91Ó°ÊÓ

According to Kepler's third law, it can be understood that the square of the time period is inversely equal to cube of the orbit's semi major axis.

The time period of orbit is given by,

$T=\frac{2{\mathrm{Ï€°ù}}^{{\displaystyle \frac{3}{2}}}}{\sqrt{GM}}$... (i)

Here,M is the mass of the planet.

From equation (i), the period of the orbit can be given as,

$T\mathrm{Î\pm }\frac{1}{\sqrt{GM}}$

Here, G is the universal gravitational constant, and M is the mass.

If we increase the mass three times, the time period will change. It has an inverse proportion between time period and mass if we increase the mass by three times.

$T^{â€\mu }=\frac{2{\mathrm{Ï€°ù}}^{{\displaystyle \frac{3}{2}}}}{\sqrt{3GM}}$

Here, r is the radius.

Divide $T^{*}$with T,we get,

$$\begin{gathered} \frac{T^{*}}{T}=\left(\frac{2{\mathrm{Ï€°ù}}^{{\displaystyle \frac{3}{2}}}}{\sqrt{3GM}}\right)\left(\frac{\begin{array}{c}\\ \sqrt{GM}\end{array}}{2{\mathrm{Ï€°ù}}^{\frac{3}{2}}}\right) \\ \frac{T^{*}}{T}=\frac{1}{\sqrt{3}} \\ {T}^{*}=\frac{T}{\sqrt{3}} \end{gathered}$$

Thus, the new time period will be $\frac{T}{\sqrt{3}}$, if we increase the mass of the planet by three-times. Option (d) is correct.