation of this formula but
make sure you do understand how the units work.
a. Using this formula, work out the average temperature at the Earth’s surface and
b. Using the formula, work out how far from the Sun a planet the size of Earth may
be, as a multiple of de, and still have a mean temperature greater than freezing.
c. Using the formula, work out how close to the Sun a planet the size of Earth may
be, as a multiple of de, and still have a mean temperature below boiling.
d. Optional: If you know the planets’ orbital radii, which ones are then candidates
for water-based life, using this rather oversimplified criterion?

2. Feb 1, 2006

### cepheid

Staff Emeritus
What have you done so far on the question? Post your attempts, and we can help point you in the right direction.

3. Feb 1, 2006

### justinbaker

for a.) i know i need to solve for T.

so is the radiated heat flux the same as incident energy flux Ie? If so then how do i solve for alpha, in the equation

total pwr absorbed = pi x R^2 x alpha x I

4. Feb 1, 2006

### justinbaker

can anyone help me?

5. Feb 1, 2006

### chroot

Staff Emeritus
Does the problem really not give you a value for alpha (alpha is known as the "albedo")?

The equation for energy in = energy out is:

$\pi r^2 \alpha I = \sigma A T^4$

Note that &alpha; on the left hand side is the albedo, while A on the right hand side is the surface area of the planet. They are not the same.

Of these, you know pi, r, I, sigma, and A. You need to know alpha; too, then it's just a simple matter of solving for T.

- Warren

6. Feb 1, 2006

### justinbaker

no it does not give a value for alpha, everything in the problem is in my orginal post.

i am still lost on the other ones as well

7. Feb 1, 2006

### chroot

Staff Emeritus
Well, if I were you, I'd just pick a value for alpha, say 0.5, and explicitly write your assumption on your paper.

Part b uses the same formula, using I = Ie(de/d)^2, T = 273 K (the freezing point of water), solved for d.

- Warren

8. Feb 1, 2006

### justinbaker

but how do i get from the alpha x sigma x T^4 equation to the equation I = Ie(de/d)^2?

9. Feb 1, 2006

### chroot

Staff Emeritus
The equation is the same for every problem in this set. Energy in = energy out:

$\pi r^2 \alpha I = \sigma A T^4$

where

$I = I_e \left( \frac{d_e}{d} \right)^2$

Thus, the equation is, in full generality:

$\pi r^2 \alpha I_e \left( \frac{d_e}{d} \right)^2 = \sigma A T^4$

- Warren