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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Astronomical Unit (AU)
The Astronomical Unit (AU) is a fundamental measurement used in astronomy to describe vast distances within our solar system. Specifically, one AU represents the mean distance from the Earth to the Sun, which is approximately 149.6 million kilometers or precisely, \(1.4960 \times 10^{11}\) meters.

This unit simplifies the description of distances in space, where traditional units like meters and kilometers would be cumbersome due to the sheer scale. The AU makes it easier to compare distances between celestial objects, such as planets and other entities within our solar system.

By using the AU, astronomers can effectively communicate and calculate orbital parameters of planets. In the context of Kepler's Third Law, the AU helps to determine the semi-major axis of Earth's orbit, which is essential for computing the mass of the Sun.
Gravitational Constant (G)
The Gravitational Constant, often denoted as \(G\), is a key component of Newton's Law of Universal Gravitation. It quantifies the strength of the gravitational force between two masses. The value of \(G\) is \(6.6743 \times 10^{-11} \text{ m}^3\text{ kg}^{-1}\text{ s}^{-2}\).

This constant allows us to calculate the gravitational attraction between two masses, a distance apart. It is crucial for understanding and predicting celestial phenomena, such as the orbits of planets, moons, and artificial satellites.

In the exercise, \(G\) is used in Kepler鈥檚 Third Law to relate the orbital period and the semi-major axis of a planet orbiting the Sun to derive the mass of the Sun. The constant provides the necessary balance in the equation, enabling accurate cosmic calculations on such grand scales.
Mass of the Sun
The mass of the Sun is an essential astronomical parameter that impacts the orbits of planets in our solar system. Through the application of Kepler's Third Law and the previously discussed parameters鈥擜stronomical Unit and Gravitational Constant鈥攚e can derive the mass of the Sun.

In the specified exercise, by simplifying the relationship given by Kepler鈥檚 Third Law \(T^2 \approx \frac{4\pi^2}{GM_1} \cdot a^3\), where \(T\) is the orbital period, \(a\) is the semi-major axis and \(M_1\) is the mass of the Sun, we can solve for \(M_1\).

Using the provided values, the resulting mass of the Sun calculated is approximately \(1.989 \times 10^{30}\) kilograms. This immense mass plays a critical role not only in maintaining the planets' orbits but also in the functioning of solar phenomena and energy emission.

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Most popular questions from this chapter

Determine the minimum amount of energy that a projectile of mass \(100.0 \mathrm{~kg}\) must gain to reach a circular orbit \(10.00 \mathrm{~km}\) above the Earth's surface if launched from (a) the North Pole or from (b) the Equator (keep answers to four significant figures). Do not be concerned about the direction of the launch or of the final orbit. Is there an advantage or disadvantage to launching from the Equator? If so, how significant is the difference? Do not neglect the rotation of the Earth when calculating the initial energies.

The distances from the Sun at perihelion and aphelion for Pluto are \(4410 \cdot 10^{6} \mathrm{~km}\) and \(7360 \cdot 10^{6} \mathrm{~km},\) respectively. What is the ratio of Pluto's orbital speed around the Sun at perihelion to that at aphelion?

Eris, the largest dwarf planet known in the Solar System, has a radius \(R=1200 \mathrm{~km}\) and an acceleration due to gravity on its surface of magnitude \(g=0.77 \mathrm{~m} / \mathrm{s}^{2}\) a) Use these numbers to calculate the escape speed from the surface of Eris. b) If an object is fired directly upward from the surface of Eris with half of this escape speed, to what maximum height above the surface will the object rise? (Assume that Eris has no atmosphere and negligible rotation.

Satellites in low orbit around the Earth lose energy from colliding with the gases of the upper atmosphere, causing them to slowly spiral inward. What happens to their kinetic energy as they fall inward?

A satellite of mass \(m\) is in an elliptical orbit (that satisfies Kepler's laws) about a body of mass \(M,\) with \(m\) negligible compared to \(M\) a) Find the total energy of the satellite as a function of its speed, \(v\), and distance, \(r\), from the body it is orbiting. b) At the maximum and minimum distance between the satellite and the body, and only there, the angular momentum is simply related to the speed and distance. Use this relationship and the result of part (a) to obtain a relationship between the extreme distances and the satellite's energy and angular momentum. c) Solve the result of part (b) for the maximum and minimum radii of the orbit in terms of the energy and angular momentum per unit mass of the satellite. d) Transform the results of part (c) into expressions for the semimajor axis, \(a\), and eccentricity of the orbit, \(e\), in terms of the energy and angular momentum per unit mass of the satellite.
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