Question

A particle is leaving the Moon in a direction that is radially outward from both the Moon and Earth.

Answer 1

Answer:

$2780m/s$

Explanation:

Essentially, Kinetic energy of the particle must equal the combined potential energies of earth and the moon when the object is on the moon's surface, meaning the full equation is

$\frac{1}{2} mv^2=\frac{G(M_E)m}{r_E} +G\frac{M_mm}{r_m}$

$M_E$=Mass of Earth=$5.97*10^2^4$

$M_m$=Mass of Moon=$7.4*10^2^2kg$

$r_E$=distance from earth's center to the moon's=$3.84*10^8m$

$r_m$=radius of moon=$1.738*10^6m$

After some algebra, the equation simplifies to

$v=\sqrt{2G*(\frac{M_E}{r_E+r_m}+\frac{M_m}{r_m})}$

Plugging in the values of G, which is $6.67*10^-^1^1 \frac{m^3}{kg*s^2}$, should yield the proper answer of 2780m/s.