If a space station must
"provide one sixth the magnitude of the acceleration due to gravity on Earth"
Can I use 9.81ms^-2/6
or is that too simple?
Thanks.
No. You wnat 1/6 OF 9.8 or
1/6 * 9.8 m/s^2
I think if the station were located on the surface of the moon it would come out about right :)
To calculate the acceleration required on the space station to provide one-sixth the magnitude of the acceleration due to gravity on Earth, you can use the following steps:
Step 1: Determine the magnitude of the acceleration due to gravity on Earth, which is approximately 9.81 m/s^2.
Step 2: Divide this value by 6 to find one-sixth of the acceleration due to gravity on Earth.
9.81 m/s^2 / 6 ≈ 1.64 m/s^2
Therefore, the acceleration required on the space station to provide one-sixth the magnitude of the acceleration due to gravity on Earth is approximately 1.64 m/s^2.
So, your calculation of 9.81 m/s^2 / 6 is correct, and it yields approximately 1.64 m/s^2.