Question Video: Understanding Speed and Radius in the Orbital Speed Equation Physics • 9th Grade

A satellite follows a circular orbit around Earth at a radial distance 𝑅 and with an orbital speed 𝑣. At what radius, in terms of 𝑅, would the satellite have to orbit in order to have an orbital speed of 𝑣/2?

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Video Transcript

A satellite follows a circular orbit around Earth at a radial distance 𝑅 and with an orbital speed 𝑣. At what radius, in terms of 𝑅, would the satellite have to orbit in order to have an orbital speed of 𝑣 divided by two?

We’re considering a satellite in circular orbit. So let’s get started by recalling the orbital speed formula, 𝑣 equals the square root of 𝐺𝑀 divided by π‘Ÿ. And although we do not have any strictly numeric values to substitute into this formula, we can still use it to consider how orbital speed 𝑣 and orbital radius π‘Ÿ relate to each other. In this question, these two are the only quantities that change. So moving forward, we’ll consider them to be the only variables in this formula. To understand why, recall that 𝐺 represents the universal gravitational constant, a value that never changes. And although mass 𝑀 can assume the mass value of whatever large object is at the center of orbit, here 𝑀 represents the mass of Earth, which isn’t going to suddenly change. So we’ll treat 𝑀 as a constant as well.

Now, to observe how 𝑣 and π‘Ÿ change with respect to each other, we’re going to use the formula to write a proportion. To devise our proportion, we’ll start by copying the formula and we will replace the equal sign with this symbol, which means β€œis proportional to.” And since a proportion tells how variables change with respect to each other, we’ll ignore the unchanging constant values 𝐺 and 𝑀. And we’ll use a one to hold their place in the numerator. Now, after distributing the radical and simplifying, we have that 𝑣 is proportional to one over the square root of π‘Ÿ. Another way to say this is that 𝑣 is inversely proportional to the square root of π‘Ÿ because, unlike 𝑣, the square root of π‘Ÿ is in the denominator.

So as one quantity increases, the other must decrease and the other way around in order to maintain this proportionality. So what changes between the initial and final orbits of the satellite? Well, we know that the speed 𝑣 is originally equal to 𝑣 and in its final orbit is equal to 𝑣 divided by two. So this decrease in 𝑣 must correspond to an increase in the square root of π‘Ÿ. The initial orbital radius is 𝑅. So we can expect that the final orbit has a radius greater than 𝑅. Now, given the factor by which 𝑣 decreases, the proportion will tell us the exact factor by which π‘Ÿ must increase. We know that speed decreases from 𝑣 to 𝑣 divided by two. So the entire left-hand side of the proportion is decreasing by a factor of two.

Now, at this point, we could just take a guess and think that because 𝑣 decreases by a factor of two, 𝑅 must increase by a factor of two. However, one very important thing to note is that π‘Ÿ appears under a radical. Therefore, whatever factor gets stuck with π‘Ÿ is going to have its square root taken. And ultimately, it’s that entire right-hand side, that one over the square root of π‘Ÿ, that maintains the balance of their proportion. So the correct new factor of π‘Ÿ has the square root two. Orbital radius then must increase by a factor of four. So in terms of the original radius 𝑅, we have found that for the satellite to maintain circular orbit with the speed of 𝑣 over two, it must have an orbital radius of four times 𝑅.

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