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?gmars= m/s2

Answer 1

Answer:

3.7 m/s^2

Explanation:

The period of a 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.

In this problem, we know that the period of the pendulum on Earth is:

$T_e = 1.50 s$

while the period of the same pendulum on Mars is

$T_m = 2.45 s$

And since the length of the pendulum L does not change, we can write:

$\frac{T_e}{T_m}=\frac{2\pi \sqrt{\frac{L}{g_e}}}{2\pi \sqrt{\frac{L}{g_m}}}=\sqrt{\frac{g_m}{g_e}}$

where

$g_e = 9.8 m/s^2$ is the free-fall acceleration on Earth

$g_m = ?$ is the free-fall acceleration on Mars

Re-arranging the equation and substituting numbers, we find:

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