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

3.7 m/s^2

Step-by-step explanation:

The period of a simple 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.

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

`T_e = 2 \pi \sqrt{(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{(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

`(T_e)/(T_m)=\sqrt{(g_m)/(g_e)}`

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

`g_m = g_e (T_e^2)/(T_m^2)=(9.8 m/s^2)((1.50 s)^2)/((2.45 s)^2)=3.7 m/s^2`