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

3.7 m/s^2

Step-by-step explanation:

The period of a pendulum is given by:

`T=2\pi \sqrt{(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:

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