Step 1: Kepler's Third Law: For objects orbiting the same central body, the square of the orbital period (\(T\)) is proportional to the cube of the semi-major axis (radius \(r\) for circular orbits). \[ \frac{T^2}{r^3} = \text{constant} \]
Step 2: Set up a ratio for two satellites. Let \(T_1, r_1\) and \(T_2, r_2\) represent the period and radius for the first and second satellite, respectively. \[ \frac{T_1^2}{r_1^3} = \frac{T_2^2}{r_2^3} \]
Step 3: Solve for the unknown period, \(T_2\). \[ T_2^2 = T_1^2 \left( \frac{r_2}{r_1} \right)^3 \]
\[ T_2 = T_1 \left( \frac{r_2}{r_1} \right)^{3/2} \]
Step 4: Substitute values and calculate \(T_2\). - \( T_1 = 3 \) hours - \( r_1 = 12000 \) km - \( r_2 = 48000 \) km The radii ratio: \( \frac{r_2}{r_1} = \frac{48000}{12000} = 4 \). \[ T_2 = 3 \times (4)^{3/2} = 3 \times (\sqrt{4})^3 = 3 \times (2)^3 = 3 \times 8 = 24 \text{ hours} \]
The height from Earth's surface at which acceleration due to gravity becomes \(\frac{g}{4}\) is \(\_\_\)? (Where \(g\) is the acceleration due to gravity on the surface of the Earth and \(R\) is the radius of the Earth.)