Back to Home

Blackholes Forum Message

Forums: Atm · Astrophotography · Blackholes · Blackholes2 · CCD · Celestron · Domes · Education
Eyepieces · Meade · Misc. · God and Science · SETI · Software · UFO · XEphem
RSS Button

Home | Discussion Forums | Blackholes I | Post
Login

Be the first pioneers to continue the Astronomy Discussions at our new Astronomy meeting place...
The Space and Astronomy Agora
Re: Tanx To U All....

Forum List | Follow Ups | Post Message | Back to Thread Topics | In Response To
Posted by Jim Bergquist on February 24, 2001 05:04:37 UTC

In regards to the article that you refer to, for an object in circular orbit the radius of the orbit remains constant. Hence the radial velocity and acceleration are both zero even though the x-y motion is sinusoidal and therefore exhibits changes in linear velocity and linear acceleration. The relation between the radius and the x-y coordinates is r = x + y. For a circular orbit, x = r*cos(theta) and y = r*sin(theta). theta = w*t where w is dependent on the radius of the circular orbit. If r = 6400 km, the Earth's radius, at theta = 0, then x = 6400 km and y = 0 km. A quarter of the way around the Earth, theta = 90 degrees and x = 0 km and y = 6400 km. But for both positions, r = 6400 km; there is no change in the radius.

Galileo showed that the path of a projectile over short distances, where the curvature of the Earth's surface is negligible, was that of a parabola (ignoring the drag produced by the Earth's atmosphere). The path that it would actually take (with no atmosphere) is that of an elliptical orbit. If we fire a projectile horizontally to the Earth's surface at relatively low velocity, the height that we fire from will be at the point of the ellipse that is farthest from the center of the Earth. It will travel along the path of the ellipse until it hits the Earth's surface. As initial velocity of the projectile increases, the ellipse that the projectile follows will become wider and the point of impact will be moved farther away along the Earth's surface. As we increase the projectile's initial velocity a point will eventually be reached where the path becomes circular and the projectile will miss the Earth's surface at a point a quarter of the distance of the way around the Earth and travel completely around it. The velocity that this occurs at (assuming no atmosphere) is the 8000 m/sec referred to.

The elliptical paths referred to above are those of a suborbital flight around the Earth. Ballistic missles travel along these paths. If we deny this possibility, then there must be a limit to how fast something can travel. It would not be a physical limit though. If you would like to verify that objects can orbit the Earth and exceed the given velocity, wait until the International Space Station can be seen from your location and time the interval (T) between consecutive appearances. The orbital velocity (v) would then be v = 2πr/T. The result should be slightly greater that what we would get if we used r = 6400 km, the radius of the Earth.

Follow Ups:

Login to Post
Additional Information
Google
 
Web www.astronomy.net
DayNightLine
About Astronomy Net | Advertise on Astronomy Net | Contact & Comments | Privacy Policy
Unless otherwise specified, web site content Copyright 1994-2018 John Huggins All Rights Reserved
Forum posts are Copyright their authors as specified in the heading above the post.
"dbHTML," "AstroGuide," "ASTRONOMY.NET" & "VA.NET"
are trademarks of John Huggins