The time period of a satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 9R is :

<p>Kepler&#39;s Third Law states that the square of the period of a satellite&#39;s orbit is proportional to the cube of the semi-major axis of its orbit. This relationship can be written as:</p> <p>$T^2 \propto r^3$</p> <p>where:</p> <ul> <li>$T$ is the orbital period</li> <li>$r$ is the radius of the circular orbit (which serves as the semi-major axis in this case)</li> </ul> <p>Considering two satellites, one with period $T$ and radius $R$, and another with unknown period $T&#39;$ and radius $9R$, we can form an equation:</p> <p>$\frac{{T&#39;}^2}{T^2} = \frac{(9R)^3}{R^3}$</p> <p>This simplifies to:</p> <p>$\frac{{T&#39;}^2}{T^2} = 729$</p> <p>Taking the square root of both sides, we get:</p> <p>$T&#39; = T \times \sqrt{729} = T \times 27$</p> <p>So, the period of another satellite in a circular orbit of radius $9R$ is $27T$.</p>