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A satellite follows a circular orbit around Earth at a radial distance ๐
and with an orbital speed ๐ฃ.
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If the satellite were moved closer to Earth so that it followed a circular orbit with the radius of ๐
divided by nine, at what speed, in terms of ๐ฃ, would it have to move in order to maintain its orbit?
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In this question, weโre going to consider two different setups for this system, the satellite orbiting Earth.
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In the first setup, the satellite has an orbital radius of ๐
and an orbital speed ๐ฃ.
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Then, in the second setup, weโre imagining that the satellite is moved much closer to Earth to a radial distance one-ninth of ๐
.
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And itโs our job to figure out what speed it needs to stay in circular orbit.
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Letโs begin by recalling a formula involving orbital radius and speed in a special case of circular orbit like we have here.
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Itโs the orbital speed formula ๐ฃ equals the square root of ๐บ๐ divided by ๐, where ๐ฃ is orbital speed.
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๐บ is the universal gravitational constant.
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๐ is the mass of the large body being orbited, which is Earth in this case.
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And ๐ is orbital radius.
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Notice that we havenโt been given any numeric values for orbital speed or orbital radius.
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However, we can still use this formula to learn how ๐ฃ responds to a change in ๐.
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Another thing to note is that while ๐บ and ๐ represent real constant quantities, we wonโt need to know or use their exact values either.
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Since we wonโt be calculating a numeric answer, itโs not necessary to write out their entire values.
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And itโll be simpler to just leave them written as ๐บ and ๐.
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Now, letโs get organized and write out what we know, beginning with the first orbit and using the subscript one to refer to this initial setup.
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We know the initial orbital radius ๐ one equals ๐
and the initial orbital speed ๐ฃ one equals ๐ฃ.
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And since we also know that the satellite is in circular orbit here, we know that these values must satisfy this formula.
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In other words, if we substitute these values into the orbital speed formula, we know that the resulting relationship must be true.
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Therefore, we know that the initial orbital speed ๐ฃ equals the square root of ๐บ๐ divided by the initial orbital radius ๐
.
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Now letโs move on to the proposed new orbit, which weโll indicate using the subscript two.
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We know that the new orbital radius ๐ two equals ๐
divided by nine.
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And we wanna find the new orbital speed ๐ฃ two that meets the conditions of circular orbit and therefore satisfies this formula.
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So letโs substitute these values in and see what happens.
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We still donโt know the satelliteโs final speed, so weโll leave it written as ๐ฃ two, which equals the square root of ๐บ๐ divided by ๐
over nine.
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We can simplify the math, first by writing it as the square root of nine ๐บ๐ divided by ๐
.
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And then we can take the square root of nine to move it out from under the radical.
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So now we know the final speed ๐ฃ two equals three times the square root of ๐บ๐ divided by ๐
, the initial radius.
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And remember, weโve already established that the square root of ๐บ๐ divided by ๐
equals ๐ฃ, the satelliteโs initial orbital speed.
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So letโs make this substitution and write the result up here.
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The satelliteโs final speed ๐ฃ two equals three times ๐ฃ, the satelliteโs initial speed.
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And so we have our answer.
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In terms of ๐ฃ, we found that for the satellite to maintain circular orbit at radius ๐
over nine, it must move at orbital speed three ๐ฃ.