The work done in lifting a body of mass ‘m’ to a height equal to the radius of earth is

Solution:

We have from concept gravitational P.E. $U=\frac{-GMm}{r}$ using this we can

get

Method – I:

${\text{P.E}}_i={\text{U}}_i=-\frac{\text{GMm}}{\text{R}}$ (On the surface of earth)

${\text{P.E}}_f={\text{U}}_f=-\frac{\text{GMm}}{\text{2R}}$ (Certain height from the surface of earth)

W =  Gain in P.E. =${\text{U}}_i-(-{\text{U}}_f)=U_i-U_f$   

  $=GMm[\frac{1}{r_1}-\frac{1}{r_2}]$                                          

$=GMm[\frac{1}{R}-\frac{1}{2R}]$ $=\frac{GMm}{2R}(\because g=\frac{GM}{R^2})$    

$=\frac{{\text{gR}}^{\text{2}}\text{m}}{\text{2R}}=\text{mg}\frac{\text{R}}{\text{2}}$                                                          

Method – II:

we know change in P.E

$\text{W}=\Delta \text{P.E}=\frac{\text{mgh}}{1+\text{h/R}}$

$=\frac{\text{mgR}}{\text{2}}(\because \text{h}=\text{R})$

So option (a) is correct conclusion.