A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s–1. If 20% of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4× 1023 kg; radius of mars = 3395 km; G = 6.67× 10-11 N m2 kg–2.
Initial velocity of the rocket, v
= 2 km/s = 2 × 103
Mass of Mars, M
= 6.4 × 1023
Radius of Mars, R
= 3395 km = 3.395 × 106
Universal gravitational constant, G = 6.67× 10–11
Mass of the rocket = m
Initial kinetic energy of the rocket = mv2
Initial potential energy of the rocket =
Total initial energy = mv2 -
If 20 % of initial kinetic energy is lost due to Martian atmospheric resistance, then only 80 % of its kinetic energy helps in reaching a height.
Total initial energy available =
= (80/100) × (1/2) mv2
Maximum height reached by the rocket = h
At this height, the velocity and hence, the kinetic energy of the rocket will become zero.
Total energy of the rocket at height h =
Applying the law of conservation of energy for the rocket,
= 495 × 103
= 495 km.