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An object orbits a planet at a distance π from the planetβs center of mass.
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The planet has a mass π.
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If the object is moved to an orbit three π from the planetβs center of mass, by what factor does the local acceleration due to gravity at its orbital radius change?
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So letβs start with a diagram.
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Hereβs an object in orbit around the planet.
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And we know that the distance between the object and the planetβs center of mass is some distance π.
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We then move this object to a new orbit, at which point its distance from the planetβs center of mass is three π.
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We need to calculate the factor by which the local acceleration due to gravity changes.
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So letβs recall the equation for the local acceleration due to gravity.
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That is, π equals πΊπ over π squared, where π is the acceleration due to gravity, πΊ is the universal gravitational constant, π is the mass of the planet, and π is the orbital radius.
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Now, in this situation, the mass of the planet is not changing.
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The universal gravitational constant πΊ never changes.
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So the only quantity thatβs changing is the orbital radius π.
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And π is in the denominator of this fraction, which means that if π increases, π will decrease.
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So we know that the acceleration due to gravity once the orbital radius is increased will be lower.
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And we just have to work out by how much.
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So letβs start with the acceleration due to gravity at the original orbital radius, which weβll call π sub one.
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And this is equal to πΊπ over π squared.
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The local acceleration due to gravity after the object has moved, which weβll call π sub two, is equal to πΊπ over three π squared.
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Now, here itβs useful to recall that any time we have two quantities multiplied together, say π₯ times π¦, all squared, this is equivalent to π₯ squared times π¦ squared.
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So here we can say that π sub two is equal to πΊπ divided by three squared times π squared.
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And three squared is equal to nine.
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So we have π sub two is equal to πΊπ over nine π squared.
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Now, itβs useful to note here that πΊπ over π squared is the same as π sub one.
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So we can separate the one over nine and write that π sub two is equal to one divided by nine times πΊπ over π squared.
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And πΊπ over π squared is equal to π one.
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So we found that the local acceleration due to gravity at this point after the object has been moved is one-ninth of the value that it had before the object was moved.
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So the factor by which the local acceleration due to gravity at the orbital radius of the object has changed is one-ninth.