A star has a circular orbit around the centre of a galaxy.
(a) The diagram shows an outline of the galaxy and the star’s position in the galaxy.
Draw an arrow on the diagram to show the force on the star that keeps the star in a circular orbit.
(b)
(i) The speed of light is \(3.0\times10^8\,\mathrm{m\,s^{-1}}\). One light year is the distance light travels in one year. Show that one light year is approximately \(10^{16}\,\mathrm{m}\).
[one year = \(3.2\times10^7\,\mathrm{s}\)]
(ii) The star is \(29000\) light years away from the centre of the galaxy and has an orbital speed of \(220\,\mathrm{km\,s^{-1}}\). Calculate the time period of the star’s orbit around the centre of the galaxy. Give your answer in standard form.
Syllabus Topic Codes (Edexcel International GCSE Physics 4PH1):
(a) Force on the star
The force must act towards the centre of the galaxy. This inward force provides the centripetal force needed to keep the star moving in a circular orbit.
The arrow should therefore point from the star towards the centre of the galaxy.
(b)(i) One light year
Use:
\(\mathrm{distance}=\mathrm{speed}\times\mathrm{time}\)
\(\mathrm{distance}=(3.0\times10^8)(3.2\times10^7)\)
\(\mathrm{distance}=9.6\times10^{15}\,\mathrm{m}\)
This is approximately:
\(\boxed{1\,\mathrm{light\ year}\approx10^{16}\,\mathrm{m}}\)
(b)(ii) Orbital time period
Convert the orbital radius from light years to metres:
\(r=29000\times10^{16}\,\mathrm{m}\)
For a circular orbit:
\(v=\dfrac{2\pi r}{T}\)
Rearranging:
\(T=\dfrac{2\pi r}{v}\)
Convert the speed:
\(220\,\mathrm{km\,s^{-1}}=220000\,\mathrm{m\,s^{-1}}\)
Therefore:
\(T=\dfrac{2\pi(29000\times10^{16})}{220000}\)
\(T\approx8.3\times10^{15}\,\mathrm{s}\)
\(\boxed{T=8.3\times10^{15}\,\mathrm{s}}\)
