Question

On the straight line path from earth to moon. the gravitational energy of the projectile of mass m increases as the distance from the earth increases, reaches a maximum at point X somewhere between the two both., and then decreases again as the projectile approaches the surface of the moon Find the distance of point X from the earth's center. With what minimum kinetic energy must we launch a I-kg projectile from the moon's surface if we want it to reach the earth

Answer 1

The Newton's law of universal gravitation  takes the form:

(1) $F=G\frac{Mm}{R^2}$ , where $G=6.674\cdot 10^{−11} m^3⋅kg^{−1}⋅s^{−2}$ - is the gravitational constant.

The maximum of the gravitational energy is reached when the force from Moon and Earth compensate each other. Thus we have an equation

(2) $F_{\oplus}=F_{)} -> \frac{M_{\oplus}}{(R-d)^2}=\frac{M_{)}}{d^2}$ where $M_{\oplus}=5.97\cdot 10^{24} kg$ is mass of Earth, $M_{)}=7.35\cdot 10^{22} kg$ is mass of Moon, and $R=3.84\cdot 10^8 m$ . From (2) we get

(3) $d=\frac{R}{1+\sqrt{M_{\oplus}/M_{)}}}=\frac{3.84\cdot 10^{8}m}{1+\sqrt{0.81\cdot 10^2}}=\frac{3.84\cdot 10^{8}m}{1+9}=3.84\cdot 10^7m$ - is the distance from the Moon center. From the Earth center the distance of point X is $R-d=3.46\cdot 10^8m$ .

The minimum kinetic energy we should launch a projectile from the moon's surface depends on many factors, for example, from which point on the surface of the moon we launch the projectile. We will assume that the projectile is launched from a point on the moon's surface that connects the centers of the two masses. It is obvious that in order for the projectile to hit the earth in the simplest case, we need to extinguish the speed of the moon relative to the earth, and reach the point X at least with zero speed. The speed of rotation of the moon around its axis is very small, about one revolution per month, so we will not take it into account. The average speed of the moon in its orbit is $V_{\tau}=1.0 km/s$ . (It is the component of projectile velocity along the moon's surface) We just need to calculate the difference in gravitational potentials between the Lunar surface and the point X. As the potentials added we get. On the Lunar surface

$U_L=-G(\frac{M_{\oplus}}{R}+\frac{M_{)}}{R_{)}})=-6.67\cdot 10^{-11}(\frac{5.97\cdot 10^{24}}{3.84\cdot 10^8 m}+\frac{7.35\cdot 10^{22}}{1.74\cdot 10^6 m})= -3.85\cdot 10^6J/kg$ , where $R_{)}=1.74\cdot 10^6m$ - radius of the moon. It is much smaller than the distance from the moon to the earth, so we do not take it into account in the first term. At point X we have

$U_X=-G(\frac{M_{\oplus}}{R-d}+\frac{M_{)}}{d})=-6.67\cdot 10^{-11}(\frac{5.97\cdot 10^{24}}{3.46\cdot 10^8 m}+\frac{7.35\cdot 10^{22}}{3.84\cdot 10^7 m})= -1.28\cdot 10^6J/kg$

The potential difference is

(4) $\Delta U_p=U_X-U_L=(3.85-1.28)\cdot 10^6 J/kg=2.57\cdot 10^6 J/kg$

Thus the kinetic energy will be (The two velocity components are orthogonal to each other)

(5) $E_k=m\frac{V^2}{2}+\Delta U_p\cdot m=m\cdot (\frac{(10^3m/s)^2}{2}+2.57\cdot 10^6)J/kg=m \cdot 3.07\cdot10^6J$

Answer:  The distance of point X from the earth's center is $3.46\cdot 10^8m$. The minimum kinetic energy must we launch a $l$ -kg projectile from the moon's surface if we want it to reach the earth is $l \cdot 3.07\cdot10^6J$