# Question Video: Finding the Orbital Period from the Radius and Velocity for Circular Orbits Physics • 9th Grade

The table shows some data about Saturn’s moon, Titan. Using this data, calculate the orbital period of Titan around Saturn. Assume that Titan has a circular orbit. Give your answer in days to the nearest day.

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### Video Transcript

The table shows some data about Saturn’s moon, Titan. Using this data, calculate the orbital period of Titan around Saturn. Assume that Titan has a circular orbit. Give your answer in days to the nearest day.

Okay, so in this question we’ve been given a table regarding Titan, which is a moon of Saturn. Specifically, we’ve been given Titan’s diameter, its orbital radius, its orbital speed, and its mass. We need to use this information whether we use all of it or some of it to calculate the orbital period of Titan around Saturn or, in other words, the time taken for Titan to complete one orbit around Saturn. Now, to answer this question, let’s quickly draw a small diagram. Let’s say that this is Saturn. We know that it’s Saturn because we’ve drawn its rings around it. And let’s say that this pink circle represents Titan’s orbit. Now, the reason we’ve drawn a circle is because we’ve been told to assume in the question that Titan has a circular orbit.

So now that we’ve drawn Titan’s orbit, let’s actually draw Titan itself. And then let’s label on this diagram all the information we’ve been given about Titan. Firstly, we’ve been told Titan’s diameter. Now, this is the diameter of the moon itself. So let’s call this 𝑑 and let’s write down that it’s 5150 kilometers. Secondly, we’ve been given Titan’s orbital radius, in other words, the distance between Saturn center and the orbit of Titan. Let’s call this orbital radius 𝑟. And let’s write down the value of 𝑟 the radius, which is 1220000 kilometres. Thirdly, we’ve been told Titan’s orbital speed. In other words, if we assume that Titan is going this way around in its orbit around Saturn, let’s say that its orbital speed, which we will call 𝑣, is 5.57 kilometers per second.

And then finally, we’ve been given Titan’s mass. So let’s say that the mass of Titan, which we will call 𝑚 is 1.35 times 10 to the power of 23 kilograms. So we’ve got four pieces of information and we’ve written all of them down on our diagram. We need to use these pieces of information to work out the period of Titan’s orbit around Saturn. So we need to work out how long it takes for Titan to go all the way around its orbit once. Now, in order to answer this question, we first need to recall that an object’s speed is defined as the distance it travels divided by the time taken for it to travel that distance.

Now, the reason that we’re recalling this equation is because, first of all, we’ve been given the orbital speed of Titan. We know that it moves at 5.57 kilometers per second, and secondly, because we know Titan’s orbital radius, we can work out the distance that Titan travels when completing one orbit. That distance is simply the distance around the circle that represents its orbit because that’s the distance covered by Titan in one orbit.

Now, that particular distance actually happens to be the circumference of the circle, which we will call 𝑐. And we can also recall that the circumference of a circle is found by multiplying two 𝜋 by the radius of the circle. So using these two equations, we can work out everything we need. We know the speed of Titan’s orbit. And we can work out the distance traveled by Titan in one orbit, using the relationship between the circumference of a circle and its radius. And this will therefore allow us to work out the time taken for Titan to travel around its orbit once or, in other words, the orbital period of Titan. So let’s start by saying that Titan’s speed, which we’ve called 𝑣, is equal to the distance it covers in one orbit, which is the circumference of the circle divided by the time taken for that distance to be covered, which we will call 𝑡. And this is what we’re trying to work out.

So to workout 𝑡, we will need to rearrange this equation. We can do this by multiplying both sides by 𝑡 divided by 𝑣. This way on the left-hand side, the 𝑣s cancel and on the right-hand side the 𝑡s cancel. And so what we’re left with is the period of Titan’s orbit on the left. And on the right, we’ve got the circumference of the circle 𝑐 divided by Titan’s orbital speed 𝑣. We can then replace 𝑐 with two 𝜋𝑟 because, once again, the circumference of a circle is equal to two 𝜋 multiplied by the radius of that circle. And then at this point, we can plug in some values. We can say that 𝑡 is equal to two 𝜋 multiplied by the radius of the orbit which is 1220000 kilometers. And we divide this by 5.57 kilometers per second, which is Titan’s orbital speed.

Now at this point, we could have converted the radius of the orbit to meters and the orbital speed to meters per second as well. But we don’t need to do this here because we can see that the unit of kilometers in the numerator cancels with the unit of kilometers in the denominator. So even if we had converted the numerator to meters and the denominator to meters per second, we would have still got the same answer because the units of meters would have canceled and so would the conversion factor in both cases. And so what we’re left with here is a fraction that has no units in the numerator because, remember, the kilometers have canceled. And because the kilometers in the denominator of cancelled as well, the unit in the denominator is one divided by seconds.

And overall then, this becomes a unit of seconds for the entire fraction. And so when we evaluate this fraction, that’s going to give us the numerical value in seconds of the period of Titan’s orbit. So when we do this, we find that Titan has a period of 1376209.349 dot dot dot seconds. However, this is not a final answer. Remember, we need to give our answer in days to the nearest day. So the first thing that we need to do is to convert this quantity into days. To do this, we need to remember the conversion that one day is equivalent to 24 hours. But then we multiply 24 by 60 because there are 60 minutes in every hour. And then we multiply that once more by 60 because there are 60 seconds in every minute.

So 24 hours in a day, each hour has 60 minutes in it, and each minute has 60 seconds in it. Therefore, in one day, there are 24 times 60 times 60 seconds. Or in other words, there are 86400 seconds in a day. So to convert from seconds to days, we need to divide our number by 86400. And we can write this numerically as the period of Titan’s orbit is equal to 1376209.349 dot dot dot seconds, divided by 86400 seconds per day. And this works out correctly in terms of units because the unit of seconds cancels in the numerator and denominator. And we’re left with a unit one divided by days in the denominator, which gives a unit of days for the fraction overall. So evaluating the fraction then, we find that the period of Titan’s orbit is 15.9283 dot, dot dot days.

But remember, we need to give our answer to the nearest day. Therefore, upon completing the rounding, we found the answer to a question. Assuming a circular orbit, Titan has an orbit of 16 days around Saturn to the nearest day. And it’s worth noting, by the way, that we didn’t use all the pieces of information given to us in the table. We had to realize that the diameter of Titan was irrelevant here and so was the mass of Titan. We only needed to use the orbital radius and the orbital speed. And sometimes, we will need to be wary of this. We need to be able to pick out the correct information in order to be able to solve a problem.