What is the temperature of objects in Low Earth Orbit?
Consider LEO to be 600km to 800km.
[Physics] What are the temperatures of objects in Low Earth Orbit (LEO)
orbital-motionsatellitesspacetemperature
Related Solutions
The Earth receives a certain amount of radiation from the Sun, and also generates a certain amount of heat from radioactivity in the core. It must radiate this amount of energy because otherwise it would heat up until the thermal emission matched the rate of energy input. It's this heat flux that you'd use when calculating how much your plate was heated by the Earth. The sums seem simple enough and I'm sure Google would be able to retrieve the values of heat received from the Sun and generated internally.
Given the heat flux from the Earth you can calculate a temperature using the Stefan Boltzmann law. This temperature wouldn't match the temperature of the Earth's surface, but then why should it? As you say, between the surface and space is a layer of insulating gas. The Stefan Boltzmann temperature would be some kind of average for the various parts of the Earth from which radiation is received. Increasing CO$_2$ or other greenhouse gases would change the temperature profiles between the surface and space, but not the overall amount of energy being emitted.
You can define an emissivity for the Earth, but I'm not sure how helpful this is. If you compare the temperature at the surface with the Stefan Boltzmann temperature you'd get an emissivity less than one, but then this isn't an especially useful comparison.
Intuitive explanation
Suppose you drill two, perpendicular holes through the center of the Earth. You drop an object through one, then drop an object through the other at precisely the time the first object passes through the center.
What you have now are two objects oscillating in just one dimension, but they do so in quadrature. That is, if we were to plot the altitude of each object, one would be something like $\sin(t)$ and the other would be $\cos(t)$.
Now consider the motion of a circular orbit, but think about the left-right movement and the up-down movement separately. You will see it is doing the same thing as your two objects falling through the center of the Earth, but it is doing them simultaneously.
caveat: an important assumption here is an Earth of uniform density and perfect spherical symmetry, and a frictionless orbit right at the surface. Of course all those things are significant deviations from reality.
Mathematical proof
Let's consider just the vertical acceleration of two points, one inside the planet and another on the surface, at equal vertical distance ($h$) from the planet's center:
- $R$ is the radius of the planet
- $g$ is the gravitational acceleration at the surface
- $a_p$ and $a_q$ are just the vertical components of the acceleration on each point
If we can demonstrate that these vertical accelerations are equal, then we demonstrate that the differing horizontal positions have no relevance to the vertical motion of the points. Then we can free ourselves to think of vertical and horizontal motion independently, as in the intuitive explanation.
Calculating $a_q$ is simple trigonometry. It's at the surface, so the magnitude of its acceleration must be $g$. Just the vertical component is simply:
$$ a_q = g (\sin \theta) $$
If you have worked through the "dropping an object through a tunnel in Earth" problem, then you already know that in the case of $p$, its acceleration linearly decreases with its distance from the center of the planet (this is why the "uniform density" assumption is important):
$$ a_p = g \frac{h}{R} $$
$h$ is equal for our two points, and finding it is again simple trigonometry:
$$ h = R (\sin \theta) $$
So:
$$ \require{cancel} a_p = g \frac{\cancel{R} (\sin \theta)}{\cancel{R}} \\ a_p = g (\sin \theta) = a_q $$
Q.E.D.
This also gives some insight to an unfortunate consequence: this method can be applied only to orbits on or inside the surface of the planet. Outside of the planet, $p$ no longer experiences an acceleration proportional to the distance from the center of mass ($a_p \propto h$), but instead proportional to the inverse square of distance ($a_p \propto 1/h^2$), according to Newton's law of universal gravitation.
Best Answer
It depends which object. If the object is a human being in a spaceship, the temperature is approximately 37 °C. If the object is a spherical object that absorbs and re-emits the solar radiation, it's closer to -15 °C in average, like the Earth without the greenhouse effect, and maybe even lower because the solar radiation is shielded by the Earth much of the time. If it's totally shielded from solar radiation, it may acquired the temperature 2.7 K from the cosmic microwave background. Did you expect some universal temperature? Just to be sure, the atmosphere at those high altitudes is de facto non-existent, so this non-existent atmosphere doesn't enforce any unified "air temperature" over there.