Question

Find the centre of gravity of a uniform vertical rod height = R/3, where R is the radius of Earth. The lower end of the rod is touching the surface of the Earth.

Answer 1

Correct Answer - `y=(int_(0)^((R )/(3))(yg_(0)R^(2)lambda dy)/((y+R)^(2)))/(int_(0)^((R )/(3))(g_(0)R^(2)lambda dy)/((y+R)^(2)))=4R ln((4)/(3))-R` from surface
`y=(int_(0)^((R )/(3))(yg_(0)R^(2)lambda dy)/((y+R)^(2)))/(int_(0)^((R )/(3))(g_(0)R^(2)lambda dy)/((y+R)^(2)))=4R ln((4)/(3))-R` from surface