Question

The Apollo Lunar Module was used to make the transition from the spacecraft to the moon's surface and back. Consider a similar module for landing on the surface of Mars. Use conservation of mechanical energy to answer these questions.
(a) As the lander is descending, if the pilot decides to shut down the engine when the lander is at a height of 1.3 m, (this may not be a safe height to shut down the engine) and the velocity of the lander (relative to the surface of the planet) is 1.3 m/s what will be velocity of the lander at impact? Note: g on the surface of mars is about 0.4 times that on the surface of the Earth.

(b) In the case of the lunar module an impact velocity of 3.0 m/s or less was essential for a safe landing. Assuming this to be the case for the Mars lander as well, at what maximum height could the pilot shut down the engines to ensure a safe landing. Assume the velocity v0 at the time the engine is shut down is 1.2 m/s.

Answer 1

Answer:

Explanation:

a. Landing height is

H=1.3m

Velocity of lander relative to the earth is, i.e this is the initial velocity of the spacecraft

u=1.3m/s

Velocity of lander at impact, i.e final velocity is needed

v=?

The acceleration due to gravity is 0.4 times that of the one on earth,

Then, g on earth is approximately 9.81m/s²

Then, g on Mars is

g=0.4×9.81=3.924m/s²

Then using equation of motion for a free fall body

v²=u²+2gH

v²=1.3²+2×3.924×1.3

v²=1.69+10.2024

v²=11.8924

v=√11.8924

v=3.45m/s

The impact velocity of the spacecraft is 3.45m/s

b. For a lunar module, the safe velocity landing is 3m/s

v=3m/s.

Given that the initial velocity is 1.2m/s²

We already know acceleration due to gravity on Mars is g=3.924m/s²

The we need to know the maximum height to have a safe velocity of 3m/s

Then using equation of motion

v²=u²+2gH

3²=1.2²+2×3.924H

9=1.44+7.848H

9-1.44=7.848H

7.56=7.848H

H=7.56/7.848

H=0.963m

The the maximum safe landing height to obtain a final landing velocity of 3m/s is 0.963m