The earth has a radius of 6380 km and turns around once on its axis in 24 h. (b) If arad at the equator is greater than g, objects will fly off the earth's surface and into space. (We will see the reason for this in Chapter 5.) What would the period of the earth's rotation have to be for this to occur?

Video transcript

Welcome back everybody. We are making observations about MArs and we are told that the gravitational acceleration on MArs is 3.7 m per second squared. We're also told that the radius on MArs is 3389. kilometers. Now, if we think about it, the radial acceleration if it is fast enough or if the rotational period is great enough then items say this is mars, items can start leaving the surface of the planet of MArs. So we are tasked with finding what rotational period is required for that to happen. So here's what we're going to do. We're actually gonna use a formula for that radial acceleration related to that rotational period. So we have the acceleration is equal to four pi squared times the radius all over T squared. Now, if I divide both sides by A and I multiply both sides by T squared. This will cancel out. I get that he squared is equal to four pi squared are all over A in order to get rid of the radical or sorry, the exponents on this side, I will take the square root of both sides, which will effectively get rid of this radical. And now we have a formula or are isolated variable of the period. Now, before plugging our values in here because we're gonna use our gravitational acceleration for this acceleration right? We need to make sure everything is in the correct units. So here's what I'm going to do. I'm going to multiply our radius by 1000 here because we know that in one kilometer there's 1000 m, which gives us a radius of 3,389, m. Great. So now that we have that, let's go ahead and find our rotational period here. Let's see. So we have four pi squared times 3,389,500. All divided by our radial acceleration or our gravitational acceleration. We can use since they're pointing in the same direction of 3.7. When you plug this into your calculator, we get 6014 seconds, but we need our final answer in hours. So we know that per hour there are 3600 seconds. These units are going to cancel out. And we get that the rotational period for objects to float off of the surface of Mars will be 1.67 hours corresponding to our answer choice of. D Thank you all so much for watching. Hope this video helped. We will see you all in the next one