a geostationary satellite is orbiting the earth at a height of 5 r above that surface of the earth , r being the radius of the earth the time period of another satellite in hours at a height of 2r from the surface of the Earth is :
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From Kepler's 3rd law:
T is directly proportional to R^3/2..
Applying this formula for these two satellites we have:
T'/T = (R+2R/R+5R)^3/2 ...( take height from the centre to satellite )
As time period or geo-satellite is 24hrs, substituting the value of T:
T'/24=( 3R/6R ) ^3/2
T'/24 = 1/2^3/2
T' = 24/2^3/2
T' = 24/2√2
On solving, T' = 6√2 hours
Hope it helps....
T is directly proportional to R^3/2..
Applying this formula for these two satellites we have:
T'/T = (R+2R/R+5R)^3/2 ...( take height from the centre to satellite )
As time period or geo-satellite is 24hrs, substituting the value of T:
T'/24=( 3R/6R ) ^3/2
T'/24 = 1/2^3/2
T' = 24/2^3/2
T' = 24/2√2
On solving, T' = 6√2 hours
Hope it helps....
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