The rings of Saturn are composed of chunks of ice that orbit t...

Question

The rings of Saturn are composed of chunks of ice that orbit the planet. The inner radius of the rings is 73,000 km, and the outer radius is 170,000 km. Find the period of an orbiting chunk of ice at the inner radius and the period of a chunk at the outer radius. Compare your numbers with Saturn’s own rotation period of 10 hours and 39 minutes. The mass of Saturn is 5.7 x 10²⁶ .

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve for this problem. Astronauts are finally on Mars, the planet has two different moons demos, which is at a distance of 23,000 kilometers from Mars and Phobos, which is a distance of 9400 kilometers from Mars. Given that Mars rotates about itself once every 24 hours and 39 minutes for the astronauts, would the moons appear to revolve around Mars faster than the sun? Assume that the mass of Mars is 6.4 multiplied by 10 to the power of 23 kg. OK. So that's our final answer that we're trying to solve for us. We're trying to figure out if the moons would appear to revolve around Mars faster than the sun. So that's what we're ultimately trying to solve for is we're trying to figure out whether or not these moons will appear to revolve around Mars faster than the sun. Yes or no. OK. With that in mind, let's read off. Our multiple choice answers to see what our final answer might be. A is Phobos apparently revolves slower than the sun and Demos apparently revolves faster than the Sun. B is Phobos apparently revolves faster than the Sun and Demos apparently revolve slower than the sun. C as Phobos and Demos, both apparently revolve faster than the sun and D is Phobos and Demos. Both apparently revolve slower than the sun. OK. So first off, let us write down all of our known variables. So we're, we know that the mass of Mars, which I'm gonna denote as capital M subscript lowercase M. So the mass of Mars is equal to 6.4 multiplied by 10 to the power of 23 and its units will be kilograms. Awesome. And we also know the orbital period of Mars. So let's call this T subscript M. Yes, this is the period for Mars, the orbital period to be exact. And this is equal to 24 hours and 39 minutes. So we have a bit of a problem here. We need to convert 24 hours and 39 minutes to seconds in order to get everything in the correct units to help us solve for our final answer in the proper units, which all of our multiple choice answers don't have units. But we're gonna have to figure out numerically the different values to determine whether or not phobos or demos or both of them will move faster. Or slower then the sun. OK. So with that in mind, let us use dimensional analysis to convert 24 hours and 39 minutes to seconds. So using dimensional analysis, we should be able to recall that there is 3600 seconds in one hour. And then we'll need to add 39 minutes, which we need to recall that in one minute, there is 60 seconds. So as you can see all of our unwanted units will cancel out, which in this case, we want minutes to cancel out and we want hours to cancel out leaving us with just seconds. So when we plug that into our calculator, we will get 88,740 seconds. And we know that TM. In this case, TM or orbital period, let's write this in scientific notation in round to one decimal place. So when we write this in scientific notation in round to one decimal place, we will get that TM is equal to 8.9 multiplied by 10 to the power of four seconds. Awesome. And that is it. We did it. So now moving right along here, let us use the subscript lowercase D to correspond with the values of quantities for demos. And let us use the subscript P to correspond to the values of phobos. Let us use capital R to denote the radius of the orbits measured from the center of Mars. And let us use capital T to denote the orbital periods. Thus, we could write that RD. So the radius of Demos is equal to. And we're told that this is equal to 23,000 kilometers. But let's quickly convert this to meters. You can use dimensional analysis, but I'm just gonna give it to you to save some time. So it's 2.3 multiplied by 10 to the power of 7 m. And we know that the radius of fouls which is gonna be RP is equal to 9.4 multiplied by 10 to the power of six meters. Once again, we're given it in kilometers, but we need to convert to units of meters. I've gone ahead and done it for you which once again, you can use dimension analysis or you could just quickly look it up. OK. Moving right along here. Now, we need to recall and use the equation to solve for the velocity of an orbiting object which we let's call this equation one which states that V is equal to the square root of capital G multiplied by capital M multiplied by lower case M actually my bad. It's capital G multiplied by capital M subscript M because we're focusing on the mass of Mars all divided by capital R. We can also write and let's call this equation two, that V is equal to two multiplied by pi multiplied by capital R all divided by capital T Awesome. And then this is the linear speed is V. So moving right along here, now if we set equations one and two equal to each other, we will get that the square root of G capital G to be exact multiplied by capital mm divided by capital R is equal to two multiplied by pi multiplied by. So we need to take two multiplied by pi and we need to multiply that by capital R and then we need to divide by T. So now we need to rearrange this equation to isolate and solve for T. So let me do that and let's call this equation three. We will get that T is equal to where T once again is our orbiting period which is two multiplied by pi multiplied by the square root of R cubed divided by capital G multiplied by mm OK. The mass of Mars. So now we need to use equation three to solve for TD and TP. So let's start with TD first. So TD is equal to two multiplied by pi. So we're gonna be using equation three, but we're gonna be plugging in our values for TD specifically. So let's plug in all of our known values. So we know two multiplied by pi multiplied by the square root of the radius of demos, which we know to be two point three multiplied by 10 to the power of 7 m. And don't forget that this value is cubed divided by the gravitational constant, which is 6.67 multiplied by 10 to the power of negative 11. So times 10 to the power of negative 11. And don't forget that its units are newtons multiplied by meters squared per kilograms squared. Awesome. And then this is multiplied by our mass of Mars which is 6.4 multiplied by 10 to the power of 23 kilograms. Awesome. So let me plug that into our calculator. We will find that this is equal to 1.06 multiplied by 10. And this is when we round the two decimal places multiplied by 10 to the power of five seconds. So following the same methodology to solve for the period of fools, we will find that TP is equal to two multiplied by pi multiplied by the square root of our mass of VOS which is 9.4 multiplied by 10 to the power of 6 m to the power of three all divided by 6.67 multiplied by 10 to the power of negative 11 newtons multiplied by meters squared per kilograms squared, multiplied by the mass of Mars which is 6.4 multiplied by 10 to the power of 23 kg. Awesome. So with that in mind, don't forget to write your units. Don't forget all your parentheses. So when we plug that into our calculator, we will get 2.77 rounding the two decimal places again times 10 to the power of four seconds. So as we can see the mass or sorry. So the period of Phobos, so TP is less than the period of Mars and the period of decimals is greater than the period for Mars. Therefore, we need to note that because of Mars own rotational motion, the astronauts on Mars would see the sun rotating with the same rotational speed of Mars. So as the period of Phobos is less than Mars, it should be appear to be moving faster than the sun on Mars. Whereas the period on Mars is less than that of demos and demos should appear to be moving slower than the sun on Mars. Therefore, Phobos will apparently revolve faster than the sun and demos will appear to revolve slower than the sun. So once again, Phobos will apparently revolve faster than the sun and decimals will apparently revolve slower than the sun. So with that in mind, that will mean that our final answer has to be the multiple choice answer. Letter B. Bobos apparently revolve faster than the sun and demos apparently revolve slower than the sun. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

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