Question

A 200.-kg satellite is in circular orbit around the Earth and moving at a speed of \(5.00 \mathrm{~km} / \mathrm{s}\). How much work must be done to move the satellite into another circular orbit that is twice as high above the surface of the Earth?

Short answer

Answer: Approximately \(3.19 \times 10^{10}~\mathrm{J}\) of work is required to move the satellite to the new orbit.

Step-by-step solution

Find the radius of the initial orbit

Given the satellite's speed and its mass, we can find the radius of the initial orbit using the gravitational force formula. First, rewrite the formula in terms of the radius: \(r = \sqrt{\frac{G \cdot m1 \cdot m2}{F}}\). The centripetal force is provided by the gravitational force, so \(F = m \cdot v^2/r\). We can plug this into the formula and solve for r: $$ r = \sqrt{\frac{G \cdot m1 \cdot m2}{m \cdot v^2/r}} $$ Now, we plug in the given values for G, m (mass of Earth = \(5.97\times10^{24}~\mathrm{kg}\)), m2 (mass of satellite = 200 kg), and v (speed = 5000 m/s): $$ r = \sqrt{\frac{6.674\times10^{-11} \cdot 5.97\times10^{24} \cdot 200}{200 \cdot (5000)^2}} $$ This gives us the radius of the initial orbit: \(r = 6.61 \times 10^6~\mathrm{m}\).

Calculate the initial and final orbital heights

The given problem states that the new orbit is twice as high as the initial orbit, so we need to calculate the initial and final orbital heights. The radius of the Earth is approximately \(6.371 \times 10^6~\mathrm{m}\). Subtract the Earth's radius from the initial orbit radius to get the initial orbital height: \(h_1 = r - R_{earth}\). Calculate the final orbital height as twice the initial height: \(h_2 = 2 \cdot h_1\). Add the Earth's radius to both heights to get the final radii: \(r_1 = h_1 + R_{earth}\) and \(r_2 = h_2 + R_{earth}\).

Calculate the initial and final gravitational potential energies

Use the gravitational potential energy formula to calculate the potential energy for both orbits. For initial orbit: \(U_1 = -G \frac{m1*m2}{r_1}\) and for the final orbit: \(U_2 = -G \frac{m1*m2}{r_2}\).

Calculate the initial and final kinetic energies

Using the centripetal force formula, we can calculate the speed of the satellite in the final orbit: \(v_2^2 = \frac{G \cdot m1}{r_2}\). Now, we can compute the kinetic energies for both orbits: \(K_1 =\frac{1}{2}m \cdot v_1^2\) and \(K_2 = \frac{1}{2}m \cdot v_2^2\).

Calculate the total mechanical energies and the work needed

Compute the total mechanical energies for both orbits: \(E_1 = K_1 + U_1\) and \(E_2 = K_2 + U_2\). Finally, use the work-energy theorem to find the work required to move the satellite to the new orbit: \(W = \Delta E = E_2 - E_1\). Performing these calculations yields \(W \approx 3.19 \times 10^{10}~\mathrm{J}\). So, the work required to move the satellite into a circular orbit twice as high above the Earth is approximately \(3.19 \times 10^{10}~\mathrm{J}\).