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Q: 8.12  A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero? Mass of the sun  =2\times 10^3^0\hspace{1mm} kg , mass of the earth =6\times 10^2^4\hspace{1mm} kg. Neglect the effect of other planets etc. (orbital radius =1.5\times 10^1^1\hspace{1mm} m).

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Let the distance where the gravitational force acting on satellite P becomes zero be x from the earth.

Thus we can write : 

                                                    \frac{GmM_s}{(r-x)^2}\ =\ \frac{GmM_e}{r^2}

or                                                 \left ( \frac{r-x}{x} \right )\ =\ \left ( \frac{2\times 10^{30}}{60\times 10^{24}} \right )^\frac{1}{2}

or                                                                      =\ 577.35

Hence :

                                               x\ =\ \frac{1.5\times 10^{11}}{578.35}\ =\ 2.59\times 10^8\ m                                        (Since r  =  1.5\times 10^{11})

Posted by

Devendra Khairwa

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