Finding the period (T) of a pendulum on Mars

[asilvester635]

1. Homework Statement
The first astronauts to visit Mars are each allowed to take along some personal items to remind them of home. One astronaut takes along a grandfather clock, which, on earth, has a pendulum that takes 1 second per swing, each swing corresponding to one tick of the clock. When the clock is set up on Mars, will it run fast or slow?

It shows you that the clock on Mars will have a slower period on Mars (T = 2.45 sec.) than on Earth (T = 1 sec.). There's a section called "My Work Version #1", where I tried to solve the problem. The formula is pretty basic. Plug in acceleration of gravity (9.8 m/s^2) and the length of the string (1 m). With that approach I got T = 3.26.

On the section called "My Work Version #2", I pretty much set T Earth and T Mars and tried to solve for T mars. At the end the value that I got under the square root is 0.166/1, but the answer key that they had is 1/0.166.

So the bottom line is, I have no idea how they got the value of 0.166 how they solved for the period, T, of Mars.

Homework Equations

T = 2pi(sqrt(L/g))

The Attempt at a Solution

To view the image, please click on the link and click download. Thanks.
https://myhpu-my.sharepoint.com/:i:...BGv3IJUrIPZ4EBcg9-xeH1d4WYptA04lKCKA?e=dA4YzU

 

[haruspex]

Neither of your links work for me. They produce error messages.

 

[asilvester635]

Neither of your links work for me. They produce error messages.

Hey, I fixed it. It should let you to be able to download the image. I would post the image to the website but the image comes out to be low quality. It's better to download it. Thanks.

 

[haruspex]

Their answer is wrong because they confused Mars with the moon. Our moon's surface gravity is one sixth of Earth's.

You answer is wrong because you plugged in "1" for L. The clock's pendulum length is not 1 m. It is the right length for timing one second on Earth, whatever that is.