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Ch 13: Gravitation
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 13, Problem 42

You decide to visit Santa Claus at the north pole to put in a good word about your splendid behavior throughout the year. While there, you notice that the elf Sneezy, when hanging from a rope, produces a tension of 395.0 N in the rope. If Sneezy hangs from a similar rope while delivering presents at the earth's equator, what will the tension in it be? (Recall that the earth is rotating about an axis through its north and south poles.)

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1
Understand that the tension in the rope is affected by the gravitational force acting on Sneezy and any additional forces due to the Earth's rotation.
At the North Pole, the tension in the rope is equal to the gravitational force acting on Sneezy, which is given as 395.0 N. This is because there is no centrifugal force acting on Sneezy at the poles.
At the equator, Sneezy experiences a centrifugal force due to the Earth's rotation. This force acts outward and reduces the effective gravitational force, thus affecting the tension in the rope.
Calculate the centrifugal force using the formula: Fcf = mω²r, where m is Sneezy's mass, ω is the angular velocity of the Earth, and r is the radius of the Earth.
Determine the new tension in the rope at the equator by subtracting the centrifugal force from the gravitational force: T = mg - mω²r. Use the known values to find the tension.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Tension in a Rope

Tension is the force exerted along a rope or string when it is pulled tight by forces acting from opposite ends. It is a vector quantity, meaning it has both magnitude and direction. In this scenario, the tension in the rope is equal to the gravitational force acting on Sneezy, assuming he is in equilibrium and not accelerating vertically.
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Centripetal Force and Earth's Rotation

Centripetal force is required for an object to move in a circular path and is directed towards the center of the circle. At the equator, the Earth's rotation provides a centripetal force that slightly reduces the effective gravitational force experienced by objects. This means the tension in the rope at the equator will be slightly less than at the poles due to this reduction in effective weight.
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Gravitational Force Variation

Gravitational force on an object is given by F = mg, where m is mass and g is the acceleration due to gravity. However, g is not constant across the Earth's surface; it is slightly less at the equator than at the poles due to the Earth's rotation and its equatorial bulge. This variation affects the tension in the rope when Sneezy is at the equator compared to the North Pole.
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Related Practice
Textbook Question

Consider the ringshaped body of Fig. E13.35. A particle with mass m is placed a distance x from the center of the ring, along the line through the center of the ring and perpendicular to its plane. (a) Calculate the gravitational potential energy U of this system. Take the potential energy to be zero when the two objects are far apart. (b) Show that your answer to part (a) reduces to the expected result when x is much larger than the radius a of the ring. (c) Use Fx = -dU/dx to find the magnitude and direction of the force on the particle (see Section 7.4). (d) Show that your answer to part (c) reduces to the expected result when x is much larger than a. (e) What are the values of U and Fx when x = 0? Explain why these results make sense.

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Textbook Question

Astronomers have observed a small, massive object at the center of our Milky Way galaxy. A ring of material orbits this massive object; the ring has a diameter of about 15 light-years and an orbital speed of about 200 km/s. Observations of stars, as well as theories of the structure of stars, suggest that it is impossible for a single star to have a mass of more than about 50 solar masses. Can this massive object be a single, ordinary star?

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Textbook Question

A thin, uniform rod has length L and mass M. A small uniform sphere of mass m is placed a distance x from one end of the rod, along the axis of the rod (Fig. E13.34). Calculate the gravitational potential energy of the rod–sphere system. Take the potential energy to be zero when the rod and sphere are infinitely far apart. Show that your answer reduces to the expected result when x is much larger than L.

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Textbook Question

A thin, uniform rod has length L and mass M. A small uniform sphere of mass m is placed a distance x from one end of the rod, along the axis of the rod (Fig. E13.34). Use Fx = -dU/dx to find the magnitude and direction of the gravitational force exerted on the sphere by the rod (see Section 7.4). Show that your answer reduces to the expected result when x is much larger than L.

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Textbook Question

In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting around once every 27 hours and moving at 30,000 km/s. What is the mass of this black hole, assuming circular orbits? Express your answer in kilograms and as a multiple of our sun's mass.

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Textbook Question

In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting around once every 27 hours and moving at 30,000 km/s. How far are these clumps from the center of the black hole?

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