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A rocket is fired vertically from the surface of Mars with a speed of 2 kilometre per second . If 20 % of its initial energy is loose due to Martine atmosphere resistance , how far will the rocket go from the surface of Mars before returning to it ? Mass of Mars 6.5 × 10 power 23 kg , Radius of mass = 3395 km, G = 6.67 × 10 minus 11 newton meter square/ KG square. |
| Answer» EXPLANATION:speed of the ROCKET ( v) = 2 km/smass of the MARS ( Mm) = 6.4 × 10²³ Kgradius of the Mars (RM ) = 3395 km = 3.395 × 10^6 minital K.E of the rocket = 1/2 mv²due to Martian resistance 20% of KE of the rocket is lost .so, available K.E = 80% of 1/2mv²= 2/5 mv²--------(1)Let the rocket be reach at height h from the surface of the Mars .so, INCREASE PE ° = PEf - PEi= -GMm.m/(Rm + h) - [ - GMm.m/Rm]= -GMm.m × h/Rm.(Rm + h)now, according to LAW of conservation of energy .increase in PE = available total KEGMm.m×h/Rm(Rm+h) = 2/5 mv²GMm/Rm = 2/5 v².(Rm + h)/h5GMm/2Rm.v² = Rm/h + 1after putting values of G, Mm, Rm and v5×6.67×10^-11×6.4×10²³/2×3.395×10^6×(2×10³)² = Rm/h +17.85 = Rm/h +1Rm/h = 6.85h = Rm/6.85 = 3395/6.85 ≈ 495 km | |