InterviewSolution
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InterviewSolution
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A rocket of mass `m` is field vertically from the surface of Mars of mass `M`, radius `R` with a sped `upsilon`. If `20%` of its initial energy is lost due to Martain atmosheric resistance, how far will the rocket go from the surface of Mars before returing to it? Let `G` be the gravational constant. |
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Answer» Total energy of rocket is `E = KE + PE = (1)/(2) m upsilon^(2) + (-(Gmm)/(R ))` Since `20%` energyis lost, hence energy remained with rocket `= 80%` of `E` `= (80)/(100) E = (3)/(5) E = (4)/(5) [(1)/(2) m upsilon^(2) - (GMm)/(R )]` Let `h` be the heighest distance of rocket from the surfcace of Mars upto which rocket can go. At this location the total energy will its potential energy. Hence `(4)/(5) [(1)/(2) m upsilon^(2) - (GMm)/(R )] = - (GMm)/((R + h))` or `(2)/(5) [upsilon^(2) - (GMm)/(R )] = - (GMm)/((R + h))` or `(2)/(5) [(upsilon^(2) R - 2 GM)/(R )] = - (GM)/(R + h)` or `(R + h) = (5 R GM)/(2 (2 GM - upsilon^(2) R))` or `h = (5R GM)/(2(2 GM - upsilon^(2)R)) - R` |
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