A planet of mass M and a single satellite of mass M/10 revolve in circular orbits about their stationary centre of mass, being held together by their gravitational attraction. the distance between their centres is D. the period of this orbital motion is ( IGNORE SPIN OFF PLANET AND SATELLITE)

Dear student,

Please find below the solution to the asked query:





The distance between their masses are separated by D

as given we have a two mass system where on mass is M and the other mass is M/10 separated by D distance apart

so the position of the centre of mass , as we know is inversely proportional to the masses and lies on the line joining them

so over here the position of the center of mass from the M kg mass is  $$\begin{gathered} \frac{{\displaystyle \frac{M}{10}}}{M+{\displaystyle \frac{M}{10}}}D= \frac{{\displaystyle \frac{M}{10}}}{{\displaystyle \frac{11M}{10}}}D = D/11 \\ while the position of the center of mass from the M/10 mass is \\ D-D/11 = 10D/11 \\ now as both are moving about their common center of mass so we have their radius of motion as found above \\ For M radius is = D/11 while for mass M/10 radius is 10D/11 \\ now both the masses are moving due to force F = \frac{G \hspace{0.17em}M\times M/10}{D^2} \\ so this force provides each of them their required centripetal force \\ so for mass M \\ we have \frac{G \hspace{0.17em}M\times M/10}{D^2} = \frac{M{v}^2}{R} \\ so V = \sqrt{GMR/10D^2} \\ so the time period of ocillation T = 2\mathrm{\pi R} /\mathrm{V} \\ \mathrm{so} \mathrm{T} = 2\mathrm{\pi R}/\sqrt{\mathrm{GMR}/10{\mathrm{D}}^2} \\ \mathrm{or} \mathrm{T} = \frac{2\mathrm{\pi D}\sqrt{10\mathrm{R}}}{\sqrt{\mathrm{GM}}}\mathrm{where} \mathrm{R} = \mathrm{D}/11 \\ \mathrm{soT} = \frac{2\mathrm{\pi D}\sqrt{10\mathrm{D}/11}}{\sqrt{\mathrm{GM}}} = 2\mathrm{\pi }\sqrt{\frac{10{\mathrm{D}}^3}{11\mathrm{GM}}} \end{gathered}$$

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.