question_answer

Prove that acceleration due to gravity on the surface of the earth is given by \[g=\frac{4}{3}\pi \rho GR,\], where \[G\] is gravitational constant, \[\rho \] is mean density and R is the radius of the earth.                                    

Answer:

                 Mass of the earth, \[M=\frac{4}{3}\pi {{R}^{3}}\rho \] Acceleration due to gravity on the earth's surface, \[g=\frac{GM}{{{R}^{2}}}=\frac{G}{{{R}^{2}}}\times \frac{4}{3}\pi {{R}^{3}}\rho =\frac{4}{3}\pi \rho \mathbf{GR}.\]