A 1200 kg communications satellite is in a circular geostationary orbit. That is

Question

A 1200 kg communications satellite is in a circular geostationary orbit. That is it keeps the same position in the sky as viewed from the earth's surface by orbiting the earth one revolution every 23 hrs 56 min 4 s. The satellite has an altitude of 3.578 times 10^7 m above the surface of the earth. The radius of the earth is 6.38 times 10^7 m, the earth's mass is 5.98 times 10^24 kg, and G = 6.67 times 10^-11 N m^2/kg^2. a) What is the radius of the satellites orbit? b) From the given time for one revolution, find the angular velocity omega in units of rad/s for the satellite. c) From omega find the velocity v of the satellite. d) Find the gravitational force of the earth on the satellite.

Explanation / Answer

2. (a) r = Re + h

= (6.38 x 10^6) + (3.578 x 10^7)

= 4.216 x 10^7 m

(B) w = 2 pi / T

=2 pi / [ (23 x 3600 s) + (56 x 60 s) + 4s ]

w = 7.29 x 10^-5 rad/s

(C) v = w r

v = 3073.5 m/s

(D) F = m v^2 / r

= 1200 x 3073.5^2 / 4.216 x 10^7
= 268.8 N

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