WEBVTT
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A planet orbits a star.
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And the planet is orbited by a moon.
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At a particular time in the moon’s orbit around the planet, the gravitational force on the moon from the planet acts perpendicularly to the gravitational force on the moon from the star.
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The force on the moon from the planet, 𝐹 sub one, equals 1.21 times 10 to the 19th newtons.
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And the force on the moon from the star, 𝐹 sub two, equals 5.00 times 10 to the 19th newtons.
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The moon’s mass is 1.13 times 10 to the 22nd kilograms.
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What is the magnitude of the moon’s resultant acceleration?
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Let’s start on our solution by drawing a diagram of this situation.
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In this scenario, we have a star.
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And orbiting around that star is a planet.
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And orbiting around the planet is a moon.
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And we’re told that there comes a particular time in the orbit of the moon.
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We’re told that there’s a particular time in the moon’s orbit when the force on the moon from the star and the force on the moon from the planet are at right angles or perpendicular to one another.
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If we draw out an expanded sketch of the forces acting on the moon, we can say that the force of the planet acting on the moon is 𝐹 sub one.
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And the force of the star on the moon is 𝐹 sub two.
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In the problem statement, we’re told the magnitude of each one of those forces, 𝐹 sub one and 𝐹 sub two.
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And we’re also told the mass of the moon, which we can call 𝑚 sub 𝑚.
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Given all this information, we want to solve for the magnitude of the acceleration the moon will experience when 𝐹 sub one and 𝐹 sub two are perpendicular to one another.
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We can call that acceleration 𝑎.
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And to get started solving for it, let’s consider again our two forces and what the resultant force on the moon is.
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Finding the resultant force acting on the moon due to 𝐹 one and 𝐹 two involves adding those two forces together.
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Adding these vectors graphically, we see that the resultant force, 𝐹 sub 𝑅, is the hypotenuse of a right triangle whose sides are 𝐹 sub one and 𝐹 sub two in magnitude.
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That means that, to solve for the magnitude of the resultant force on the moon, we can use the Pythagorean theorem.
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This theorem tells us that the magnitude of 𝐹 sub 𝑅 squared equals the magnitude of 𝐹 sub one squared plus the magnitude of 𝐹 sub two squared.
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And if we take the square root of both sides of this equation, we see we now have an expression for the magnitude of 𝐹 sub 𝑅 by itself.
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We’re given the magnitudes of 𝐹 sub one and 𝐹 sub two and can plug those into this expression now.
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Before entering this value on our calculator, we can recall that we wanna solve for the moon’s acceleration magnitude 𝑎.
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By Newton’s second law of motion, this acceleration 𝑎 is equal to the net force acting on an object divided by its mass.
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So in our case, we take our resultant force magnitude, divide it by the mass of the moon whose value we’re given.
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And that fraction will be equal to our acceleration magnitude 𝑎.
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When we enter this entire expression on our calculator, we find that 𝑎, to three significant figures, is 4.55 times 10 to the negative third meters per second squared.
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That’s the moon’s acceleration magnitude under the influence of the planet and the star.