International space station revolves around the earth in an orbit with tangential velocity 7.66 km/s In which orbit does it revolve?

Explanation

Given : Tangential velocity `v_(c)=7.66` km/s
=7660 m/s
We know that , Gravitational constant (G)
`=6.67xx10^(-11)Nm^(2)//kg^(2)`
mass of earth (M)= `6xx10^(24)"kg"` ,
radius of earth (R)=6400 km
= `6400xx10^(3)`
To find : type of orbit [i.e ., Height of the orbit above the earth 's surface(h)]
Formula : `v_(c)=sqrt((GM)/(R+h))`
Calculation : From formual, `v_(c)^(2)=(GM)/(R+h)`
`therefore" " R+h=(GM)/(v_(c)^(2))`
`=(6.67xx10^(-11)xx5xx10^(24))/((7660)^(2))`
`=(6.67xx6)/(7660xx7660)xx10^(13)`
`=(40.02)/(7660xx7660)xx10^(13)`
`=6.820xx10^(6)`
`therefore" " h=(6820xx10^(3))-R`
`=(6820-6400)xx10^(3)`
`=420xx10^(3)m=420"km"`
Hence , the height of the orbit is 420 km above the earth's surface . This implies that , the international space station revolves in low earth orbit.
The international space station revolves in low earth orbit.