Question

acceleration due to gravity on Earth. But what about on theMoon? If an astronaut on the surface of the moon swings a 1.meter long pendulum, and it has a period of 4.9 seconds,what is the acceleration due to gravity, g, on the moon?

Answer 1

Given the formula:

`T=2\pi\sqrt{(L)/(g)}`

Where:

• Period, T = 4.9 seconds

,

• Length, L = 1 meter

Let's find the acceleration due to gravity, g, on the moon.

To solve for g, let's first rewrite the formula for g.

Take the following steps:

• Square both sides:

`\begin{gathered} T^2=(2\pi\sqrt{(L)/(g)})^2 \\ \\ T^2=4\pi^2*(L)/(g) \\ \\ T^2=(4\pi^2L)/(g) \end{gathered}`

• Cross multiply:

`g=(4\pi^2L)/(T^2)`

To solve for g, substitute 4.9 for T and 1 for L:

`\begin{gathered} g=(4\pi^2*1)/(4.9^2) \\ \\ g=(4\pi^2)/(24.01) \\ \\ g=1.64\approx1.6\text{ m/s}^2 \end{gathered}`

Therefore, the acceleration due to gravity on the moon is 1.6 m/s².

ANSWER:

C. 1.6 m/s²