A spaceship is launched into a circular orbit close to the earth's surface . What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull. Radius of earth `= 6400 km`, `g = 9.8m//s^(2)`.
Text Solution
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The orbital velocity in a circular orbit close to the earth is `r=sqrt(gR)` The velocity required to escape `v_(e)=sqrt(2gR)` Hence additional velocity required is `v _(e)-v=(sqrt(2)-1)sqrt(gR)` therefore, `v_(e)-v=0.414xxsqrt(9.8xx6400xx10^(3))` `=3778.71m//s=3.278kms^(-1)`
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