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.
Answers
Answer:
Explanation:
speed of the rocket ( v) = 2 km/s
mass of the Mars ( Mm) = 6.4 × 10²³ Kg
radius of the Mars (RM ) = 3395 km = 3.395 × 10^6 m
inital K.E of the rocket = 1/2 mv²
due 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 KE
GMm.m×h/Rm(Rm+h) = 2/5 mv²
GMm/Rm = 2/5 v².(Rm + h)/h
5GMm/2Rm.v² = Rm/h + 1
after putting values of G, Mm, Rm and v
5×6.67×10^-11×6.4×10²³/2×3.395×10^6×(2×10³)² = Rm/h +1
7.85 = Rm/h +1
Rm/h = 6.85
h = Rm/6.85 = 3395/6.85 ≈ 495 km
Answer:
refer to the attachment........,
