What is the ratio of the kinetic energy of the planet to the kinetic energy of t

Question

What is the ratio of the kinetic energy of the planet to the kinetic energy of the star?

A planet with a mass of 3.06E+27 kg orbits a star of mass ms = 2.28E+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 8.20E+11 m away from the center of the star moving with a speed of 1.36E+4 m/s with ? = 59 degrees as shown. The star has a radius of rs= 7.28E+8 m and rotates about its own axis with a period of 21.8 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=2/5msrs2 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 kinetic energy of the planet to the kinetic energy of the star?

Explanation / Answer

Note that for the planet,

Lplanet = m v r sin(theta)

Thus,

Lplanet = 2.925*10^43 kg*m^2/s

For the star,

Lstar = (2/5 ms rs^2) (2 pi /T)

where T = 21.8 days = 1.883*10^6 s

Thus,

Lstar = 1.612*10^42 kg*m^2/s

THUS, THE PLANET HAS GREATER ANGULAR MOMENTUM!

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For the KE,

KEplanet = 1/2 m v^2 = 2.83*10^35 J

KEstar = 1/2 I w^2 = 1/2 (2/5 ms rs^2) (2 pi /T)^2

KEstar = 2.69*10^36 J

Thus,

KEplanet/KEstar = 0.105   [ANSWER]

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