InterviewSolution
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InterviewSolution
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A satellite orbits the earth at a height of `3.6xx10^(6)m` from its surface. Compute it’s a kinetic energy, b. potential energy, c. total energy. Mass of the satellite `=500kg` mass of the earth `=6xx10^(24)`kg, radius of the earth `=6.4xx10^(6), G=6.67xx10^(-11)Nm^(2)kg^(-2)`. |
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Answer» Here `r=R+h=6.4xx10^(6)+3.6xx10^(6)=10^(7)m` Orbital velocity of the satellite around the earth is given by `v_(0)=sqrt((GM)/((R+h)))=sqrt((6.67xx10^(-11)xx6xx10^(24))/(10^(7)))` `=sqrt(6.67xx6xx10^(6))ms^(-1)` a. `KE` of the satellite `=1/2mv_(0)^(2)` `=1/2xx500xx6.67x6xx10^()6=10^(10)J` b. `PE` of the satellite `=-(GMm)/(R+H)` `=-(6.67xx10^(-11)xx(6xx10^(24))xx500)/(10^(7))` `=-2xx10^(10)J` c. Total energy `=KE+PE=10^(10)-2xx10^(10)=-10^(10)J` |
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