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To: RightWhale
Ok, so is there a minimum height for orbital space flight? Or can you get to sub orbital and then only lateral speed counts?
29 posted on 06/02/2004 12:13:36 PM PDT by taxcontrol (People are entitled to their opinion - no matter how wrong it is.)
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To: taxcontrol

The tenable atmosphere ends at about 50 miles height, defined as the air being too thin to act on wings and control surfaces. Beyond that is space. Any ballistic trajectory, from a softball pitch to a cannon shot, to an ICBM flight is an orbital trajectory. Suborbital means that the orbit will not go all the way around the planet because it intersects the body of the planet so the flight will be a segment of a complete orbit. It's all orbital, and it is free ballistic flight in space, so it is orbital space flight and also the special case of suborbital flight.


30 posted on 06/02/2004 12:18:58 PM PDT by RightWhale (Theorems link concepts; proofs establish links)
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To: taxcontrol

Height requirements = 100 km.

 

Orbital speed requirements:

The orbital speed of a body, generally a planet, a natural satellite, an artificial satellite, or a multiple star, is the speed at which it orbits around the barycenter of a system, usually around a more massive body. It can be used to refer to either the mean orbital speed, the average speed as it completes an orbit, or instantaneous orbital speed, the speed at a particular point in its orbit.

The orbital speed can be described by Kepler's second law, which states that as a body moves around its orbit during a fixed amount of time, the line from the barycenter to the body sweeps a constant area of the orbital plane, regardless of which part of its orbit the body traces during that period of time. This means that the body moves faster near its periapsis than near its apoapsis, because at the smaller distance it needs to trace a greater arc to cover the same area. This law is usually stated as "equal areas in equal time."

The mean orbital speed can be derived either from observations of the orbital period and the semimajor axis of its orbit, or from knowledge of the masses of the two bodies and the semimajor axis.

v_o = {2 \pi r \over T}
v_o = \sqrt{m G \over r}

where vo is the orbital velocity, r is the length of the semimajor axis, T is the orbital period, m is the mass of the other body, and G is the gravitational constant. Note that this is only an approximation that holds true when the orbiting body is of considerably lesser mass than the central one.

More precisely,

v_o = \sqrt{m_2^2 G \over (m_1 + m_2) r}

where m1 is now the mass of the body under consideration, m2 is the mass of the body being orbited, and r is specifically the radius between the two bodies, ignoring the barycenter. This is still a simplified version; it doesn't allow for elliptical orbits, but it does at least allow for bodies of similar masses.

 

You then have orbital space flight.

 

Simple!!!!!


31 posted on 06/02/2004 12:28:35 PM PDT by Lokibob (All typos and spelling errors are mine and copyrighted!!!!)
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