A satellite moves in a circular orbit around the earth at a height h=R/2 above the ath surface, where R is the radius of the earth. The time period of revolution nearly is
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Hey dear,
◆ Answer -
T = 2.6 hr
◆ Explaination -
# Given -
h = R/2
# Solution-
Period of revolution is given by -
T = 2π √[(R+h)^3 / (GM)]
But we know , h = R/2 & GM = gR^2
T = 2π √[(R+R/2)^3 / (gR^2)]
T = 2π √[(3/2)^3 R/g]
T = 2 × 3.142 × √[(27/8)×6.4×10^6/9.8]
T = 9329 s
T = 2.6 hr
Approximate time period of revolution is 2.6 hr.
Hope this helps...
◆ Answer -
T = 2.6 hr
◆ Explaination -
# Given -
h = R/2
# Solution-
Period of revolution is given by -
T = 2π √[(R+h)^3 / (GM)]
But we know , h = R/2 & GM = gR^2
T = 2π √[(R+R/2)^3 / (gR^2)]
T = 2π √[(3/2)^3 R/g]
T = 2 × 3.142 × √[(27/8)×6.4×10^6/9.8]
T = 9329 s
T = 2.6 hr
Approximate time period of revolution is 2.6 hr.
Hope this helps...
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