Lowest possible altitude for a Satellite

[Tom Kunich]

The top of the stratopause (60 km below the 50th parallel or so) has an atmospheric pressure of 7 x 10E-11. This is most assuredly enough drag to decay an orbit rapidly. But the major problem doesn't come from the density of the atmosphere but because an orbit this low requires a rather large velocity forcing the satellite to generate a great deal of drag from hitting a lot of molecules at this speed. What is it? - v = sqrt /Gravity x mass of satellite / actual radius of orbit.

So in truth we do have satellites as low as 100 miles but with the capacity to boost you CAN retain a LEO as low as the stratopause. The latest NASA weather satellite GOES has the capacity to boost its orbit but it has an elliptical orbit with a low of about 8,000 km putting it into the area of medium Earth orbit.

The normal definition of a low Earth orbit is anything below 3,000 km.

Presently SpaceX is planning on planting thousands of satellites into LEO around 1,200 km for more rapid world wide Internet access.

I have been unable to find the latest suggestions for very low Earth orbit weather satellites that would require constant renewal since they would be falling out of the sky almost like rain.

 

[jbriggs444]

v = sqrt /Gravity x mass of satellite / actual radius of orbit

The mass of the satellite does not enter in (*). One can obtain the orbital velocity by equating the centripetal acceleration for a circular orbit $a=\frac{v^2}{r}$ with the centripetal acceleration provided by gravity $a=\frac{GM}{r^2}$.
$$\frac{v^2}{r} = \frac{GM}{r^2}$$
$$v^2 = \frac{GM}{r}$$
$$v = \sqrt{\frac{GM}{r}}$$
The M in the numerator is the mass of the primary (e.g. the mass of the Earth), not the mass of the satellite.

For an orbit at an altitude of less than a few hundred km, the orbital radius will be approximately equal to the radius of the Earth and the orbital velocity will be approximately independent of altitude. About 8 kilometers per second. [Rule of thumb: orbital velocity = escape velocity divided by the square root of two]

(*) The mass of the satellite is irrelevant unless the satellite has significant mass compared to the primary. Then one has to consider that both primary and satellite are orbiting their combined center of mass.

 

[Tom Kunich]

Thanks, that was from memory and I probably was think it had to do with the moon's mass. A little difficult to concentrate when you're about to go to the dentist. And since my waiting 9 months for an implant to heal which just extremely painfully pulled out, at the moment it's even more difficult to concentrate.