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 m/s
Mass of Mars, M = 6.4 × 1023 kg
Radius of Mars, R = 3395 km = 3.395 × 106 m
Universal gravitational constant, G = 6.67× 10–11 N m2 kg–2
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 -
= 0.4mv2 -
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,
0.4mv2 - =
= 495 × 103 m
= 495 km.
What is Heliocentric theory?
According to the Heliocentric theory, the sun is at the centre and various planets revolve around the sun at their axis.
At night, if we see the planets and the stars in the sky, all appear to lie in the hemisphere (rest of the hemisphere we are unable to see because of being on the other side of the earth). This sphere is called the celestial sphere.