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The approximate height from the surface of earth at which the weight of the body becomes \mathrm{\frac{1}{3}} of its weight on the surface of earth is :
\mathrm{\left [ Radius\, of\, earth\, R= 6400\, km\; and\; \sqrt{3}= 1.732 \right ]}

Option: 1

\mathrm{3840\, km}


Option: 2

\mathrm{4685\, km}


Option: 3

\mathrm{2133\, km}


Option: 4

\mathrm{4267\, km}


Answers (1)

best_answer

Acceleration due to gravity at height h from earth's surface is ,
\mathrm{g_{h}= \frac{GM}{\left ( R+h \right )^{2}}= \frac{9R^{2}}{\left ( R+h \right )^{2}}}

\mathrm{W_{h}= \frac{Ws}{3}\quad (Given)}
\mathrm{mg_{h}= \frac{mg}{3}}
\mathrm{g_{h}= \frac{9}{3}= \frac{9R^{2}}{\left ( R+h \right )^{2}}}

\mathrm{R+h= \sqrt{3}R}
          \mathrm{h= \left ( \sqrt{3} -1\right )R}
               \mathrm{= 0.732\times 6400}
           \mathrm{h= 4685\, km}
 
The correct option is (2)


 

Posted by

rishi.raj

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