Question

While on the moon, the Apollo astronauts enjoyed the effects of a gravity much smaller than that on

Answer 1

Answer:

1.628 $\frac{m}{sec^{2} }$

Explanation:

Anywhere in the universe, In a closed system, Conservation of energy is applicable.

In this case

Neil is initially on the surface of moon and has a velocity of 1.51 $\frac{m}{sec}$ in upward direction.

⇒He has Kinetic energy= $K_{i}$ = $\frac{1}{2} m{v^{2} }$ J

But with respect to the surface of the moon,

where m=mass of moon

           v=velocity of Neil

He has Potential energy= $P_{i}$=0 J

At the highest point of his jump, his velocity =0

⇒ Kinetic energy=$K_{f}$=0 J

His Potential energy with respect to the surface of moon=$P_{f}$=$m \times g\times h$

where m=mass of moon

           g= gravitational acceleration on moon

           h=height from moon's surface

By Conservation Energy Principle

$K_{i}$+$P_{i}$=$K_{f}$+$P_{f}$

$K_{i}$+0=0+$P_{f}$

$\frac{1}{2} m{v^{2} }$ = $m \times g\times h$

$\frac{v^{2} }{2}$ =  $g\times h$

$\frac{1.5^{2} }{2}$ J= $g\times 0.7 m$

⇒ g = $\frac{1.14}{0.7}$ = 1.628 $\frac{m}{sec^{2} }$