Questions: How much work would be required to move a satellite of mass m from a circular orbit of radius r1=2 rE about the Earth to another circular orbit of radius r2=3 rE ? ( rE is the radius of the Earth.) Express your answer in terms of the variable m and constants rE, ME, and G.

Solution Steps

The gravitational potential energy \( U \) of a satellite of mass \( m \) at a distance \( r \) from the center of the Earth is given by:

\[ U = -\frac{G M_{\mathrm{E}} m}{r} \]

where \( G \) is the gravitational constant, and \( M_{\mathrm{E}} \) is the mass of the Earth.

The initial potential energy \( U_1 \) when the satellite is at radius \( r_1 = 2r_{\mathrm{E}} \) is:

\[ U_1 = -\frac{G M_{\mathrm{E}} m}{2r_{\mathrm{E}}} \]

The final potential energy \( U_2 \) when the satellite is at radius \( r_2 = 3r_{\mathrm{E}} \) is:

\[ U_2 = -\frac{G M_{\mathrm{E}} m}{3r_{\mathrm{E}}} \]

The work done \( W \) to move the satellite from the initial orbit to the final orbit is the change in potential energy:

\[ W = U_2 - U_1 \]

Substituting the expressions for \( U_1 \) and \( U_2 \):

\[ W = \left(-\frac{G M_{\mathrm{E}} m}{3r_{\mathrm{E}}}\right) - \left(-\frac{G M_{\mathrm{E}} m}{2r_{\mathrm{E}}}\right) \]

Simplifying the expression:

\[ W = \frac{G M_{\mathrm{E}} m}{2r_{\mathrm{E}}} - \frac{G M_{\mathrm{E}} m}{3r_{\mathrm{E}}} \]

\[ W = \frac{G M_{\mathrm{E}} m}{r_{\mathrm{E}}} \left(\frac{1}{2} - \frac{1}{3}\right) \]

\[ W = \frac{G M_{\mathrm{E}} m}{r_{\mathrm{E}}} \left(\frac{3}{6} - \frac{2}{6}\right) \]

\[ W = \frac{G M_{\mathrm{E}} m}{r_{\mathrm{E}}} \cdot \frac{1}{6} \]

\[ W = \frac{G M_{\mathrm{E}} m}{6r_{\mathrm{E}}} \]

Final Answer