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Ch 13: Gravitation
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 13, Problem 31

On October 15, 2001, a planet was discovered orbiting around the star HD 68988. Its orbital distance was measured to be 10.5 million kilometers from the center of the star, and its orbital period was estimated at 6.3 days. What is the mass of HD 68988? Express your answer in kilograms and in terms of our sun's mass.

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Start by understanding Kepler's Third Law, which relates the orbital period and distance of a planet to the mass of the star it orbits. The formula is: \( T^2 = \frac{4\pi^2}{G M} r^3 \), where \( T \) is the orbital period, \( r \) is the orbital radius, \( G \) is the gravitational constant, and \( M \) is the mass of the star.
Convert the given orbital distance from kilometers to meters. Since 1 kilometer equals 1000 meters, multiply 10.5 million kilometers by 1000 to get the distance in meters.
Convert the orbital period from days to seconds. There are 86400 seconds in a day, so multiply 6.3 days by 86400 to get the period in seconds.
Rearrange Kepler's Third Law to solve for the mass of the star: \( M = \frac{4\pi^2 r^3}{G T^2} \). Substitute the values for \( r \), \( T \), and \( G \) (which is approximately \( 6.674 \times 10^{-11} \) m³/kg/s²) into the equation.
To express the mass of HD 68988 in terms of our sun's mass, divide the calculated mass by the mass of the sun, which is approximately \( 1.989 \times 10^{30} \) kg.

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Key Concepts

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

Kepler's Third Law

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. This law is crucial for determining the mass of a star when the orbital characteristics of a planet are known. It can be expressed as T^2 ∝ a^3, where T is the orbital period and a is the semi-major axis.
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Gravitational Force and Orbital Motion

The gravitational force between a star and a planet provides the centripetal force necessary for the planet's circular motion. This relationship is described by Newton's law of universal gravitation, which states that the force is proportional to the product of the two masses and inversely proportional to the square of the distance between them. Understanding this helps in calculating the mass of the star based on the planet's orbit.
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Mass of the Sun as a Reference

The mass of the sun is often used as a reference unit in astrophysics, known as the solar mass (M☉). It provides a convenient way to express the mass of other stars in a comparative manner. Knowing the mass of HD 68988 in terms of solar masses allows for easier comparison and understanding of its size relative to our sun.
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Related Practice
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A uniform, spherical, 1000.0 kg1000.0\(\text{ kg}\) shell has a radius of 5.00 m5.00\(\text{ m}\). Sketch a qualitative graph of the magnitude of the gravitational force this sphere exerts on a point mass m as a function of the distance rr of mm from the center of the sphere. Include the region from r=0r = 0 to rr\(\to\]\infty\).

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A uniform, solid, 1000.0-kg sphere has a radius of 5.00 m. Find the gravitational force this sphere exerts on a 2.00-kg point mass placed at the following distances from the center of the sphere: (i) 5.01 m, (ii) 2.50 m.

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Textbook Question

A uniform, spherical, 1000.0-kg shell has a radius of 5.00 m. Find the gravitational force this shell exerts on a 2.00-kg point mass placed at the following distances from the center of the shell: (i) 5.01 m, (ii) 4.99 m, (iii) 2.72 m.

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