A planet with a mass of 3.05E+27 kg orbits a star of mass m_s = 1.42E +30 kg in
ID: 1501205 • Letter: A
Question
A planet with a mass of 3.05E+27 kg orbits a star of mass m_s = 1.42E +30 kg in an elliptical orbit. The figure shows a view from above the system. At the time shown in the figure, the planet is 5.90E+11 m away from the center of the star moving with a speed of 1.27E+4 m/s with theta = 54.5 degrees as shown. The star has a radius of r_s = 6.22E+8 m and rotates about its own axis with a period of 28.8 days. The following question asks you to compare the angular momentum of the planet orbiting around the center of the star to the angular momentum of the rotating star. Assume that the star is a uniform sphere with Icm = 2/5mstj and that the planet is a point mass (i.e., you can ignore the rotation of the planet around its own axis.) What is the ratio of the magnitude of the angular momentum of the planet to the magnitude of the angular momentum of the star? What is the ratio_of the kinetic energy of the planet to the kinetic energy of the star?Explanation / Answer
A) Angular momentum of planet = m * v * r * sin(theta)
= 3.05 * 1027 * 1.27 * 104 * 5.90 * 1011 * sin54.4
= 1.858 * 1043 kg. m2/s
Angular momentum of star = I * w
= (2/5 * M * R2) * w
= (2/5 * 1.42 * 1030 * (6.22 *108)2) * (2 * 3.14/28.8 * 24 * 60 *60)
= 5.546 * 1041 kg. m2/s
=> Ratio of Angular momentum of planet to Angular momentum of star = (1.858 * 1043 )/(5.546 * 1041)
= 33.501
B) Kinetic energy of planet = 1/2 * m * v2
= 1/2 * 3.05 * 1027 * (1.27 * 104)2
= 2.4596 * 1035 J
Kinetic energy of star = 1/2 * (2/5 * M * R2) * w2
= 1/2 * (2/5 * 1.42 * 1030 * (6.22 *108)2) * (2 * 3.14/28.8 * 24 * 60 *60)2
= 7 * 1035 J
=> Ratio of Kinetic energy of planet to Kinetic energy of star = (2.4596 * 1035)/(7 * 1035 )
= 0.3513
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.