Using Kepler's laws of planetary motion, we can derive the following expression for a circular orbit: where T2 = 4π²a³ .3 fl Torbital period a= orbital radius in km = distance from the center of the earth to the orbit = Kepler's constant = 3.986004418 x 105 km³/s² The earth rotates once per sidereal day of 23 h 56 min 4.09 s (a) Determine the orbital radius of a GEO satellite (b) Assuming an earth radius of 6 370 km what is the orbit height h (1)
Using Kepler's laws of planetary motion, we can derive the following expression for a circular orbit: where T2 = 4π²a³ .3 fl Torbital period a= orbital radius in km = distance from the center of the earth to the orbit = Kepler's constant = 3.986004418 x 105 km³/s² The earth rotates once per sidereal day of 23 h 56 min 4.09 s (a) Determine the orbital radius of a GEO satellite (b) Assuming an earth radius of 6 370 km what is the orbit height h (1)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Using Kepler's laws of planetary motion, we can derive the following expression
for a circular orbit:
where
T² =
=
4π²a³
μl
Torbital period
a= orbital radius in km distance from the center of the earth to the orbit
= Kepler's constant =
3.986004418 x 105 km³/s²
The earth rotates once per sidereal day of 23 h 56 min 4.09 s
(a) Determine the orbital radius of a GEO satellite
(b) Assuming an earth radius of 6,370 km, what is the orbit height h
(1)
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