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A particle is fired vertically upwards from the urface of earth and reaches a height 6400 km. Find the initial velocity of the particle if R = 6400 km and g at the surface of earth is 10 m//s^2. |
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Answer» SOLUTION :`1/2m upsilon^2=(MGH)/([1+(h//R)])` here h=R=6400 km and g=10 `m//s^2` so `upsilon^2=gh`, i.e., `upsilon =sqrt(10xx6400xx10^3)`= 8 km/s |
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