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Question Video: Determining the Circumference and Period of an Orbit Physics • 9th Grade

A satellite orbits Earth at an orbital radius of 42,200 km and moves at a speed of 3.1 km/s. The satellite has a circular orbit. What is the circumference of the orbit of the satellite? Give your answer in kilometers, using scientific notation to one decimal place. What is the period of the satellite’s orbit? Give your answer in hours to the nearest hour.

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Video Transcript

A satellite orbits Earth at an orbital radius of 42,200 kilometers and moves at a speed of 3.1 kilometers per second. The satellite has a circular orbit. What is the circumference of the orbit of the satellite? Give your answer in kilometers using scientific notation to one decimal place. What is the period of the satellite’s orbit? Give your answer in hours to the nearest hour.

Okay, this question is about a satellite that orbits Earth. And in the first part of the question, we are asked to work out the circumference of the satellite’s orbit. We are told that the satellite has an orbital radius of 42,200 kilometers. So if this blue blob here represents the Earth and this pink blob represents the satellite, then this orange dashed line represents the trajectory of the satellite’s orbit, which we are told is circular. And this orbit has a radius 𝑟 measured from the center of mass of Earth, where 𝑟 is equal to 42,200 kilometers. So we have a circular orbit that we know the radius of, and we are asked to find its circumference.

Let’s recall that the circumference of a circle 𝑐 is equal to two 𝜋 multiplied by its radius 𝑟. Taking this equation for the circumference of a circle and substituting in the radius of our satellite’s orbit, we get that the circumference of this orbit is equal to two 𝜋 multiplied by 42,200 kilometers. Doing the multiplication gives us that the circumference of the orbit is 265,150.4 kilometers, where the ellipses indicate that there are further decimal places. The question asks us to give our answer in kilometers using scientific notation to one decimal place.

Now our answer is already in units of kilometers, but we still need to convert it to scientific notation. To get this number into scientific notation, we move the decimal point one, two, three, four, five places to the left. So, our result becomes 2.651504 etcetera times 10 to the five still with units of kilometers. Finally, rounding this to one decimal place, our result rounds up to give 2.7 times 10 to the five kilometers. And so our answer to the first part of the question is that the circumference of the satellite’s orbit is equal to 2.7 times 10 to the five kilometers.

If we now look at the second part of the question, we see that we are asked to find the period of the satellite’s orbit. We know that the radius of the satellite’s orbit, which we’ve already labeled 𝑟, is equal to 42,200 kilometers. The question also tells us that the satellite moves in this orbit at a speed of 3.1 kilometers per second. Let’s label the speed as 𝑣. We can recall that for an object moving in a circular orbit, the orbital period capital 𝑇 is equal to two 𝜋 multiplied by the radius of the orbit 𝑟 divided by the speed 𝑣. Now it’s worth pointing out that this equation is really nothing more than a different way of writing the perhaps-more-well-known equation speed equals distance over time.

If we take this speed–distance–time equation and multiply both sides by time, then on the right-hand side we can cancel the time in the numerator with the time in the denominator. Then, if we divide both sides by speed, on the left-hand side, we can cancel the speed in the numerator with the speed in the denominator. Now for an object moving in a circular orbit, then in a time equal to one period capital 𝑇, the object completes one full orbit. In other words, it covers a distance equal to the circumference 𝑐 of the orbit. We know that 𝑐 is equal to two 𝜋 multiplied by the radius 𝑟 and that we’ve labeled the speed as 𝑣.

So, if we look now at this equation, we see that starting with the simple speed–distance–time equation, we have got back to our equation for the period of a circular orbit. We know the values of the radius 𝑟 and the speed 𝑣 of the satellite, so we can substitute those values into this equation in order to calculate the period capital 𝑇. When we do this, we have that capital 𝑇 is equal to two 𝜋 multiplied by 42,200 kilometers divided by 3.1 kilometers per second. Evaluating this expression gives a result of 85,532.39 seconds, where the ellipses indicate that there are further decimal places. Notice that since the radius 𝑟 was measured in units of kilometers and the speed was in kilometers per second, then our period capital 𝑇 comes out in units of seconds.

However, the question asks us to give our answer in hours. We know that there are 60 seconds in each minute and that there are 60 minutes in each hour. Equivalently, we can say that there is one sixtieth of a minute in each second and one sixtieth of an hour in each minute. So, to convert our result to units of hours, we take the value in units of seconds and multiply it by one over 60 minutes per second and then again by one over 60 hours per minute. If we track what’s going on with the units, we see that we have seconds which cancel with the per second and that we have minutes that cancel with the per minute. This leaves us with units of hours. Doing the multiplication gives us the period of the orbit in units of hours as 23.759 hours where again the ellipses indicate that there are further decimal places.

The final step is that the question wants the result given to the nearest hour, so we need to round this value to the nearest whole number of hours. Doing this gives our final answer to the question that, to the nearest hour, the period of the satellite’s orbit is equal to 24 hours.

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