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An artificial satellite is moving in a circular orbit around the earth with a speed equal to 3//8 times of th emagnitude of escape velocity from the earth. If the satellite is stopped suddely in its orbit and allowed to fall freely onto the earth, then the speed with whcih it hits the surface of the earth is (take g = 10 m//sand R_(e ) = 6400 km) |
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Answer» `SQRT((29)/(4)) Km//s` `V_(s) = (3)/(8) sqrt(2gR_(e) ) = sqrt((9gR_(e ))/(32)) = sqrt((R_(r )^(2)g)/(R_(e ) + h))` `9R_(e ) + 9h = 32 R_(e )` `h = (23)/(9) R_(e )` Now TOTAL ENERGY at heigh'h' `=` total energy at the surface of the earth `0 - (GM_(e )m)/(R_(e )) = (1)/(2)mv^(2) - (GM_(e )m)/(R_(e ))` `(1)/(2)mv^(2) = (GM_(e )m)/(R_(e )) - (GM_(e )m)/(R_(e ) +(23)/(9)R_(e ))` `(1)/(2)mv^(2) = (23)/(32) (GM_(e )m)/(R_(e ))` `v = sqrt((23)/(16)(GM_(e ))/(R_(e ))), sqrt(gR_(e) ) = 8 Km//sec` `v = sqrt((23)/(16)gR_(e) ) = sqrt((23)/(2)) Km//s` |
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