Electronics and Communication Engineering - Satellite Communication - Discussion
Discussion Forum : Satellite Communication - Section 2 (Q.No. 2)
2.
Assuming earth to be a sphere of radius 6400 km and height of a geosynchronous satellite above Earth as 36000 km, the velocity of a geosynchronous satellite is __________ km/hr.
Discussion:
5 comments Page 1 of 1.
Dineu said:
7 years ago
Raduis of earth = 6400,
ht abve earth= 36000,
Total ht of Satlte from centre of earth = 6400+36000=42400km,
Distance traversed = perimeter=2πR = 2*3.14*42400.
Time= 24 hrs.
Speed= Distance/Time= (2*3.14*42400)/24 km/hr.
ht abve earth= 36000,
Total ht of Satlte from centre of earth = 6400+36000=42400km,
Distance traversed = perimeter=2πR = 2*3.14*42400.
Time= 24 hrs.
Speed= Distance/Time= (2*3.14*42400)/24 km/hr.
(1)
Rafael said:
1 decade ago
Why 11100?
Gyananshu said:
1 decade ago
For a geosynchronous satellite the time period = 24 hrs.
Total height from the center of the earth = (36000 + 6400) = 42400 km.
Hence velocity = (2*pi*42400) km/24 hrs = 11100 km/hr.
Total height from the center of the earth = (36000 + 6400) = 42400 km.
Hence velocity = (2*pi*42400) km/24 hrs = 11100 km/hr.
Vgh said:
1 decade ago
Can you explain in detail?
MANIKUMAR said:
1 decade ago
Almost circular orbit so total distance is parameter of circle.
V=S*T
V=S*T
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