A man is positioned in a circular orbit at a height of `1.6xx10^(5)`m above the surface of the earth. The radius of the earth is `6.37xx10^6`m and the mass is `5.98xx10^(24)` kg. What would be the orbital speed of the man ? `(G=6.67xx10^(-11)N*m^2*kg^(-2))`

Explanation

Radius of the earth `R=6.37xx10^6m =63.7xx10^5m`
height ,h `=1.6xx10^5m`
Distance of the man from the centre of the earth r =R+h
`=63.7xx10^5+1.6xx10^5=65.3xx10^5m`
`therefore` Orbital speed,
`v=sqrt((GM)/(R+h))=sqrt((GM)/r)=sqrt((6.67xx10^(-11)xx5.98xx10^(24))/(65.3xx10^5) `
`=0.78xx10^4m*s^(-1)=7.8 km*s^(-1)`.