Difference between revisions of "Gravity and notes about Kittinger's jump"
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g at North Pole: 9.832 m / s^2 | g at North Pole: 9.832 m / s^2 | ||
| − | == Edge of space -- | + | == Edge of space -- “Kármán Line” == |
| − | + | The definition of "space" is arbitrary. The atmosphere doesn't just end and space begins. It's all rather fuzzy. Theodore von Kármán (or Karman) came up with a satisfying definition in the 1950's. His idea was to ask, "At what altitude is the air so thin that an aircraft would have to fly '''faster''' than orbital velocity to gain any aerodynamic lift?" He calculated that this altitude is roughly 100 kilometers (62 mi). Basically, this is the altitude at which a plane would have to fly so fast to stay up that it would basically be in orbit anyway, so you might as well stop worrying about aerodynamics at this point. At this altitude, the atmospheric pressure is 3.2×10<sup>−2</sup> Pa and the density is 5.6×10<sup>−7</sup> kg/m<sup>3</sup>. So the '''Kármán Line''' is a point where the air is so thin that aerodynamic lift is not useful. This is a nice definition for the '''edge of space'''. | |
| + | |||
| + | In practice air drag are aerodynamic forces are still significant higher than the '''Kármán Line'''. That is, it's not practical to actually orbit at the '''Kármán Line''' because your orbit would decay so quickly that you wouldn't make a single complete orbit. This line is the edge between '''flying''' and orbiting. Obviously, the design of the aircraft will effect this altitude limit, so it's still a fuzzy line (each aircraft should have it's own '''Kármán Line'''), but this idea is a good general starting place for defining what it means to be '''in space'''. | ||
| + | |||
| + | You still get some aerodynamic control above the '''Kármán Line''' -- control surfaces can effectively alter the orientation of your craft, but you don't get much useful lift above 100 kilometers. NASA uses 122 kilometers (75.8 miles) as their definition of "re-entry" for the Space Shuttle. This is the altitude where it's time for the pilots to start paying attention to aerodynamics and start using aerodynamic control surfaces for maneuvering. Above this altitude the pilots rely on attitude control rocket thrusters to control their orientation. | ||
Karman Line altitude: 100 kilometers, 62 miles | Karman Line altitude: 100 kilometers, 62 miles | ||
Revision as of 15:05, 23 December 2011
Contents
Circular Earth orbits
Velocity required for orbit is independent of mass and size. A lead bullet orbits at the same velocity as a feather.
The force of the Moon's gravity is 83.3% (or 5/6) less than Earth's force of g. It is less dense, so difference in size does not correspond to difference in g. So if you weigh 175 pounds on Earth you would weigh about 30 pounds on the Moon.
The Moon is 1/4 of the diameter of the Earth. mass of the Moon = 7.36 * 10^22 kilograms mass of Earth = 5.97 * 10^24 kilograms Radius of Earth = 6378 kilometers, 3963 miles
The force of the Earth's gravity (g) at various places:
g at San Francisco: 9.800 m / s^2 g at Denver: 9.796 m / s^2 g at North Pole: 9.832 m / s^2
Edge of space -- “Kármán Line”
The definition of "space" is arbitrary. The atmosphere doesn't just end and space begins. It's all rather fuzzy. Theodore von Kármán (or Karman) came up with a satisfying definition in the 1950's. His idea was to ask, "At what altitude is the air so thin that an aircraft would have to fly faster than orbital velocity to gain any aerodynamic lift?" He calculated that this altitude is roughly 100 kilometers (62 mi). Basically, this is the altitude at which a plane would have to fly so fast to stay up that it would basically be in orbit anyway, so you might as well stop worrying about aerodynamics at this point. At this altitude, the atmospheric pressure is 3.2×10−2 Pa and the density is 5.6×10−7 kg/m3. So the Kármán Line is a point where the air is so thin that aerodynamic lift is not useful. This is a nice definition for the edge of space.
In practice air drag are aerodynamic forces are still significant higher than the Kármán Line. That is, it's not practical to actually orbit at the Kármán Line because your orbit would decay so quickly that you wouldn't make a single complete orbit. This line is the edge between flying and orbiting. Obviously, the design of the aircraft will effect this altitude limit, so it's still a fuzzy line (each aircraft should have it's own Kármán Line), but this idea is a good general starting place for defining what it means to be in space.
