If the total mechanical energy of a particle of mass m is E, its speed can be given as v = sqrt((2(E-U))/(m)). where U= potential energy function of the particle. Using the above expression find the speed of the particle at earth.s surface when the particle falls freely in gravity from rest from a height =R, where R = radius of earth.

If the total mechanical energy of a particle of mass m is E, its speed

1.	If the total mechanical energy of a particle of mass m is E, its speed can be given as v = sqrt((2(E-U))/(m)). where U= potential energy function of the particle. Using the above expression find the speed of the particle at earth.s surface when the particle falls freely in gravity from rest from a height =R, where R = radius of earth.
Answer» Solution : The gravitational POTENTIAL ENERGY of the particle m is given as `U = - (GMm)/(r ) = E = U_0 + K_0 = -(GMm)/(r_0) + 0 =- (GMm)/(r_0), r_0 = 2R` Then, SUBSTITUTING E and U in the EXPRESSION ` v = sqrt((2(E - U))/(m))` we have` v= sqrt((2{(-(GMm)/(r_0)) - (-(GMm)/(r )) })/(m) )` , this gives , ` v = sqrt((2GM(r_0-r))/(r r_0))` where `r_0 - r = h = R , because r = R ` , then , `v = sqrt((GM)/(R )) = sqrt(gR) `, where `g = (GM)/(R^2)`