InterviewSolution
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InterviewSolution
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A rocket is fired vertically upwards with a speed of `upsilon (=5 km s^(-1))` from the surface of earth. It goes up to a height `h` before returning to earth. At height `h` a body is thrown from the rocket with speed `upsilon_(0)` in such away so that the body becomes a satellite of earth. Let the mass of the earth, `M = 6 xx 10^(24) kg`, mean radius of the earth, `R = 6.4 xx 10^(6)m, G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) , g = 9.8 ms^(-2)`. Answer the following questions: The value of `h` isA. `1.5 xx 10^(5) m`B. `3.2 xx 10^(5)m`C. `3.2 xx 10^(6)m`D. `1.6 xx 10^(6)m` |
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Answer» Correct Answer - D According to law of conservation of total mechanical energy, total energy of rocket at the surface of earth = total energy of rocket at the highest point or `(1)/(2) m upsilon^(2) + ((-GMm)/(R )) = 0 + ((-GMm)/(R + h))` or `(upsilon^(2))/(2) = (GM)/(R ) - (GM)/((R + h)) = (gR^(2))/(R ) - (gR^(2))/((R + h))` `= gR (1-(R )/(R + h)) = gR ((h)/(R + h))` or `upsilon^(2) (R + h) = 2g Rh` or `R upsilon^(2) = 2g R h - upsilon^(2)h` `= (2g R - upsilon^(2))h` or `h = (R upsilon^(2))/((2gR - upsilon^(2)))` `= (6.4 xx 10^(6) xx (5 xx 10^(3))^(2))/((2 xx 9.8 xx 6.4 xx 10^(6)) - (5 xx 10^(3))^(2))` `= 1.6 xx 10^(6)m` |
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