Q: 8.17 A rocket is fired vertically with a speed of 5 km s-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth $=6.0 \times10^2^4\hspace {1mm}kg$; mean radius of the earth $=6.4 \times10^6\hspace {1mm}m$; $G=6.67 \times10^-^1^1\hspace {1mm}Nm^2kg^-^1$.

The total energy is given by :

Total energy  =  Potential energy  +  Kinetic energy

$=\ \left ( \frac{ -GmM_e}{R_e} \right )\ +\ \frac{1}{2}mv^2$

At the highest point velocity will be zero.

Thus the total energy of the rocket is :

$=\ \left ( \frac{ -GmM_e}{R_e\ +\ h} \right )\ +\ 0$

Now we will use the conservation of energy :

Total energy initially (at earth's surface)    =    Total energy at height h

$\left ( \frac{ -GmM_e}{R_e} \right )\ +\ \frac{1}{2}mv^2\ =\ \left ( \frac{ -GmM_e}{R_e\ +\ h} \right )$

or                                                       $\frac{1}{2}v^2\ =\ \frac{ gR_eh}{R_e\ +\ h}$

or                                                     $h\ =\ \frac{R_e v^2}{2gR_e\ -\ v^2}$

or                                                           $=\ \frac{6.4\times 10^{6}\times (5\times 10^3)^2}{2g\times 6.4\times 10^6\ -\ (5\times 10^3)^2}$

or                                                           $=\ 1.6\times 10^6\ m$

Hence the height achieved by the rocket from earth's centre =   R  +  h

$=\ 6.4\times 10^6\ +\ 1.6 \times 10^6$

or                                                                                             $=\ 8\times 10^6\ m$

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