Temperature of a spinning and non spinning asteroid

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SUMMARY

The discussion centers on calculating the equilibrium temperature of a rapidly rotating asteroid with an albedo of 0.05 orbiting a 3 solar mass star, yielding a luminosity of 70.15 solar luminosities. The equilibrium temperature calculated using the formula T4=(L(1-A))/(16*pi*sigma*D2) is 205.48 K. The conversation highlights the distinction between the temperatures of a spinning asteroid and a gravitationally locked one, emphasizing that the spinning asteroid maintains a constant temperature while the locked asteroid experiences significant temperature variation on its star-ward side.

PREREQUISITES
  • Understanding of thermal equilibrium and temperature calculations
  • Familiarity with the Stefan-Boltzmann Law
  • Knowledge of albedo and its effect on temperature
  • Basic concepts of orbital mechanics and semi-major axis
NEXT STEPS
  • Study the Stefan-Boltzmann Law in detail
  • Research the effects of albedo on planetary temperatures
  • Learn about gravitational locking and its implications for temperature distribution
  • Explore the concept of luminosity in relation to stellar mass
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Astronomy students, astrophysicists, and anyone interested in planetary science and thermal dynamics of celestial bodies.

eckerm
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Homework Statement


Consider a rapidly rotating asteroid with an albedo (reflectivity) averaged over the solar spectrum of 0.05 that orbits around a 3 solar mass star that has the same surface temperature as the Sun. The asteroid’s orbit has a semi-major axis of 15 AU. What is its equilibrium temperature? Consider a similar asteroid that does not rotate. What would the average temperature be on its star-ward side?

Homework Equations



T4=(L(1-A))/(16*pi*sigma*D2)

L/Lsun=1.5*(M/Msun)3.5

The Attempt at a Solution



I plugged the mass into the second equation and got that the luminosity of the star in the problem is 70.15 solar luminosities. If I assume the semi-major axis is the average distance from the star, plugging all those numbers into the first equation gives a temperature of 205.48 K. I don't know which temperature this gives me, though, the spinning or stationary. And whichever it gives me, I'm not sure how I should go about finding the other.
 
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eckerm said:
I plugged the mass into the second equation
I'm not familiar with that equation, so I do not know how it is applied. Can you post a link? But it seems to me that the star is the same temperature as the sun and has only a slightly larger radius, so its power output should only be slightly more.
eckerm said:
If I assume the semi-major axis is the average distance from the star
That's a bit of an assumption. Won't it spend rather longer at the greater distances? But you are not given the eccentricity, so I guess you have no choice there.
eckerm said:
a similar asteroid that does not rotate.
I think the question means one that is gravitationally locked, so one day=one year.
eckerm said:
I don't know which temperature this gives me, though, the spinning or stationary.
Because the emission law is to the fourth power, the average emission power over the surface will not be the same as the emission power at the average temperature over the surface. In the spinning case you can take the temperature as constant; in the gravitationally locked case it will be very different.
 

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