To determine the time period of a spy satellite that orbits the Earth at a height of 6400 km, we can use Kepler's Third Law of Planetary Motion. This law states that the square of the orbital period of a satellite is directly proportional to the cube of the semi-major axis of its orbit.
The formula is given by:
\(T^2 \propto R^3\)
Where:
First, we calculate the total distance for both the geostationary and spy satellites.
According to Kepler's Third Law:
\(\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{R_2}{R_1}\right)^3\)
Substituting the known values:
\(\left(\frac{T_2}{24}\right)^2 = \left(\frac{12800}{42400}\right)^3\)
Solving this, we find:
\(T_2^2 = 24^2 \times \left(\frac{12800}{42400}\right)^3\)
\(T_2^2 = 576 \times \left(\frac{1}{3.3125}\right)^3\)
\(T_2^2 = 576 \times (0.301)^3\)
\(T_2^2 = 576 \times 0.0272529\)
\(T_2^2 = 15.6984944\)
\(T_2 = \sqrt{15.6984944} \approx 4 \, \text{hours}\)
Thus, the time period of the spy satellite is approximately 4 hours. Therefore, the correct answer is 4 h.
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.)