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Chapter 12: Problem 71

The astronomical unit (AU, equal to the mean radius of the Earth's orbit) is \(1.4960 \cdot 10^{11} \mathrm{~m},\) and a year is \(3.1557 \cdot 10^{7}\) s. Newton's gravitational constant is \(G=\) \(6.6743 \cdot 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2} .\) Calculate the mass of the Sun in kilograms. (Recalling or looking up the mass of the Sun does not constitute a solution to this problem.)

Short Answer

Expert verified
Answer: The approximate mass of the Sun is \(1.989 \cdot 10^{30}\) kilograms.

Step by step solution

01

Write down Kepler's Third Law of Planetary Motion

Kepler's Third Law states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (a) of its orbit, with the constant of proportionality being determined by the masses of the two celestial bodies involved and Newton's gravitational constant (G): \(T^2 = \frac{4\pi^2}{G(M_1 + M_2)} \cdot a^3\) Since the mass of the Earth (\(M_2\)) is much smaller than the mass of the Sun (\(M_1\)), for this problem, we can approximate the equation as: \(T^2 \approx \frac{4\pi^2}{GM_1} \cdot a^3\)
02

Insert the given values in the equation

Now let's insert the given values for the astronomical unit (AU), the length of a year in seconds (T), and Newton's gravitational constant (G) into the equation we derived in Step 1: \((3.1557 \cdot 10^{7} \mathrm{~s})^2 \approx \frac{4\pi^2}{(6.6743 \cdot 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2})(M_1)} \cdot (1.4960 \cdot 10^{11} \mathrm{~m})^3\)
03

Solve for the mass of the Sun (M_1)

Now we need to solve the equation for the mass of the Sun (\(M_1\)). We first simplify the equation: \(M_1 \approx \frac{4\pi^2 (1.4960 \cdot 10^{11} \mathrm{~m})^3}{(6.6743 \cdot 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2})(3.1557 \cdot 10^{7} \mathrm{~s})^2}\) Then we calculate the value of \(M_1\): \(M_1 \approx 1.989 \cdot 10^{30} \mathrm{~kg}\) So the mass of the Sun is approximately \(1.989 \cdot 10^{30}\) kilograms.

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Most popular questions from this chapter

A man inside a sturdy box is fired out of a cannon. Which of following statements regarding the weightless sensation for the man is correct? a) The man senses weightlessness only when he and the box are traveling upward. b) The man senses weightlessness only when he and the box are traveling downward. c) The man senses weightlessness when he and the box are traveling both upward and downward. d) The man does not sense weightlessness at any time of the flight.

A planet is in a circular orbit about a remote star, far from any other object in the universe. Which of the following statements is true? a) There is only one force acting on the planet. b) There are two forces acting on the planet and their resultant is zero. c) There are two forces acting on the planet and their resultant is not zero. d) None of the above statements are true.

Newton was holding an apple of mass \(100 . \mathrm{g}\) and thinking about the gravitational forces exerted on the apple by himself and by the Sun. Calculate the magnitude of the gravitational force acting on the apple due to (a) Newton, (b) the Sun, and (c) the Earth, assuming that the distance from the apple to Newton's center of mass is \(50.0 \mathrm{~cm}\) and Newton's mass is \(80.0 \mathrm{~kg}\).

With the usual assumption that the gravitational potential energy goes to zero at infinite distance, the gravitational potential energy due to the Earth at the center of Earth is a) positive. b) negative. c) zero. d) undetermined.

A 1000.-kg communications satellite is released from a space shuttle to initially orbit the Earth at a radius of \(7.00 \cdot 10^{6} \mathrm{~m}\). After being deployed, the satellite's rockets are fired to put it into a higher altitude orbit of radius \(5.00 \cdot 10^{7} \mathrm{~m} .\) What is the minimum mechanical energy supplied by the rockets to effect this change in orbit?
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