Question

Astronauts on the first trip to Mars take along a pendulumthat has a period on earth of 1.50 {\rm s}. The period on Mars turns out to be 2.45{\rm s}.What is the free-fall acceleration onMars?

Answer 1

Answer:

3.7 m/s^2

Explanation:

The period of a simple pendulum is given by:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum and g is the free-fall acceleration on the planet.

Calling L the length of the pendulum, we know that:

$T_e = 2 \pi \sqrt{\frac{L}{g_e}}=1.50 s$ is the period of the pendulum on Earth, and $g_e = 9.8 m/s^2$ is the free-fall acceleration on Earth

$T_m = 2 \pi \sqrt{\frac{L}{g_m}}=2.45 s$ is the period of the pendulum on Mars, and $g_m = ?$ is the free-fall acceleration on Mars

Dividing the two expressions we get

$\frac{T_e}{T_m}=\sqrt{\frac{g_m}{g_e}}$

And re-arranging it we can find the value of the free-fall acceleration on Mars:

$g_m = g_e \frac{T_e^2}{T_m^2}=(9.8 m/s^2)\frac{(1.50 s)^2}{(2.45 s)^2}=3.7 m/s^2$