In class the other day we calculated the minimum acceleration we need to stay on the earth…what we found was fascinating!

First, we need to discover the diameter of the Earth as well as the time, in seconds, it takes the Earth to rotate once:

d=2*pi*radius=2*pi*6.3781 x 10^6

d=4.007 x 10^7 m

t=24hrs=1440min=8.64 x 10^4 sec

Now we have a change in displacement and time to calculate velocity:

v=(distance)/(time)=(4.007 x 10^7 m)/(8.64 x 10^4 sec)

v=463.77 m/s

We know the acceleration in circular motion is a=(v^2)/r, so plugging in the velocity and radius will give us the minimum acceleration of one point on the Earth without anything flying off!

v^2=2.1508 x 10^5 m^2/s^2

a=(v^2)/r=(2.1508 x 10^5)/(6.3781 x 10^6)

a=0.0337 m/s^2

So, we can see the minimum acceleration needed for us to stay on the Earth’s surface is much smaller than the 9.81 m/s^2 acceleration we experience every day. This is the exact reason why we are able to fall down.

Imagine if the centripetal acceleration were only .0337! We would merely hover over the surface of the earth — how crazy!?

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Posted by Frank Noschese on May 10, 2009 at 7:27 pm

Follow up question:

What is the shortest length of day possible for us to still be barely held to the surface of the earth by gravity?

Posted by rebbieg on May 15, 2009 at 1:11 am

That’s a great question! I’m working on a post in response :-) Thanks for visiting!