Question

If it has enough kinetic energy, a molecule at the surface of the Earth can "escape the Earth's gravitation", in the sense that it can continue to move away from the Earth forever. Using the principle of conservation of energy, show that the minimum translation kinetic energy needed for "escape" is mgRE, where m is the mass of the molecule, g is the free-fall acceleration at the surface of the Earth, and RE is the radius of the Earth. (Do this on paper. Your instructor may ask you to turn in this work.)

Answer 1

Answer:

$K_E=mgr_E$

Explanation:

By conservation of energy, the sum of the kinetic and gravitational potential energies at the surface of the Earth must be equal than their sum at infinity, so we have:

$K_E+U_E=K_\infty+U_\infty$

$\frac{mv_E^2}{2}-\frac{GM_Em}{r_E}=\frac{mv_\infty^2}{2}-\frac{GM_Em}{r_\infty}$

Where $G=6.67\times10^{-11}Nm^2/kg^2$ is the gravitational constant,$M_E=5.97\times10^{24}kg$ and $r_E=6371000m$ are the mass and radius of the Earth, m is the mass of the particle, $v_E$ its velocity at the surface of the Earth (which would be its escape velocity) and $v_\infty$ and $r_\infty$ are the velocities and distance at infinity, which would be null and infinity respectively, so the right hand side of our equation is 0J, which leaves us with:

$\frac{GM_Em}{r_E}=\frac{mv_E^2}{2}=K_E$

Also, since the force the molecule experiments is the force of gravity (disregarding drag), we can write its weight in terms of Newton's Law of Gravitation:

$F=mg=\frac{GM_Em}{r_E^2}$

Which means that:

$\frac{GM_Em}{r_E}=mgr_E$

So finally putting all together we can write:

$K_E=\frac{GM_Em}{r_E}=mgr_E$