does anyone know any equations to work out the pressure at 300km above the earths surface?
2006-07-16 23:43:27 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
300 km, or 300,000 metres, is in space, so there is effectively no pressure.
Air pressure falls off exponentially with height; if we call the sea-level pressure 1 Atmosphere (Atm), then at 3000m (about 10,000 feet) the pressure is about two-thirds of an Atm; by the time you've got to the top of Everest (8848m) you are looking at 30% of sea level pressure.
The former US spy plane, the Blackbird, used to fly up to 30,000m, where the pressure is just 1.7% of an Atm. At the same exponential reduction, the proportion of sea level pressure at 300km is 2.47x10e-18 Atm. That is 1 divided by 4 followed by 17 zeroes! So, no air pressure at all.
2006-07-17 01:39:24 · answer #1 · answered by Paul FB 3 · 4⤊ 1⤋
Be carefull with Paul FB's answer. It makes the assumption of uniform composition and temperature. Modelling the atmospheric pressure as an exponential decay is a good approximation at all normal heights but by an altitude of 300Km it will have broken down.
The pressure in low earth orbit varies significantly due to solar activity causing the outermost part of the atmosphere to expand.
2006-07-17 12:25:05 · answer #2 · answered by m.paley 3 · 1⤊ 0⤋
satellite at 350km of m = 10,000kg and crossection of s = 10m^2 looses about h = 50km = 50,000m height in a year without orbit correction. It looses the energy E= m*g*h = 5*10^8 Joules.
This equals to kinetic energy of encountered air: E = m*V^2/2 --> m = sqrt(2E/V^2) ~= 8 kg air was met on the way. (I assumed inelastic scattering, which may be argued reducing the number up to a factor of 2)
At the orbital speed of ~8000 m/secit travels during 1 year l = t*v = 365*24*3600sec*8000 m/sec = 2.4*10^11m, covering the volume V = s*l = 2.4*10^12meters, which contained about 8kg of air, yielding the density of 3*10(-12) kg /m^3 or rougl 4*10^11 less than at the sea level.
2016-03-02 00:01:58 · answer #3 · answered by Ilya 1 · 0⤊ 0⤋
The Chemical Rubber Company (CRC) Handbook of Chemistry and Physics tells you what the pressure at that altitude is and how they obtained it.
2006-07-17 14:46:01 · answer #4 · answered by rb42redsuns 6 · 0⤊ 0⤋
Determine the rate at which the pressure decreases and you have the distance calculate it.
2006-07-17 01:29:48 · answer #5 · answered by tej 2 · 0⤊ 0⤋
very thin atmosphere and so negligible pressure.
2006-07-17 05:57:16 · answer #6 · answered by raj 7 · 0⤊ 0⤋
yes ....been there done that, no t shirts up there tho'
2006-07-16 23:48:03 · answer #7 · answered by Anonymous · 0⤊ 0⤋
no
2006-07-16 23:46:00 · answer #8 · answered by JayClutch 2 · 0⤊ 0⤋
go up and see...........
2006-07-16 23:46:33 · answer #9 · answered by paulrb8 7 · 0⤊ 0⤋