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There are two planets in orbit around a star.
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Both planets have circular orbits.
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Planet A has a speed of π£, and planet B has a speed of two π£.
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Give your answers to these questions as a decimal if necessary: what is the ratio of the orbital radii of the two planets, π sub B over π sub A?
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What is the ratio of the orbital periods of the two planets, π sub B over π sub A?
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Letβs start by drawing a sketch of the situation.
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In this sketch, planets A and B are both in orbit with planet A moving in a circular orbit with instantaneous speed π£ and planet B also in a circular orbit with instantaneous speed two π£.
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As we look into solving for the ratio of their orbital radii, letβs recall the relationship for the speed of an object in a circular orbit.
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In the case of a circular orbit, a satelliteβs speed is equal to the square root of πΊ, the universal gravitational constant, times the mass of the body around which the satellite orbits divided by the radius; that is, the distance from the center of the orbital mass to the satellite.
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In our diagram, πB and πA start at the center of the orbited mass.
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Letβs use this relationship for circular orbit speed to solve for this ratio of radii, the speed of planet A, π£ sub A, equals the square root of capital πΊ times the mass of the star divided by π sub A.
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And weβre told that that speed is equal to π£.
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π£ sub B, the speed of planet B, is equal to the square root of capital πΊ times the mass of the star divided by π sub B.
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Then weβre told that this is equal to two times π£.
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If we divide π£ sub A by π£ sub B, then significant cancelation happens in this ratio.
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We can crossout πΊ and π, the universal gravitational constant and the mass of the star.
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We can also cancel out the orbital speed, π£.
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This leaves us with a simplified expression; that one over the square root of π sub A divided by one over the square root of π sub B is equal to one divided by two.
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If we then square both sides of the equation, this cancels out the square roots in the left side of our equation and gives us the result one over π sub A divided by one over π sub B equals one divided by four.
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If we multiply both the numerator and the denominator on the left-hand side of the equation by π sub B, this cancels out π sub B in the denominator, leaving us with a simplified fraction π sub B divided by π sub A which is equal to one-fourth or, in decimal form, 0.25.
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That concludes part one.
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Now we want to move on to solve for the ratio of the orbital periods π sub B over π sub A.
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As we start here, letβs recall the relationship for orbital period for a circular orbit.
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Orbital period, π, for a circular orbit equals two π times the square root of the orbital radius cubed all over πΊ times the mass of the planet or object being orbited.
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If we write this expression for our scenario, π sub A, the orbital period of planet A, is equal to two π times the square root of π sub A cubed divided by πΊ times π; and likewise, π sub B, the orbital period of planet B, equals two π rimes the square root of π sub B cubed over πΊπ.
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Just like with the ratio of radii, if we divide π sub A by π sub B, then we see the factors of two π cancel out as well as the factors of one over the square root of πΊ times π.
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Weβre left with a ratio π sub A cubed divided by π sub B cubed all raised to the one-half power.
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Now rather than π sub A divided by π sub B, we want the inverse of that ratio.
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We can invert both sides of the equation and simplify the exponent so that the right side is π sub B divided by π sub A raised to the to the three-halves power.
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Our solution earlier show that this ratio, π sub B to π sub A, is equal to one-fourth.
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When we raise one-fourth to the three-halves power, we find the result of one-eighth or, in decimal notation, 0.125.
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This is the ratio of the orbital period of planet B to the orbital period of planet A.