# Escape velocity to intersection of two gravitational fields

Find the minimal velocity needed for a meteorite of mass $m$ to get to earth from the moon. Hint: the distance between the center of earth and the center of moon is $\approx 60 R_E$, and the meteorite should reach a certain point $O$ on that distance, where the gravitational forces of the moon and the earth are canceling each other out. It is located $\approx 6 R_E$ from the center of the moon.

I thought that the best way to solve this question is to find the difference between the initial enegy of the body and the final one at $O$. Beyond that point, the gravitational pull of the earth will 'overcome' the gravitational pull of the moon, and the object will gather speed itself by reducing the GPE in the earth's gravitational field. Therefore:

$U_{G,i}=-G \frac{M_{moon}m}{R_{moon}}$

$U_{G,f}=-G \frac{M_{moon}m}{6R_E}-G \frac{M_E m}{54R_E}$

$U_{G,f}-U_{G,i} \approx m \cdot 1.51 \cdot 10^6 \text{J}$

(there's no mistake in the calculation)

$E_k=m \frac{v^2}{2}=m \cdot 1.51 \cdot 10^6$

Therefore $v \approx 1739.97 \text{m/s}$

However this answer is wrong. The correct magnitude of the speed should be somewhere around $2.26 \text{km/s}$ which is close to the escape velocity from the moon ($\approx 2.37 \text{km/s}$).

Where am I wrong? Why the difference in GPE is not satisfying the problem?

$$U_{g,i}=-G\dfrac{M_{moon}.m}{\underbrace{R_{moon}}_{(\ distance \ from \ moon\ center) }}-G\dfrac{M_{Earth}.m}{\underbrace{(60R_{earth}-R_{moon})}_{(initial \ distance\ from\ Earth)} }$$
You missed $U_i$ due to Earth!