What minimum energy is needed to transfer an artificial…

What minimum energy is needed to transfer an artificial Earth satellite of mass $m = 2\ \mathrm{t}$ from a circular orbit at height $h_1 = R_Z$ to one at height $h_2 = 2R_Z$? The Earth's mean radius is $R_Z = 6370\ \mathrm{km}$ and its mean density $\rho = 5500\ \mathrm{kg/m^3}$.

Explanation of Method

Energy to raise a satellite to a higher orbit.

Step 1: Idea

The minimum energy needed is the difference between the satellite's total mechanical energy (kinetic plus potential) in the two orbits. So we first find the total energy of a satellite in a circular orbit of radius $r$.

Step 2: Energy in a circular orbit

In a circular orbit gravity provides exactly the centripetal force: $m\frac{v^2}{r} = \gamma\frac{Mm}{r^2}$.

\[ mv^2 = \frac{\gamma Mm}{r}. \]

So the kinetic energy is $E_k = \gamma Mm/(2r)$. The potential energy (zero at infinity) is $E_p = -\gamma Mm/r$. Adding them: $E = \frac{\gamma Mm}{2r} - \frac{\gamma Mm}{r} = -\frac{\gamma Mm}{2r}$.

The heights are measured from the surface, so the orbit radii are $r_1 = R_Z + R_Z = 2R_Z$ and $r_2 = R_Z + 2R_Z = 3R_Z$.

\[ \Delta E = E_2 - E_1 = -\frac{\gamma Mm}{6R_Z} + \frac{\gamma Mm}{4R_Z}, \]

\[ \Delta E = \frac{\gamma Mm}{12R_Z}. \]

The result is positive: the higher orbit has more energy, even though the satellite moves more slowly there. The kinetic energy decreases, but the potential energy increases twice as much.

Step 3: Earth's mass

The mass is not given directly, but the density and radius are. The Earth is treated as a uniform sphere: $M = \tfrac43\pi R_Z^3\rho = \tfrac43\pi (6.37 \cdot 10^6)^3 \cdot 5500$.

\[ M \approx 5.95 \cdot 10^{24}\ \mathrm{kg}. \]

Step 4: Numbers

\[ \Delta E = \frac{6.67 \cdot 10^{-11} \cdot 5.95 \cdot 10^{24} \cdot 2000}{12 \cdot 6.37 \cdot 10^6}. \]

\[\Delta E \approx 1.04 \cdot 10^{10}\ \mathrm{J}\]

This is the theoretical minimum; a real transfer (two engine burns along an elliptical transfer orbit) needs at least this much, plus the fuel needed to accelerate the fuel itself.

Check

Using $ \gamma M/R_Z = gR_Z $ with $g = 9.79\ \mathrm{m/s^2}$: $ \Delta E $ $= mgR_Z/12 $ $= 2000 \cdot 9.79 \cdot 6.37 \cdot 10^6/12 $ $= 1.04 \cdot 10^{10}\ \mathrm{J}\ \checkmark $

$R$ $2R$ $3R$ $E = -\frac{\gamma M m}{2r}$ $\Delta E = \frac{\gamma M m}{12R} \approx 1.04 \cdot 10^{10}\ \mathrm{J}$

Circular orbits of radius $2R$ (height $R$) and $3R$ (height $2R$), to scale. The total energy $-\gamma Mm/2r$ is less negative on the larger orbit; the difference is the minimum energy needed.

Result

\[\boxed{\,\Delta E \approx 1.04 \cdot 10^{10}\ \mathrm{J}\,}\]