Geosynchronous orbits exist at approximately $3.60 \times 10^{4} \mathrm{~km}$ above the Earth's surface. The radius of the Earth and the mass of the Earth are $R_{\mathrm{E}}=6.37 \times 10^{3} \mathrm{~km}$ and $M_{\mathrm{E}}=5.97 \times 10^{24} \mathrm{~kg}$, respectively. The gravitational constant is
\[
G=6.67 \times 10^{-11} \mathrm{~m}^{3} /(\mathrm{kg} \cdot \mathrm{s})^{2}
\]
Consider a $395 \mathrm{~kg}$ satellite in a circular orbit at a distance of $3.05 \times 10^{4} \mathrm{~km}$ above the Earth's surface.
What is the minimum amount of work $W$ the satellite's thrusters must do to raise the satellite to a geosynchronous orbit?
\[
W=
\]
Assume the change in mass of the satellite is negligible.
\(\boxed{\text{The minimum amount of work the satellite's thrusters must do to raise the satellite to a geosynchronous orbit is approximately } 5.54 \times 10^{8} \, \text{J}}\).
Step 1 :Given that the mass of the satellite is \(m = 395 \, \text{kg}\), the initial distance from the Earth's surface is \(d_{\text{initial}} = 3.05 \times 10^{7} \, \text{m}\), and the final distance (geosynchronous orbit) is \(d_{\text{final}} = 3.60 \times 10^{7} \, \text{m}\).
Step 2 :The radius of the Earth is \(R_{E} = 6.37 \times 10^{6} \, \text{m}\), the mass of the Earth is \(M_{E} = 5.97 \times 10^{24} \, \text{kg}\), and the gravitational constant is \(G = 6.67 \times 10^{-11} \, \text{m}^{3} \, \text{kg}^{-1} \, \text{s}^{-2}\).
Step 3 :The initial and final distances from the center of the Earth are \(r_{\text{initial}} = R_{E} + d_{\text{initial}} = 3.687 \times 10^{7} \, \text{m}\) and \(r_{\text{final}} = R_{E} + d_{\text{final}} = 4.237 \times 10^{7} \, \text{m}\), respectively.
Step 4 :The gravitational potential energy of the satellite at a distance \(r\) from the center of the Earth is given by \(U = - \frac{G M_{E} m}{r}\).
Step 5 :The initial and final gravitational potential energies of the satellite are \(U_{\text{initial}} = - \frac{G M_{E} m}{r_{\text{initial}}} = -4.27 \times 10^{9} \, \text{J}\) and \(U_{\text{final}} = - \frac{G M_{E} m}{r_{\text{final}}} = -3.71 \times 10^{9} \, \text{J}\), respectively.
Step 6 :The work done to move the satellite from one orbit to another is equal to the change in gravitational potential energy of the satellite, which is \(W = U_{\text{final}} - U_{\text{initial}} = 5.54 \times 10^{8} \, \text{J}\).
Step 7 :\(\boxed{\text{The minimum amount of work the satellite's thrusters must do to raise the satellite to a geosynchronous orbit is approximately } 5.54 \times 10^{8} \, \text{J}}\).