Calculating Orbital Radius from Albedo and Temperature?

Click For Summary
To calculate the orbital radius of a hypothetical planet with an albedo of 0.1 and an atmosphere of carbon dioxide, the escape velocity must be six times the root-mean-square molecular velocity. The escape velocity equation is equated to the molecular velocity equation, leading to a theoretical maximum temperature of 889.87K. The albedo affects the planet's equilibrium temperature by determining how much incoming solar radiation is absorbed. The equilibrium temperature can then be related to the orbital radius through the Stefan-Boltzmann law. Understanding these relationships is crucial for determining the planet's orbital radius around the Sun.
spinnaker
Messages
23
Reaction score
0

Homework Statement



Assume that a planet can have an atmosphere if the escape speed of the planet is 6 times larger than the thermal speed of the molecules in the atmosphere (also known as the root-mean-square molecular velocity). Suppose a hypothetical object having the same mass and radius as Mercury, and an albedo of 0.1, orbits the Sun at just the right location for this condition to be met. Assume that its atmosphere is made of up of carbon dioxide. What is the radius of the this object's orbit around the Sun?

Homework Equations



v_esc = sqrt(2GM/R)
M = M_Merc = 3.30 x 10^23 kg
R = R_Merc = 2.439 x 10^6 m
albedo = 0.1

v_esc = 6 v_rms = 6 sqrt (3kT/M_CO2); M_CO2 = 44(1.67e-27) = 7.348e-26

The Attempt at a Solution



I equated v_esc and 6 v_rms:

sqrt(2GM/R) = 6 sqrt(3kT/M_CO2)

and solved for T, getting me a theoretical maximum temperature of 889.87K.

After this I'm completely stuck. I don't know where albedo kicks in, and I don't know how to get the radius of an orbit around the Sun from the rms velocity or the escape velocity.

Any help would be appreciated. Thanks!
 
Physics news on Phys.org
Once you have the surface temperature you don't need velocities any more. This the astronomy part: if a planet absorbs .1 of the incoming light and acts as a perfect blackbody emitter for infrared, how does its equilibrium temperature depend on the incoming radiation (which then depends on the orbital radius)?
 

Similar threads

Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
1
Views
2K