Consider a star that orbits around Sagittarius A* in a circular orbit of 530 AU. If the star's...

Question:

Consider a star that orbits around Sagittarius A* in a circular orbit of 530 AU. If the star's orbital speed is 2500 km/s, what is its orbital period?

Orbital Period

The orbital period is the time taken by an object to complete one revolution along a circular path. The orbital period depends on two factors

  1. The radius of the circular path and
  2. The orbital speed of the object.

Answer and Explanation:

Data Given

  • Radius of the circular path {eq}r = 530 \ \rm AU = 530 \ \rm AU \times 1.50 \times 10^{11} \ \rm m/AU = 7.95 \times 10^{13} \ \rm m {/eq}
  • Orbital speed of the star {eq}v = 2500 \ \rm km/s = 2.5 \times 10^6 \ \rm m/s {/eq}

The orbital period is given by

{eq}\begin{align} T = \frac{2 \pi r}{v} \end{align} {/eq}

{eq}\begin{align} T = \frac{2 \pi \times 7.95 \times 10^{13} \ \rm m}{2.5 \times 10^6 \ \rm m/s} \end{align} {/eq}

{eq}\begin{align} T = 2.0 \times 10^8 \ \rm s \end{align} {/eq}

{eq}\begin{align} \color{blue}{\boxed{T =6.33 \ \rm Years }} \end{align} {/eq}


Learn more about this topic:

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Kepler's Three Laws of Planetary Motion

from Basics of Astronomy

Chapter 22 / Lesson 12
42K

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