Could an astronaut (or anyone else) be lowered from space w/o burning up?

[flight]

Vlad, there are a lot of problems with your post.

[QUOTE}No. A Geosynchronous orbit around the equator (a geostationary orbit, which I should have said in my first post) means that the satellite is “parked” over a geographical point on the earth’s surface. As seen from the moon, the satellite moves in an orbit, but seen from the earth, the satellite is stationary. Look at satellite TV antenna dishes: they are pointed at a satellite in a geostationary orbit and don’t move.[/QUOTE]

You are both right. The satellite oribts the earth once per 24 hours, but since the Earth rotates, it appears to be stationary. Lemur866 was right though, it is in orbit and it is moving very fast.

Let’s look at basic angular (circular) motion. Angular velocity is calculated: V ang = rotation in radians / time. One complete rotation = 2(pi) radians. We can say that the earth or a geostationary satellite has an angular velocity of 2(pi) radians / 24 hrs or 0.083 radians per hour. For any point anywhere on a straight line from the earth’s center out to a geostat satellite the angular velocity is the same: 0.083 radians per hour. the radius, or distance from the center of rotation is not in the equation.

However, if we want to calculate linear velocity from angular velocity, the equation is: V lin = V ang X radius. For a satellite 22 000 km from the earth’s rotational axis, the V lin = 0.083 rad/hr X 22 000 km = 5758 km/hr. Pretty fast. If an astronaut is lowered toward the roational axis for a distance of 10 000 km, then his V lin = 0.083 X (22 000 - 10 000) = 0.083 X 12 000 = 3141 km/hr. He is still moving at the same angular velocity, but his linear velocity is slower because his radius of rotation is smaller. Simply standing on the earth’s surface, you’re crawling along at only about 1600 km/hr. Standing at the north or south pole, where your distance from the axis is 0, your V lin is 0, even though your V ang is still 0.083 radians/hr.

Coriolis force. We are not conserving angular velocity, we are conserving angular momentum. This means that when the guy gets closer to the Earth he will be rotating at a faster angular velocity. So, assuming this is lowering on a rope rather than part of a space elevator, the rope may end up wrapping around the Earth a few times before the guy touches ground. There is nothing in the lowering scenario to slow him without entering into a very odd rotation about the spacecraft, so by the time he makes groundfall he will be moving easily fast enough to burn up.

Your analogy of orbital falling is correct, but misapplied. There are two vectors to a satellite’s movement: one is toward the earth’s center due to gravity (Vg), the other is parallel to the earth’s surface (Vp). Vg = Vp for a given stable orbit, meaning that the satellite moves forward (and away from the earth) the same distance it is pulled toward the earth. In a geostationary orbit, this takes place above the same point on the earth’s equator, without any change in position or distance.

In a stable circular orbit Vg=0. There is a force and acceleration in that direction but no velocity (due to changing reference frames). There is, in fact, huge changes in position, it is just that the Earth moves along with it. There is no change in distance wrt the Earth, but that is just because the Earth is a non-inertial reference frame. In general, Lemur866 had it right.

[Tuckerfan]

@kanicbird

IIRC NASA tested an emergency ‘bubble’ that could allow a astronaut to excape from orbit unpowered, though never used. It may have been back in the skylab days.

That would be the MOOSE, and it doesn’t appear to have ever gotten off the drawing table.