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

A satellite orbits Earth at an orbital radius of 7,720 km and moves at a speed of 7.2 km/s . The satellite has a circular orbit. What is the circumference of the satellite’s orbit? 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 minutes to the nearest minute.

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Video Transcript

A satellite orbits Earth at an orbital radius of 7,720 kilometers and moves at a speed of 7.2 kilometers per second. The satellite has a circular orbit. What is the circumference of the satellite’s orbit? 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 minutes to the nearest minute.

Okay, so, in this question, we have a satellite that’s orbiting Earth. And in the first part of the question, we’re asked to work out the circumference of the satellite’s orbit. We are told that the satellite has an orbital radius of 7,720 kilometers. So, if this blue blob here represents the Earth and this pink blob is the satellite, then this orange dashed line represents the trajectory of the satellite’s orbit and this orbit has a radius of 𝑟 measured from the centre of mass of the Earth, where 𝑟 is equal to 7,720 kilometers. We know that this orbit is circular, and we’re asked to find its circumference.

We can recall that the circumference of a circle 𝑐 is equal to two 𝜋 multiplied by the radius 𝑟 of that circle. Substituting in that the value of 𝑟 for this satellite is 7,720 kilometers, we get that the circumference of the satellite’s orbit is equal to two 𝜋 multiplied by 7,720 kilometers. Doing the multiplication gives us that the circumference is equal to 48,506.19 kilometers, where the ellipses indicate that there are further decimal places.

The question asks us to give our answer in kilometres using scientific notation to one decimal place. So far, we’ve achieved one of those three things. Our answer is already in units of kilometres. Now, we just need to convert it to scientific notation and round it to one decimal place. To get this number into scientific notation, we move the decimal point one, two, three, four places to the left. Then, our result becomes 4.850619 et cetera times 10 to the four, still with units of kilometres. Finally, rounding to one decimal place, the number rounds up to give 4.9 times 10 to the four kilometers. And so, our answer to the first part of the question is that the circumference of the satellite’s orbit is equal to 4.9 times 10 to the four 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 orbit is 7,720 kilometers, and the question also tells us that the satellite moves in this orbit at a speed of 7.2 kilometers per second. Let’s label this speed as 𝑣. We can recall that the orbital period capital 𝑇 is equal to two 𝜋 multiplied by the radius 𝑟 divided by the speed 𝑣.

It’s worth pointing out that this equation isn’t telling us anything particularly strange. This two 𝜋𝑟 in the numerator is simply equal to the circumference 𝑐 of the orbit, in other words, the distance traveled by the satellite in a time equal to one period capital 𝑇. So, this equation is really just saying that the time taken to complete one orbit is equal to the distance traveled in that orbit divided by the speed of the satellite. By multiplying both sides by the speed and canceling the speeds in the numerator and denominator on the right and then dividing both sides by time and canceling the time in the numerator with the time in the denominator on the left, then we see that this formula was really nothing more than a different way of writing the equation speed equals distance over time.

We know that two 𝜋 times the radius 𝑟 is equal to the circumference 𝑐 of the orbit, and we’ve already found the value of this circumference. Substituting the circumference 𝑐 in place of two 𝜋𝑟 in this equation gives us that capital 𝑇 is equal to 𝑐 divided by 𝑣. Then putting in our values for the circumference 𝑐 and the speed 𝑣, we get that the period capital 𝑇 is equal to 4.9 times 10 to the four kilometers divided by 7.2 kilometers per second.

For ease of writing, we’ve written out the rounded value of the circumference here. But when we’re actually doing the calculation, we need to be careful to use the full precise value. When we do this division, we find that the value of the period is equal to 6,736.97 and so on with further decimal places. And since the circumference is given in units of kilometres and the speed is in kilometres per second, then our period capital 𝑇 will have units of seconds. The question asks us to give our answer in minutes to the nearest minute. Since there are 60 seconds in one minute or equivalently there is one-sixtieth of a minute in one second, then to convert our result to units of minutes we take our value in units of seconds and multiply by one over 60 minutes per second.

When we look at the units, we see that we have seconds which cancel with per second. So, we get left with units of minutes. Doing the multiplication gives us that the period capital 𝑇 is equal to 112.28 minutes, where again the ellipses are used to indicate that there are further decimal places. The final step is to round this result to the nearest minute. When we do this, we get our final answer that to the nearest minute the period of the satellite’s orbit is equal to 112 minutes.

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