Explanation Apply the Kepler’s third law in this situation, and write the relation between time period and length of semi major axis. T 2 s 3 = 4 π 2 G M SgrA (I) Here, T is the period of revolution, s is the semi major axis, G is the gravitational constant, and M SgrA is the mass of SgrA. Solve equation (I) for M SgrA . M SgrA = 4 π 2 s 3 G T 2 (II) The given situation is analogues to that shown in Figure 6-16. Write the expression for distance of closest approach. r min = s − e s (III) Here, r min is the distance of closest approach, s is the semi major axis, and e is the eccentricity. Solve equation (III) for s . r min = s ( 1 − e ) s = r min 1 − e (IV) Use equation (IV) in (II). M SgrA = 4 π 2 ( r min 1 − e ) 3 G T 2 (V) Conclusion: Given that, distance of closest approach is 123 AU , eccentricity is 0

Pearson eText -- Physics for Scientists and Engineers with Modern Physics -- Instant Access (Pearson+)

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Author: Douglas Giancoli

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Central force is the force that acts along the center of mass of one body to the center of mass of another body. Its direction from or away from a point on a body to another is called the center of the force. Such forces include gravitational force betwe…

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Chapter 6, Problem 71GP

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The mass of SgrA in kg and in terms of mass of the Sun.

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Chapter 6, Problem 70GP

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The mass of SgrA in kg and in terms of mass of the Sun.

Solution Summary: The author compares the mass of SgrA in kg with that of the Sun. They apply the Kepler’s third law and write the relation between time period and length of semi major

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The mass of SgrA in kg and in terms of mass of the Sun.

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