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.

Open in App

Solution

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 =

Initial potential energy of the rocket

Total initial energy

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

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, we can write:

Suggest Corrections

Similar questions

A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the Sun = 2 × 10³⁰ kg; mass of mars = 6.4 × 10²³ kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 10⁸kg; G= 6.67 × 10^–11 m²kg^–2.

A rocket is fired vertically with a speed of 5 km s^–1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 × 10²⁴ kg; mean radius of the earth = 6.4 × 10⁶ m; G= 6.67 × 10^–11 N m² kg^–^2.

Join BYJU'S Learning Program