A spaceship makes a circular orbit with period Taround a star. If it were to orbit, at the same distance, a star with three times the mass of the original star, would the new period (in terms of T) be $\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}\mathbf{3}\mathbf{T}\mathbf{,}\mathbf{}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{T}\sqrt{\mathbf{3}}\mathbf{,}\mathbf{}\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{,}\mathbf{}\mathbf{\left(}\mathbf{d}\mathbf{\right)}\frac{\mathbf{}\mathbf{T}}{\sqrt{\mathbf{3}}}\mathbf{,}\mathbf{}\mathbf{or}\mathbf{}\mathbf{\left(}\mathbf{e}\mathbf{\right)}\mathbf{}\frac{\mathbf{T}}{\mathbf{3}}$?

• The mass of star 2 is given by,${\mathrm{M}}_2=3\mathrm{M}$.

The rate at which a body moves via a circular route, there is a radial acceleration in the middle of the circle and the centripetal force is pointed towards the circle.

The centripetal acceleration is given by,

${\mathrm{a}}_{\mathrm{r}}=\frac{{\mathrm{v}}^2}{\mathrm{R}}\dots \left(\mathrm{i}\right)$

Here, R is distance between the ship and star, v is the velocity of the circular orbit.

Applying Newton’s second law,

$$\begin{gathered} \underset{}{\sum {\mathrm{F}}_{\mathrm{r}}={\mathrm{F}}_{\mathrm{staronspaceship}}} \\ ={\mathrm{m}}_{\mathrm{ship}}{\mathrm{a}}_{\mathrm{r}} \\ {\mathrm{F}}_{\mathrm{staronspaceship}}={\mathrm{m}}_{\mathrm{ship}}{\mathrm{a}}_{\mathrm{r}} \end{gathered}$$

Here,$F_{staronspaceship}$is the force of the star on spaceship,$a_r$ is the centripetal acceleration.

Force exerted on spaceship is given by,

${\mathrm{F}}_{\mathrm{staronspaceship}}=\frac{{\mathrm{GMm}}_{\mathrm{ship}}}{{\mathrm{R}}^2}$ … (ii)

Equate equation (ii) with above equation.

$$\begin{gathered} \frac{{\mathrm{GMm}}_{\mathrm{ship}}}{{\mathrm{R}}^2}{\mathrm{m}}_{\mathrm{ship}}{\mathrm{a}}_{\mathrm{r}} \\ \frac{\mathrm{GM}}{{\mathrm{R}}^2}={\mathrm{a}}_{\mathrm{r}} \end{gathered}$$

Equate equation (i) with above equation.

$$\begin{gathered} \frac{\mathrm{GM}}{{\mathrm{R}}^2}=\frac{{\mathrm{v}}^2}{\mathrm{R}} \\ \frac{\mathrm{GM}}{\mathrm{R}}={\mathrm{v}}^2 \end{gathered}$$

The velocity of orbital motion is given by,

$\mathrm{v}=\frac{2\mathrm{\pi R}}{\mathrm{T}}\dots \left(\mathrm{iii}\right)$

Equating equation (iii) with above equation, we get,

$$\begin{gathered} \frac{\mathrm{GM}}{\mathrm{R}}={\left(\frac{2\mathrm{\pi R}}{\mathrm{T}}\right)}^2 \\ \frac{\mathrm{GM}}{\mathrm{R}}=\frac{4{\mathrm{\pi }}^2{\mathrm{R}}^2}{{\mathrm{T}}^2} \\ {\mathrm{T}}^2=\frac{4{\mathrm{\pi }}^2{\mathrm{R}}^3}{\mathrm{GM}} \end{gathered}$$

For star (2), the time period can be given as,

${\mathrm{T}}_2^2=\frac{4{\mathrm{\pi }}^2{\mathrm{R}}^3}{{\mathrm{GM}}_2}$

Substituting the values in the above equation, we get,

$$\begin{gathered} {\mathrm{T}}_2^2=\frac{4{\mathrm{\pi }}^2{\mathrm{R}}^3}{\mathrm{G}\left(3\mathrm{M}\right)} \\ {\mathrm{T}}_2^2=\frac{1}{3}\times \frac{4{\mathrm{\pi }}^2{\mathrm{R}}^3}{\mathrm{GM}} \\ {\mathrm{T}}_2^2=\frac{1}{3}\times {\mathrm{T}}^2 \\ {\mathrm{T}}_2=\sqrt{\frac{{\mathrm{T}}^2}{3}} \\ {\mathrm{T}}_2=\frac{\mathrm{T}}{\sqrt{3}} \end{gathered}$$

Thus, the new time period is $\frac{\mathrm{T}}{\sqrt{3}}$.And option (d) is correct.