91Ó°ÊÓ

First, let's find the distance between the satellite and the Earth's center. The Earth has a radius of about \(6400 \, km\), so the total distance is \(6400 + 16000 = 22400 \, km\). Then, convert this distance into meters: \(22400 \times 10^3 \, m\). Now, we can use the gravitational force formula to find the weight of the satellite in orbit: \(F = G \frac{m_{satellite}m_{Earth}}{r^2}\), where \(m_{satellite} = 320 \, kg\), \(m_{Earth} = 5.972 \times 10^{24}\, kg\) and \(r = 22400 \times 10^3 \, m\). Plug the values into the formula to get: \(F = (6.674 \times 10^{-11} \, N\, m^2/kg^2) \frac{(320 \, kg)(5.972 \times 10^{24}\, kg)}{(22400 \times 10^3 \, m)^2}\). After calculating, the weight of the satellite in orbit is approximately: \(F_{orbit} = 772.38 \, N\).

Now let's find the weight of the satellite when it was on the Earth's surface. We will again use the gravitational force formula, but this time, the distance is equal to the Earth's radius: \(r = 6400 \times 10^3 \, m\). Plug the values into the formula to get: \(F = (6.674 \times 10^{-11} \, N\, m^2/kg^2) \frac{(320 \, kg)(5.972 \times 10^{24}\, kg)}{(6400 \times 10^3 \, m)^2}\). After calculating, the weight of the satellite on the Earth's surface is approximately: \(F_{surface} = 3135.24 \, N\).

According to Newton's third law of motion, "For every action, there is an equal and opposite reaction." Therefore, the force that the satellite exerts on the Earth while in orbit is equal in magnitude and opposite in direction to the gravitational force acting on the satellite. Since we have calculated the weight of the satellite in orbit (\(F_{orbit}\)) in Step 1, the force the satellite exerts on the Earth is also \(772.38 \, N\), but in the opposite direction. In summary: (a) The satellite's weight when in orbit is \(772.38 \, N\). (b) The satellite's weight when on the Earth's surface was \(3135.24 \, N\). (c) While in orbit, the satellite exerts a force of \(772.38 \, N\) on the Earth.