On August 2, 1971, Astronaut David Scott, while standing on the surface of the Moon, dropped a 1.3-kg hammer and a 0.030-kg falcon feather from a height of 1.6 . Both objects hit the Moon’s surface 1.4 after being released. What is the acceleration due to gravity on the surface of the Moon?
about one sixth of g on earth (to check)
height = (1/2) G t^2
3.2 = G (1.4)^2
G = 1.63 m/s^2
to check g/6 = 9.81/6 = 1.635
To find the acceleration due to gravity on the surface of the Moon, we can use the equation for free fall:
d = (1/2) * g * t^2
where:
d = distance (1.4 m)
g = acceleration due to gravity
t = time (1.4 s)
We can rearrange the equation to solve for g:
g = (2 * d) / t^2
Plugging in the given values:
g = (2 * 1.4) / (1.4)^2
g = 2.8 / 1.96
g ≈ 1.43 m/s^2
Therefore, the acceleration due to gravity on the surface of the Moon is approximately 1.43 m/s^2.
To find the acceleration due to gravity on the surface of the Moon, we can use the equation of motion for free fall:
s = u * t + (1/2) * a * t^2
where:
- s is the vertical distance traveled by the object (1.4 m in this case).
- u is the initial velocity of the object (0 m/s since the objects were dropped).
- t is the time it takes for the object to hit the surface of the Moon (1.4 s in this case).
- a is the acceleration due to gravity on the Moon's surface (the value we're looking for).
First, let's calculate the distance traveled by each object. Since both objects were dropped from the same height, their s values are the same:
1.4 m = 0 * 1.4 s + (1/2) * a * (1.4 s)^2
Simplifying the equation gives us:
1.4 m = (1/2) * a * 1.96 s^2
Now, we can solve for a:
a = (2 * 1.4 m) / (1.96 s^2)
a = 1.4286 m/s^2
Therefore, the acceleration due to gravity on the surface of the Moon is approximately 1.43 m/s^2.