Tangential speed at equator.
Posted 20 January 2009 - 03:25 PM
Is this correct? I thought it was more like 465m/s.
Posted 20 January 2009 - 03:37 PM
Posted 20 January 2009 - 03:38 PM
Posted 20 January 2009 - 04:29 PM
Good catch. Gareth56, double-check the units in the text. Is it kilometers per hour or meters per second? 2*pi/sidereal day * 6378 km = 1674 km/hr.
However, seeing that I did some approximations it is well possible that the actual value would be ~1700 km/h. In that case the college book might have screwed up the units (or you misread them).
Note well: It is 2*pi/sidereal day, not 2*pi/24 hours. There is a slight difference, probably important only to the truly pedantic. A sidereal day is 23.9344696 hours long.
Posted 21 January 2009 - 11:20 AM
"Why is the launch area for the European Space
Agency in South America and not in Europe?
Satellites are boosted into orbit on top
of rockets, which provide the large tangential speed
necessary to achieve orbit. Due to its rotation, the surface
of Earth is already traveling toward the east at a
tangential speed of nearly 1700 m/s at the equator.
This tangential speed is steadily reduced further
north, because the distance to the axis of rotation is
decreasing. It finally goes to zero at the North Pole.
Launching eastward from the equator gives the satellite
a starting initial tangential speed of 1700 m/s,
whereas a European launch provides roughly half that
speed (depending on the exact latitude)."
Posted 21 January 2009 - 11:37 AM
You could contact the publisher and point that out. They might have an errata page online. Is there a URL given in the book?
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Posted 21 January 2009 - 11:52 AM
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