Comments on: A typical GPS (Global Positioning System) satellite orbits at an altitude of 4.0 107 m.? http://www.vehiclegpstracking.org/a-typical-gps-global-positioning-system-satellite-orbits-at-an-altitude-of-4-0-107-m.html Vehicle GPS Tracking Systems and Devices Sat, 04 Feb 2012 20:23:56 +0000 hourly 1 http://wordpress.org/?v=3.3.1 By: Cheryl P http://www.vehiclegpstracking.org/a-typical-gps-global-positioning-system-satellite-orbits-at-an-altitude-of-4-0-107-m.html/comment-page-1#comment-1418 Cheryl P Tue, 13 Jul 2010 23:59:31 +0000 http://www.vehiclegpstracking.org/a-typical-gps-global-positioning-system-satellite-orbits-at-an-altitude-of-4-0-107-m.html#comment-1418 a) Period is (2pi/sqrt(G*M_E_)) (r^(3/2)) G is the constant M_E_ is mass of the earth r is altitude of the satellite+radius of the Earth Plug those numbers in to get the period in seconds. Then divide the answer be 60 (to get minutes) and 60 again (to get hours). The answer will be somewhere around 35 b. v=sqrt((G*M_E_)/r) The numbers are the same as in part a. The answer will be in m/s, so divide by 1000 to get km/s The answer will be somewhere around 2.7 a) Period is (2pi/sqrt(G*M_E_)) (r^(3/2))

G is the constant
M_E_ is mass of the earth
r is altitude of the satellite+radius of the Earth
Plug those numbers in to get the period in seconds.
Then divide the answer be 60 (to get minutes) and 60 again (to get hours).
The answer will be somewhere around 35

b. v=sqrt((G*M_E_)/r)
The numbers are the same as in part a.
The answer will be in m/s, so divide by 1000 to get km/s
The answer will be somewhere around 2.7

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