You still get some aerodynamic control above the Kármán Line -- control surfaces can effectively alter the orientation of your craft, but you don't get much useful lift above 100 kilometers. NASA uses 122 kilometers (75.8 miles) as their definition of "re-entry" for the Space Shuttle. This is the altitude where it's time for the pilots to start paying attention to aerodynamics and start using aerodynamic control surfaces for maneuvering. Above this altitude the pilots rely on attitude control rocket thrusters to control their orientation.
Karman Line altitude: 100 kilometers, 62 miles radius: 6478 kilometers, 4025 miles g at this altitude: 9.49 m / s^2
The first artificial satellite, Sputnik-1, had a perigee (minimum altitude) of 215 kilometers (133.6 miles). It had an eccentric orbit with an apogee (maximum altitude) of 939 km (583.5 miles), so it presumably suffered less air drag than a satellite in a nearly circular orbit close to 215 km. It's orbit lasted only 3 months before it reentered the atmosphere. Yuri Gagarin, the first human to orbit the Earth, had an orbit with a perigee of 169 km (105 miles).
Edge of space -- gravitational influence of the Earth
If going by nearest gravitational source then you have to go about 21 million kilometers away from the Earth for it to no longer be the most significant gravitational body. This is the distance where the strongest gravitational source is no longer the Earth. For reference, the average distance from the Earth to the Sun is 150 million kilometers (also called 1 Astronomical Unit).
Lowest practical orbit
A practical minimum altitude is depends on your definition of practical. At 180 kilometers a satellite can orbit only a few hours before atmospheric drag will bring it back to Earth. The International Space Station (ISS) orbits at a minimum of 340 kilometers. At this altitude its orbit decays about 2 km/month. The ISS has about one of the lowest altitudes of any orbiting artificial body. If you want a satellite to stay in orbit more than a few days without the need for boosts to counteract orbital decay then 200 kilometers is a good start.
altitude: 200 kilometers, 124 miles radius: 6578 kilometers, 4087 miles g at this altitude: 9.20677898 m / s^2
Other interesting orbits and the force of g
About half the satellites in orbit are under 1000 kilometers in altitude.
altitude: 1000 milometers, 621 miles
radius: 7378 kilometesr, 4585 miles
g at this altitude: 7.31843389 m / s^2
about 75% of Earth at sea level
GPS Satellite orbital altitude: 20200 km, 12551 miles
radius: 26560 km, 14018 miles
g at this altitude: 0.01 m/s^2
Geostationary orbit (orbital velocity in sync with revolution of the Earth):
geostationary altitude: 42300 km, 26284 miles g at this altitude: 0.0025 m/s^2
Moon orbit:
384403 kilometres, 238857 miles g at this altitude: not fair because the Moon's gravity comes into play.
Gravity and the force it causes
The Gravitational constant, G, is the same throughout the entire universe.
G = 6.673 × 10^-11 m^3/kg*s^2 G = 6.673 × 10^-11 m^3 kg^-1 s^-2
Note that g, force of Earth's Gravity, is calculated from your Altitude, A, as follows:
g = G * MassEarth / RadiusOrbit^2
= 6.673 * 10^-11 m^3 kg^-1 s^-2 * 5.97 * 10^24 kg / (6378 km + ALTITUDE km) ^ 2
= (ALTITUDE km + 6378 km) ^-2 * 6.673 * 10^-11 m^3 kg^-1 s^-2 * 5.97 * 10^24 kg
Also note that you can copy that formula directly into Google search and it will calculate the force of g for you. Replace ALTITUDE with the altitude you want to find. You MUST NOT forget the units; in this case "100 km", not just "100":
(100 km + 6378 km) ^-2 * 6.673 * 10^-11 m^3 kg^-1 s^-2 * 5.97 * 10^24 kg
Google will give you:
= 9.49322051 m / s^2
Joseph Kittinger
On August 16, 1960, Captain Joseph Kittinger made a jump from the Excelsior III at 31.3 kilometers, 19.4 miles, 102800 feet.
g at this altitude: 9.7 m / s^2
The force of gravity at 31 km is about 97% of what you feel at sea level. Kittinger was about 3% lighter than on the ground. So if he was 180 pounds on the ground he would be about 175 pounds at 31 kilometers.
