This figure demonstrates the perturbation forces.

First of all, Weiyong, Massimiliano and Zoltán, thank you very much for your answers. I really appreciate your gesture.

Yes Massimiliano. By "effect" I meant the decelaration a satellite suffers due to the atmospheric drag (I am aware that depends on the satellite shape and attitude. Nevertheless, thank you for your comment).

Weiyong, thank you for the figure. As Zultán said, is very informative.

Zultán, thank you for your explanantion and the references you gave me.

All of you have been of great help.

Kind regards.

From "The Effect Of Radiation Pressure On

The Secular Acceleration Of Satellites" (Stanley P. Wyatt, 1960):

"For perigee heights less than about 800 km the period changes due to radiation pressure are minor compared with those due to atmospheric drag. At greater heights and lower air densities, radiation pressure becomes increasingly

important. When a satellite is in sunshine all around its orbit, the period change

arising from the pressure of sunlight is zero. But during the weeks or months it is penetrating the earth's shadow and is therefore exposed to a photon wind only part of each circuit, the secular acceleration may attain substantial values, positive or negative depending on the orientation of the orbit relative to the sun."

Atmospheric drag however is highly variable and dependent on numerous factors including physical characteristics of the satellite itself and the environmental conditions encountered during its orbit. Two excellent reference sources, (of many) are included:

"Analysis of Satellite Drag", E.M. Gaposchkin and AJ. Coster, and "A Critical Assessment of Satellite Drag and Atmospheric Density Modeling, David A. Vallado, David Finkleman

Thank you Mark. I'll check the references you cited.

Here's an excellent reference that discusses satellite drag.

http://www.ll.mit.edu/publications/journal/pdf/vol01_no2/1.2.6.satellitedrag.pdf

Figure 2. gives all non-gravitational forces including;

(1) radiation pressure;

(2) neutral drag;

(3) charge drag;

You seem to be asking when does radiation pressure equal or exceed neutral drag?

Well, radiation pressure will be opposed to the direction at which light arrives, which isn't always against the direction of motion. Still, since the sun's intensity is fixed at 1 AU from the sun, we can calculate what this is likely to be.

So, remember always thought that the radiation pressure depends on the angle of of light relative to the surface, whether the surface is reflective or not and whether the reflective surface produces specular or diffuse reflections (mirror or matte finish).

Radiation pressure at Earth's orbit is a function of energy flux. The energy flux is 1,368 Watts/m2. This means that for absorbed light that 4.56 micropascals is produced and that for reflected light 9.12 micropascals is produced.

Now the neutral drag is given by

F/A = 1/2 * rho * V^2 * Cd

Cd is equal to 2 for high altitudes.

We know orbital velocity. Orbital velocity at zero altitude is 7.89 km/sec for Earth. In low orbit velocities range from 7.8 km/sec to 6.9 km/sec. Taking the forces above for light we can calculate what the density (rho) has to be in order for an object moving at these velocities to feel the same force as produced by light.

4.56 micropascals = 1/2 * rho * (7800 m/sec)^2 * 2

rho = 4.56 micropascals / (7800 m/sec)^2 = 7.5e-14 kg/m3

Doubling this figure yields 1.5e-13 kg/m3 to reproduce the higher figure.

A rule of thumb is that air density drops by a factor of 1/10th for every 16 km increase in altitude. So, we have

altitude = log ( P/p) * 16 = log(1.22/7.5e-14) * 16 = 211km.

altitude = log (P/p) * 16 = log(1.22/1.5e-13) * 15 = 207 km.

So, above 207 km light pressure has the potential to be a larger factor than air drag (neutral drag) for highly reflective surfaces. Above 211 km light pressure has the potential to be a largrer factor than air drag for black surfaces.

Thank you William. All this information about atmospheric drag is a new topic for me because I've been working with GPS data since I got my degree. So, any comment about atmospheric drag is really appreciated.

Since most polar orbiting satellite operate within 500 – 1000, your idea of 850km is still within the range and not absolute. This material may give you a clue to your question as there is no definite altitude where this occurs.

------ - If you give a space vehicle exactly escape velocity, it will just barely escape the gravitational field, which means that its velocity will be approaching zero as its distance from the force center approaches infinity. If, on the other hand, we give our vehicle more than escape velocity at a point near Earth, we would expect the velocity at a great distance from Earth to be approaching some finite constant value. This residual velocity the vehicle would have left over even at infinity is called hyperbolic excess velocity. We can calculate this velocity from the energy equation written for two points on the hyperbolic escape trajectory – a point near Earth called the burnout point and a point an infinite distance from Earth where the velocity will be the hyperbolic excess velocity, v∞. Solving for v∞ we obtain

Note that if v∞ = 0 (as it is on a parabolic trajectory), the burnout velocity, vbo, becomes simply the escape velocity.

It is, or course, absurd to talk about a space vehicle "reaching infinity" and in this sense it is meaningless to talk about escaping a gravitational field completely. It is a fact, however, that once a space vehicle is a great distance from Earth, for all practical purposes it has escaped. In other words, it has already slowed down to very nearly its hyperbolic excess velocity. It is convenient to define a sphere around every gravitational body and say that when a probe crosses the edge of this sphere of influence it has escaped. Although it is difficult to get agreement on exactly where the sphere of influence should be drawn, the concept is convenient and is widely used, especially in lunar and interplanetary trajectories. For most purposes, the radius of the sphere of influence for a planet can be calculated as follows:

where Dsp is the distance between the Sun and the planet, Mp is the mass of the planet, and Ms is the mass of the Sun. Equation (4.89) is also valid for calculating a moon's sphere of influence, where the moon is substituted for the planet and the planet for the Sun.

Thank you Jaiye...

After reading all the articles recommended in the discussion that followed my question, I've come to the conclusion that it's quite difficult to determine a transition height (where radiation pressure overcomes atmospheric drag), basically because these two phenomena depend on a lot of factors that vary on a daily (and hourly) basis.

However, I think that using an altitude between 850-900 km is a good rough estimation of the aforementioned transition height. But, I want to underline that this would be only a first order approximation.

It depends on multiple factors and some of which dominate only specific altitude which unfortunately make it very complex do not forget the geometry as well but this graph should be a good approximation.