# Video: Calculating the Period and Length of a Planet’s Orbit from the Obit Radius and the Speed of the Planet

Two planets, A and B, orbit a star. Both planets have circular orbits. Planet A orbits the star at a distance of 1.5 × 10⁸ km and at a speed of 30 km/s. Planet B orbits the star at a distance of 4.8 × 10⁸ km and at a speed of 17 km/s. How many times is the length of planet B’s orbit greater than planet A’s? How many times longer does it take for planet B to orbit the star than planet A? Give your answer to 1 decimal place.

07:35

### Video Transcript

Two planets, A and B, orbit a star. Both planets have circular orbits. Planet A orbits the star at a distance of 1.5 times 10 to the power of eight kilometres and at a speed of 30 kilometres per second. Planet B orbits the star at a distance of 4.8 times 10 to the power of eight kilometres and at a speed of 17 kilometres per second. How many times is the length of planet B’s orbit greater than planet A’s?

Okay, so we’ve been told that we’re dealing with two planets, A and B, which are both orbiting a star. And we also know that both planets have circular orbits. So let’s say that this in orange is the star. The blue represents the orbit of planet A, which is the blue blob here. And the pink represents the orbit of planet B, which is the pink blob.

Now, the orbits of both planets are meant to be circular. And we’ve been told that this distance, the radius of the orbit of planet A, is 1.5 times 10 to the power of eight kilometres because we’ve been told that planet A orbits the star 1.5 times 10 to the power of eight kilometres away from the star. And the radius of the orbit of planet B is 4.8 times 10 to the power of eight kilometres because, once again, that’s the distance from the start to planet B. And it orbits in a circular orbit.

Now, as well as this, we know the speeds of the two planets. We know that planet A orbits at a speed of 30 kilometres per second. And planet B orbits at a speed of 17 kilometres per second. Now, the first part of the question asks us, how many times is the length of planet B’s orbit greater than planet A’s? So to answer this question, we first need to consider the length of planet B’s orbit and the length of planet A’s orbit. Well in both cases, the length of the orbit is going to be the distance travelled by each planet in one orbit, in other words, the distance around the circle. Or, at least, that’s the case for planet A. And for planet B it’s the distance travelled by the planet all the way around this circle.

In other words, we’re trying to find the circumferences of both circles. Now, we can recall that the circumference 𝐶 of a circle is given by multiplying two by 𝜋 by the radius of the circle 𝑟. So we can find out the length of planet A’s orbit first by saying that the circumference of the circle, that is, A’s orbit, is equal to two 𝜋 times the radius of A’s orbit. And that, secondly, the circumference of B’s orbit is equal to two 𝜋 times the radius of B’s orbit where we’ve said that the distance from the star to A’s orbit is 𝑟 sub A. And the distance from the star to B’s orbit is 𝑟 sub B.

Now, we want to find out how many times greater the length of planet B’s orbit is than planet A’s orbit. In other words, what we’ve been told is that the length of planet B’s orbit 𝐶 B is, let’s say, 𝑛 times larger than the length of planet A’s orbit where this value of 𝑛 is what we’re trying to find out. How many times greater is planet B’s orbit compared to planet A’s orbit? So in order to find out 𝑛, we just divide both sides of the equation by 𝐶 sub A. When we do this, the 𝐶 sub As on the right-hand side cancel. And we’re just left with 𝑛 on the right.

At this point, we can substitute in the expressions for 𝐶 sub B and 𝐶 sub A on the left-hand side. And so we get two 𝜋𝑟 sub B divided by two 𝜋𝑟 sub A. But then at this point, the two 𝜋s in the numerator and the denominator cancel, leaving us with just 𝑟 B over 𝑟 A. Then we can sub in the values for 𝑟 B and 𝑟 A, that’s this value for 𝑟 B and this value for 𝑟 A, and then evaluate the expression on the left-hand side of the equation which, when simplified down, becomes 3.2. Therefore, our answer to this part of the question is that the length of planet B’s orbit is 3.2 times greater than the length of planet A’s orbit.

