A planet has a mass of \( 8 \times 10^{\wedge} 24 \mathrm{~kg} \) and a radius of 6800 km . A satellite is launched into an orbit at a height of \( 36,000 \mathrm{~km} \) above the planet's surface. Calculate the time period of the satellite.

To find the time period of the satellite in orbit around the planet, we can use Kepler's third law of planetary motion, which states that the square of the orbital period (T) is proportional to the cube of the semi-major axis (r) of its orbit. The equation can be expressed as: \[ T = 2\pi \sqrt{\frac{r^3}{GM}} \] Where: - \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \, \mathrm{N \cdot m^2/kg^2} \) - \( M \) is the mass of the planet - \( r \) is the distance from the center of the planet to the satellite Given the parameters: - Mass of the planet, \( M = 8 \times 10^{24} \, \mathrm{kg} \) - Radius of the planet, \( R = 6800 \, \mathrm{km} = 6.8 \times 10^{6} \, \mathrm{m} \) - Height of the satellite above the surface, \( h = 36,000 \, \mathrm{km} = 3.6 \times 10^{7} \, \mathrm{m} \) The total distance \( r \) from the center of the planet to the satellite is: \[ r = R + h = 6.8 \times 10^{6} \, \mathrm{m} + 3.6 \times 10^{7} \, \mathrm{m} = 4.24 \times 10^{7} \, \mathrm{m} \] Now we can substitute these values into the formula to find \( T \): 1. Calculate \( r^3 \): \[ r^3 = (4.24 \times 10^{7})^3 \approx 7.628 \times 10^{22} \, \mathrm{m}^3 \] 2. Substitute into the equation: \[ T = 2\pi \sqrt{\frac{7.628 \times 10^{22}}{(6.674 \times 10^{-11})(8 \times 10^{24})}} \] 3. Calculate the denominator: \[ GM \approx (6.674 \times 10^{-11})(8 \times 10^{24}) \approx 5.3392 \times 10^{14} \, \mathrm{m}^3/\mathrm{s}^2 \] 4. Now calculate the expression: \[ T = 2\pi \sqrt{\frac{7.628 \times 10^{22}}{5.3392 \times 10^{14}}} \] \[ T = 2\pi \sqrt{142.8 \times 10^{8}} \approx 2\pi \sqrt{1.428 \times 10^{10}} \approx 2\pi \times 119.5 \times 10^{5} \] \[ T \approx 2 \times 3.14159 \times 119.5 \times 10^5 \approx 7510000 \, \mathrm{s} \approx 7510 \, \mathrm{min} \approx 125.2 \text{ hours} \] Thus, the time period of the satellite is approximately 125.2 hours.