A satellite is held in orbit by a \( 2000-\mathrm{N} \) gravitational force. Each time the satellite completes an orbit of circumference 80000 km , the work done on it by gravity is: 1) \( 1.6^{\prime} 10^{8} \mathrm{~J} \). 2) \( 1.6^{\cdot} 10^{11} \mathrm{~J} \). 3) \( 6.4^{\prime} 10^{11} \mathrm{~J} \). 4) 0 . 5) None of the above.

To determine the work done on the satellite by gravity during one complete orbit, we can use the formula for work done, which is given by: \[ W = F \cdot d \cdot \cos(\theta) \] Where: - \( W \) is the work done, - \( F \) is the force applied, - \( d \) is the distance over which the force is applied, - \( \theta \) is the angle between the force and the direction of motion. In this case: - The gravitational force \( F = 2000 \, \text{N} \), - The distance \( d \) is the circumference of the orbit, which is \( 80000 \, \text{km} = 80000 \times 1000 \, \text{m} = 8 \times 10^7 \, \text{m} \), - The angle \( \theta = 0^\circ \) because the gravitational force acts towards the center of the orbit, and the satellite moves in a circular path. Since \( \cos(0^\circ) = 1 \), the formula simplifies to: \[ W = F \cdot d \] Now, substituting the values: \[ W = 2000 \, \text{N} \cdot 8 \times 10^7 \, \text{m} \] Calculating this gives: \[ W = 2000 \cdot 8 \times 10^7 = 1.6 \times 10^{11} \, \text{J} \] Thus, the work done on the satellite by gravity during one complete orbit is: \[ W = 1.6 \times 10^{11} \, \text{J} \] Therefore, the correct answer is: **2) \( 1.6 \cdot 10^{11} \, \text{J} \)**.