Estimate the total mechanical energy of the Earth in its…

Estimate the total mechanical energy of the Earth in its elliptical orbit around the Sun. The masses of the Earth and the Sun are about $M_Z = 6 \cdot 10^{24}\ \mathrm{kg}$ and $M_S = 2 \cdot 10^{30}\ \mathrm{kg}$; the greatest Earth–Sun distance is $d = 1.52 \cdot 10^{11}\ \mathrm{m}$.

Step 1: Idea

On an elliptical orbit the kinetic and potential energies both change: the Earth is faster when closer to the Sun. Their sum, the total mechanical energy, stays constant. For a circular orbit of radius $r$ we found $E = -\gamma Mm/(2r)$ (Problem 7). For an ellipse the same formula holds with $r$ replaced by the semi-major axis $a$, half the longest diameter of the ellipse:

\[ E = -\gamma\frac{M_SM_Z}{2a}. \]

Step 2: The semi-major axis

The Earth's orbit is almost a circle (eccentricity 0.017): its greatest distance, $1.52 \cdot 10^{11}\ \mathrm{m}$, and its smallest, $1.47 \cdot 10^{11}\ \mathrm{m}$, differ by only about 3%. Since only an estimate is asked for, we may take $a \approx d$.

Step 3: Numbers

\[ E \approx -\frac{6.67 \cdot 10^{-11} \cdot 2 \cdot 10^{30} \cdot 6 \cdot 10^{24}}{2 \cdot 1.52 \cdot 10^{11}}. \]

\[E \approx -2.6 \cdot 10^{33}\ \mathrm{J}\]

The energy is negative: the Earth is bound to the Sun. To free it from the Sun completely we would have to supply $2.6 \cdot 10^{33}\ \mathrm{J}$, roughly the energy the Sun radiates in 80 days.

Check

With the true semi-major axis $a = 1.496 \cdot 10^{11}\ \mathrm{m}$: $E = -2.67 \cdot 10^{33}\ \mathrm{J}$, within 2% $\checkmark$

$ E = -E_k $ with $ E_k $ $= \tfrac12 M_Zv^2 $ $= 0.5 \cdot 6 \cdot 10^{24} \cdot (2.98 \cdot 10^4)^2 $ $= 2.66 \cdot 10^{33}\ \mathrm{J}\ \checkmark $