The earth has a radius of 6380 km and turns around once on its axis in 24 h. (a) What is the radial acceleration of an object at the earth's equator? Give your answer in m/s2 and as a fraction of g.

Video transcript

Welcome back everybody. We are looking at our home planet of earth and we are told that Earth has a radius of 6371 km Now, at some point above Earth, we are going to launch a satellite and are told that the mass of this satellite is 200 kg and it will be orbiting at a distance of km away from The surface of the Earth. We are told that the period or the time for the satellite to take one full revolution is 1.64 hours. And we are tasked with finding what it's centripetal acceleration is. Well, the centripetal acceleration is equal to four pi squared times are all over the period squared. However, we are going to need this period in seconds. So let me convert that real quick. We are given that our period is 1.64 hours and I am going to multiply this by seconds, giving us our period in seconds to be 5904 Great. Now that we have all of our values, let's go ahead and plug it into this formula Right here to find our centripetal acceleration. We have four times I squared times our radius. But what radius are we looking at? We're actually going to be looking at the entire distance from the satellite to the center of the Earth. It will be 6371 kilometers plus 700 kilometers. We need to be working in meters. So I'm going to multiply this by 1000. I'm gonna divide all of this by our period of 5000, 904 seconds. And when you plug this into your calculator, you get that your centripetal acceleration is equal to eight m per second squared or responding to our final answer of C. Thank you guys so much for watching. Hope this video helped. We will see you all in the next one.