Question

Acceleration due to gravity on the moon is 1.6m/s^2 or about 16% of the value of gg on Earth. If an astronaut on the moon threw a moon rock to a height of 7.8m what would be its velocity as it struck the moon’s surface? How would the fact that the moon has no atmosphere affect the velocity of the falling moon rock? Explain your answer.

Answer 1

To solve this problem it is necessary to apply the concepts related to the conservation of Energy. Mathematically the conservation of kinetic energy must be paid in the increase of potential energy or vice versa. This expressed in algebraic terms is equivalent to

Kinetic Energy = Potential Energy

$\frac{1}{2}mv^2 = mgh$

Where

m = Mass

v = Velocity

g = Gravity

h = Height

As the mass is the same then we have to

$\frac{1}{2} v^2 = gh$

Rearrange to find v,

$v = \sqrt{2gh}$

Our values are given as

$g = 1.6m/s^2$

$h = 7.8m$

Therefore replacing we have

$v = \sqrt{2(1.6)(7.8)}$

$v = 4.99m/s$

Hence the velocity at the moon would be 4.99m/s

The only direct affectation is that concerning the Resistance or drag force generated by a fluid - such as air in the ground - that can diminish / sharpen the direct effects of gravity. Disregarding the resistance of the air, as we can see in the equation previously given, there should be no affectation because the speed depends on the gravity and height.