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A satellite follows a circular orbit around Earth at a radial distance π
and with an orbital speed π£.
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At what radius, in terms of π
, would the satellite have to orbit in order to have an orbital speed of π£ divided by two?
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Weβre considering a satellite in circular orbit.
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So letβs get started by recalling the orbital speed formula, π£ equals the square root of πΊπ divided by π.
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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 π£ and orbital radius π relate to each other.
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In this question, these two are the only quantities that change.
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So moving forward, weβll consider them to be the only variables in this formula.
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To understand why, recall that πΊ represents the universal gravitational constant, a value that never changes.
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And although mass π can assume the mass value of whatever large object is at the center of orbit, here π represents the mass of Earth, which isnβt going to suddenly change.
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So weβll treat π as a constant as well.
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Now, to observe how π£ and π change with respect to each other, weβre going to use the formula to write a proportion.
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To devise our proportion, weβll start by copying the formula and we will replace the equal sign with this symbol, which means βis proportional to.β
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And since a proportion tells how variables change with respect to each other, weβll ignore the unchanging constant values πΊ and π.
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And weβll use a one to hold their place in the numerator.
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Now, after distributing the radical and simplifying, we have that π£ is proportional to one over the square root of π.
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Another way to say this is that π£ is inversely proportional to the square root of π because, unlike π£, the square root of π is in the denominator.
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So as one quantity increases, the other must decrease and the other way around in order to maintain this proportionality.
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So what changes between the initial and final orbits of the satellite?
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Well, we know that the speed π£ is originally equal to π£ and in its final orbit is equal to π£ divided by two.
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So this decrease in π£ must correspond to an increase in the square root of π.
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The initial orbital radius is π
.
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So we can expect that the final orbit has a radius greater than π
.
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Now, given the factor by which π£ decreases, the proportion will tell us the exact factor by which π must increase.
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We know that speed decreases from π£ to π£ divided by two.
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So the entire left-hand side of the proportion is decreasing by a factor of two.
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Now, at this point, we could just take a guess and think that because π£ decreases by a factor of two, π
must increase by a factor of two.
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However, one very important thing to note is that π appears under a radical.
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Therefore, whatever factor gets stuck with π is going to have its square root taken.
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And ultimately, itβs that entire right-hand side, that one over the square root of π, that maintains the balance of their proportion.
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So the correct new factor of π has the square root two.
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Orbital radius then must increase by a factor of four.
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So in terms of the original radius π
, we have found that for the satellite to maintain circular orbit with the speed of π£ over two, it must have an orbital radius of four times π
.