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

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

1. A

$\text{mg}\frac{R}{2}$

2. B

mg R

3. C

Zero

4. D

$\frac{3}{2}\text{mgR}$

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### Solution:

We have from concept gravitational P.E.  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. =${\text{U}}_{i}-\left(-{\text{U}}_{f}\right)={U}_{i}-{U}_{f}$

$=GMm\left[\frac{1}{{r}_{1}}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}\frac{1}{{r}_{2}}\right]$

$=GMm\left[\frac{1}{R}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}\frac{1}{2R}\right]$ $=\frac{GMm}{2R}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\because \text{\hspace{0.17em}}g=\frac{GM}{{R}^{2}}\right)$

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

Method – II:

we know change in P.E

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

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

So option (a) is correct conclusion.

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