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.

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Even though the Moon does not have an atmosphere, the trajectory of a projectile near its surface is only approximately a parabola. This is because the acceleration due to gravity near the surface of the Moon is only approximately constant. Describe as precisely as you can the actual shape of a projectile's path on the Moon, even one that travels a long distance over the surface of the Moon.

Is the orbital speed of the Earth when it is closest to the Sun greater than, less than, or equal to the orbital speed when it is farthest from the Sun? Explain.

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