 What is the centripetal acceleration of a person standing at 28 degrees north of the equator? This is similar to the last problem, so a similar approach should work. The centripetal acceleration and speed are still given by the same equations, and the period will still be 24 hours for the person standing at 28 degrees north of the equator. However, the radius of the path that the person follows will be different. If we sketch out the path the person follows alongside the radius of the earth, we can see that the person will follow a different circular path with a different radius than they did in the problem before. To work out the length of the radius r, we can use trigonometry. The radius of the path and the axis of the earth's rotation meet a bright angle. If we label this angle theta, we can see that the angle theta is 90 minus 28 degrees, since the angle between the radius of the equator and the axis of the earth is 90 degrees in total. So now that we have an angle, the radius of the earth, and a right angle triangle, we can use trigonometry to work out r. From Sokotoa, we know that sine of theta will equal r. The radius of the uniform circular motion path of a person standing at 28 degrees north of the equator divided by r-earth, the radius of the uniform circular motion of the person standing at the equator, which is equal to the radius of the earth. Rearranging, we find that r is equal to r-earth times sine theta. So now we have everything we need. As before, we can substitute our equation for velocity into our equation for centripetal acceleration. Big T is still 24 hours of 3,600 seconds, and r is the radius of the earth times sine of theta, or 6.4 million meters times sine theta. Now, we could plug these numbers into a calculator to get our answer, or we could make life a bit easier by recognizing that all of these terms are the same as in our previous example, except for our sine theta. And all of these terms will multiply together to give us 0.034 meters per second squared. So we can then see that the centripetal acceleration in this example is 0.034 meters per second squared times sine of 90 minus 28 degrees. Calculating, we find that this comes to 0.030 meters per second squared. Does this look right? Let's check. If our person was back on the equator, theta would be 90 degrees minus 0 degrees, which is equal to 90 degrees, and sine of 90 degrees is 1. So we'd get the same answer as we got previously. That's a good sign.