Let’s move on then to also considering the speeds of the planets in their orbits. How many times longer does it take for planet B to orbit the star than planet A? Give your answer to one decimal place. Okay, so this time, instead of just considering the length of their orbits, we’re going to be considering the planet’s orbit times. In other words, how long does it take planet A to go all the way around its orbit once compared to how long it takes planet B to go all the way around its orbit once. To work this out, we need to remember that the speed of an object is defined as the distance travelled by that object divided by the time taken for that object to travel that distance.

Now, we’ve already seen the distances travelled by the planets around the star. Those happen to be the length of each orbit. But now, we’re also considering the speed of each planet. We’ve been told that the planet A moves at 30 kilometres per second. And planet B moves at 17 kilometres per second. So in the case of both planets, we know the speed at which each planet travels. And we know the distance travelled by each planet. That’s the circumferences of the circles we saw earlier. And so we know the speed and the distance. So we can work out the time taken for each planet to orbit once around the star.

We do this by rearranging the equation. We multiplied both sides of the equation by the time divided by the speed. This way, on the left-hand side, the speed cancels. And on the right-hand side, the time cancels. In the end, we’re left with the time taken to complete an entire orbit is equal to the distance of that orbit divided by the speed that the planet travels at. So for planet A, we can say that the time taken for an entire orbit is equal to the distance of the planet A divided by the speed of the planet A. And for the planet B, we can say that the time taken for one entire orbit is equal to the distance of one entire orbit divided by the speed of plant B.

But then we saw earlier that the distance travelled by a planet is equal to the circumference of the circle which was equal to two times 𝜋 times the radius of the circle. So instead of 𝑑 sub A, we can replace this with two 𝜋𝑟 sub A. And 𝑑 sub B can be replaced by two 𝜋𝑟 sub B. Now once again, we’re trying to find out how many times larger one of these times is compared to the other. So we can say that the time taken for planet B to orbit around the star is this time 𝑚 times larger than the time taken for planet A to orbit around the star. And so if we want to find out this value of 𝑚, we divide both sides of the equation by 𝑡 sub A. This way, 𝑡 sub A cancels on the right-hand side. And we’re just left with 𝑚.

So we can say that 𝑚 is equal to 𝑡 sub B divided by 𝑡 sub A. But instead of 𝑡 sub B and 𝑡 sub A, we’ve replaced the right-hand sides of each equation. Now, we can see that we’ve got a fraction divided by another fraction. This is equivalent to multiplying the two fractions together with the fraction that was initially in the denominator flipped on its head. And so we can see that we’ve got a two 𝜋 in the numerator and two 𝜋 in the denominator when we’re multiplying these two fractions. Those cancel out. And so all we’re left with is 𝑟 sub B, that’s this 𝑟 sub B here, multiplied by 𝑆 sub A, that’s this 𝑆 sub A, in the numerator. And in the denominator, we’ve got 𝑆 sub B multiplied by 𝑟 sub A. And so this fraction is equal to the value of 𝑚.

So when we plug in the values, we’ve got the values of 𝑟 sub B, the radius of planet B, that’s 4.8 times 10 to the power of [eight] kilometres, multiplied by the speed of planet A, 30 kilometres per second. And we divide this by the speed of planet B, 17 kilometres per second, multiplied by the radius of planet A, 1.5 times 10 to the power of eight kilometres.

Then we can see that all our units are consistent. We can see, for example, that this power of kilometres will cancel with this power of kilometres and that this power of kilometres per second will cancel with this power of kilometres per second. So we’ve got just numbers in our fraction. And there’s going to be no unit for the answer which makes sense because the answer is going to be a value of 𝑚. And of course, 𝑚 is the value that gives us how many times longer it takes planet B to orbit the star than it takes planet A.

So when we evaluate this, we find 𝑚 to be 5.64 dot dot dot, so on and so forth. But remember, we’ve been asked to give our answer to one decimal place. So here’s the first decimal place. And it’s going to be the next one, the value of four, that tells us what happens to the first decimal place. Well, four is less than five. So this value will stay exactly the same. It’s not going to round up. So to one decimal place our answer is 5.6. And hence, we can say that to one decimal place, it takes planet B 5.6 times longer than planet A to orbit the star.