WEBVTT physics/ap-physics-1-2/fullerton
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Hi everyone and welcome back to Educator.com.
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In this mini-lesson, we are going to go through the first page of the APlusPhysics worksheet on gravity and you can download that worksheet from the link right below the video here.
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I recommend taking a minute, printing that out, giving it a try and then we will come and check our answers.
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Number 1 -- A space probe is launched into space from Earth's surface.
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Which graph represents the relationship between the magnitude of the gravitational force exerted on Earth by the space probe and the distance between the space probe and the center of Earth.
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We know that the gravitational force equals (g)m1m2/r², so we have this inverse-square law relationship between the distance and the gravitational force.
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You can also see that as you have greater distances, you are going to have lower forces, so that eliminates Number 1 and Number 4.
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Because it is an inverse-square law, not a linear relationship, Number 2 must be our best answer.
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Number 2 -- The diagram shows two bowling balls (A) and (B), each having a mass of 7 kg placed 2 m apart.
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What is the magnitude of the force exerted by ball (A) on ball (B)?
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To do this I am going to go to Newton's Law of Universal Gravitation, where force of gravity is m1m2/r² -- (G) is 6.67 × 10^-11 N-m²/kg², and mass 1 is 7 kg and mass 2 is 7 kg divided by the square of the distance between them, 2².
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This implies then that we have 6.67 × 10^-11 × 7 × 7/4 (2²)...
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...and I end up with an answer of around 8.17 × 10^-10 N corresponding to answer Number 3.
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Number 3 -- A 60 kg physics student would weigh 1560 N on the surface of Planet (X).
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What is the magnitude of the acceleration due to gravity on the surface of Planet (X)?
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If the weight (mg) is 1560 N, and if we want (G), the acceleration due to gravity, that is going to be 1560 N/60 kg (mass)...
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...therefore, the acceleration due to gravity must be 1560/60 or about 26 m/s², so the correct answer is Number 4.
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Number 4 -- Earth's mass is approximately 81 times the mass of the moon.
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If the Earth exerts a gravitational force of magnitude (F) on the moon, the magnitude of the gravitational force of the moon on Earth is...?
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If Earth exerts a force (F) on the moon, the moon must exert (F) back on Earth, and that is Newton's Third Law.
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Number 5 -- An object weighs 100 N on Earth's surface. When it is moved to a point 1 Earth radius above Earth's surface, it will weigh...?
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If there is the Earth -- initially it is on Earth's surface at a distance (r) away, then if it is moved so that it is now a distance of 2(r) away, we have doubled the distance.
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Remember that the force of gravity is an inverse-square law with a distance -- so effectively, we have doubled the distance, therefore we have cut the force in four, or 1/4, so 1/4 of 100 N is going to be answer Number 1, 25 N.
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Number 6 -- A container of rocks with a mass of 65 kg is brought back from the moon's surface where the acceleration due to gravity is 1.62 m/s².
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What is the weight of the container of rocks on Earth's surface?
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Weight is just mass times the acceleration due to gravity and our mass is 65 kg, the acceleration due to gravity on Earth is 9.8 m/s², so that is going to be right around 638 N, answer Number 1.
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We did not need to know the acceleration due to gravity on the moon since we already knew what its mass was.
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Number 7 -- The graph below represents the relationship between gravitational force and mass for objects near the surface of the Earth.
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What does the slope of the graph represent?
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Slope is going to be rise/run, so we are going to have change in gravitational force divided by mass.
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If you remember, the gravitational force is also (mg), so (Fg) -- if I divide both sides by (m), Fg/m = G (the acceleration due to gravity), so our slope is going to give us the acceleration due to gravity, answer Number 1.
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Let us take a look at one more -- A person weighing 785 N on the surface of Earth would weigh 298 N on the surface of Mars.
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What is the magnitude of the gravitational field strength on the surface of Mars?
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The way I would do that is we know that on Earth, their weight (mg) is 785 N, so let us find their mass.
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If I divide both sides by (g), 9.8 m/s², then mass is going to be 785/9.8 m/s² or about 80 kg, so on Mars they weighed 298 N, so on Mars, (m) times the acceleration on Mars is 298 N.
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If we want (g) now, we will divide both sides by (m), which will be 298 N/80 kg (mass), so 298/80 = 3.72 N/kg, so the correct answer there is Number 2.
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That completes page 1. Thanks so much for your time everybody. Make it a great day!