If a satellite orbiting the Earth is 9 times closer to the Earth than the Moon, what is the time period of rotation of the satellite? Given rotational time period of Moon $=27$ days and gravitational attraction between the satellite and the moon is neglected.

<p>To find the time period of the satellite, we can use Kepler’s third law, which states that the square of the orbital period is proportional to the cube of the orbital radius. Mathematically, this is expressed as:</p> <p>$\frac{T^2}{R^3} = \text{constant}$</p> <p>Here’s how we solve the problem step by step:</p> <p>Let:</p> <p><p>The Moon's distance from Earth be $R_{\text{moon}}$.</p></p> <p><p>The Moon's orbital period be $T_{\text{moon}} = 27$ days.</p></p> <p><p>The satellite's orbital radius be $R_{\text{sat}} = \frac{R_{\text{moon}}}{9}$ (since it is 9 times closer).</p></p> <p><p>According to Kepler's law for both orbits, we have:</p> <p>$$\frac{T_{\text{sat}}^2}{R_{\text{sat}}^3} = \frac{T_{\text{moon}}^2}{R_{\text{moon}}^3}$$</p></p> <p><p>Substitute $R_{\text{sat}}$:</p> <p>$$\frac{T_{\text{sat}}^2}{\left(\frac{R_{\text{moon}}}{9}\right)^3} = \frac{27^2}{R_{\text{moon}}^3}$$</p></p> <p><p>Simplify the left side by noting that:</p> <p>$$\left(\frac{R_{\text{moon}}}{9}\right)^3 = \frac{R_{\text{moon}}^3}{9^3} = \frac{R_{\text{moon}}^3}{729}$$</p> <p>So the equation becomes:</p> <p>$$\frac{T_{\text{sat}}^2}{\frac{R_{\text{moon}}^3}{729}} = \frac{27^2}{R_{\text{moon}}^3}$$</p></p> <p><p>Multiply both sides by $\frac{R_{\text{moon}}^3}{729}$:</p> <p>$T_{\text{sat}}^2 = 27^2 \times \frac{1}{729}$</p></p> <p><p>Recognize that $27^2 = 729$:</p> <p>$T_{\text{sat}}^2 = \frac{729}{729} = 1$</p></p> <p><p>Take the square root:</p> <p>$T_{\text{sat}} = 1 \text{ day}$</p></p> <p>Thus, the satellite’s orbital period is 1 day, which corresponds to <strong>Option D: 1 day</strong>.</p>