Q.Earth revolves around the Sun at 30 km s$^{-1}$. Calculate the kinetic energy of the Earth. In the previous example you calculated the potential energy of the Earth. What is the total energy of the Earth in that case? Is the total energy positive? Give reasons.

Step 1. Earth's orbital kinetic energy is $K.E.=\dfrac12M_ev^2$, with $M_e=5.9\times10^{24}\ \text{kg}$ and $v=30\ \text{km s}^{-1}=3\times10^4\ \text{m s}^{-1}$.

Step 2. $K.E.=\dfrac12(5.9\times10^{24})(3\times10^4)^2=\dfrac12(5.9\times10^{24})(9\times10^8)=\dfrac12(5.31\times10^{33})\approx2.655\times10^{33}\ \text{J}=26.55\times10^{32}\ \text{J}$.

Step 3. From the previous exercise, $U\approx-49.84\times10^{32}\ \text{J}$.

Step 4. Total energy: $E=K.E.+U\approx26.5\times10^{32}+(-49.84\times10^{32})\approx-23.3\times10^{32}\ \text{J}$. …

Compute Earth's kinetic energy from its orbital speed, then add the pre …

• Forgetting to convert 30 km/s to m/s before squaring.
• Concluding a negative total energy means something is 'wrong', rather than …