Escape velocities from the Moon

We wish to determine two numbers, the minimum velocity with which an object can leave the moon and reach the earth, and the minimum velocity with which an object can leave the moon and escape the earth/moon system.

II. RELEVANT CLASSICAL MECHANICS

Let x, v, and a denote distance, velocity, and acceleration respectively of an object along a line. Then, from the chain rule we have:

(1) a = dv/dt = (dv/dx)*(dx/dt) = v*(dv/dx)

and hence

(2) a*dx = v*dv

For convenience let f(x) be the indefinite integral of a*dx.

Consider an object moving position x1 to position x2 with initial velocity v1 and final velocity v2. We integrate both sides of (2) to get

(3) v2**2 – v1**2 = 2 * (f(x2)-f(x1))

Now consider the case where the acceleration of the the object is due to gravitational forces from one or more masses M_i located at positions X_i. The acceleration due to mass M_i is given by

(4) a_i(x) = G*M_i/(x-X_i)**2

in the direction from x to X_i where G is the universal gravitational constant. The total acceleration a(x) is given by

(5) a(x) = sum_over_i(a_i(x))

If f_i(x) is the integral of a_i(x) then, since the accelerations combine linearly,

(6) f(x) = sum_over_i(f_i(x))

with f_i(x) being given by

(7) f_i(x) = G*M_i/(x-X_i)

We can replace the factor G*M_i by

(8) G*M_i = g_i*R_i**2

where R_i is a reference distance which, in subsequent calculations, is the mass body radius and g_i is the surface acceleration of gravity.

There is one other item that we an equation for, the crossover point. Suppose we have two masses, M_1 and M_2, separated by a distance d. The crossover point is the point at which the combined gravitational force due to the two masses is zero. Let y_i be the distance from mass i to to the crossover point (y_1 + y_2 = d). Since the net force is zero we have

(9) G*M_1/y_1**2 = G*M_2/y_2**2

Solving (9) for y_1/y_2 we get

(10) y_1/y_2 = K = sqrt(M_1/M_2)

The distances, y_1 and y_2, are given by

(11)
y_1 = d*K/(1.+K)
y_2 = d/(1.+K)

III. PHYSICAL CONSTANTS

The following values will be used for the relevant physical constants. They are taken from on line reference sources. It was not stated whether the distance to the moon was center to center or surface to surface. I assumed that it was center to center. The effect on the results due to an erroneous assumption, if any, is minor. In any event, the distance to the moon varies during the course of its orbit.

Earth Radius	6378 km
Lunar Radius	1738 km
Earth to Moon distance	384,000 km
Earth/Moon mass ratio	81.36
g_Earth	.00980 km/sec
g_Moon	.00162 km/sec

From equation (8) and these values we get

G*M_Earth = 398653. km**3/sec**2
G*M_Moon = 4893. km**3/sec**2

The distances to crossover are given by

Earth to crossover: 345,600 km
Moon to crossover: 38,400 km

IV CALCULATION OF MINIMUM LAUNCH TO EARTH VELOCITY

The minimum launch to Earth velocity occurs when an object leaves the moon with just enough velocity to reach the crossover point with zero velocity, i.e., the terminal velocity (v2) is zero. The contributions of the Earth and the Moon to squared launch velocity are given by sum of the differences of the respective f’s at the lunar surface and the crossover point. For the moon this yields

2*4893.*(1/1738. – 1/38400.) = 5.3758 km**2/sec**2

For the earth this yields

2*398653.*(1/382262. – 1/345600.) = -.2208 km**2/sec**2

for v1**2 = 5.1550 km**2/sec**2

and v1 = 2.279 km/sec

V CALCULATION OF EARTH/MOON ESCAPE VELOCITY

The minimum escape velocity occurs when an object leaves the moon from the direction opposite Earth with just enough velocity to escape the Earth/Moon system. In this case the terminal velocity is 0 and the final distance is infinite. The contribution from the moon to the squared velocity is:

2*4893.*(1/1738.) = 5.630 km**2/sec**2

and the contribution from Earth is

2*398653*(1/385738.) = 2.067 km**2/sec**2

which yields 7.697 km**2/sec**2

and an escape velocity of 2.774 km/sec

VI SUMMARY

The moon’s escape intrinsic escape velocity is 2.37 km/sec. As can be seen from the results the advantage gained by launching toward Earth is negligible, a reduction from 2.37 to 2.28 km/sec. The effect of the Earth on the escape velocity from the Earth/Moon system is somewhat greater, an increase from 2.37 to 2.77 km/sec. Launching outward requires about 20% more velocity and about 45% more energy.

There are two mistakes that are easy to make in thinking about the problem if one doesn’t look at the math. The first is to assume that velocities add. They do add for dynamic movement. The matter at hand, however, are escape velocities. Escape velocities depend upon the potential energy of gravitational fields; energies add but they varies with the square of the velocity.

The second mistake that is easy to make is to assume the ratio of the distances from the two bodies to the crossover point is the same as the mass ratio; instead it the same as the square root of the mass ratio.

At first sight it seems surprising that the effect of the Earth on system escape velocity is greater than that on launch to Earth velocity. The reason for the greater effect is that in the launch to Earth scenario the Earth’s gravitational field is only operating over a relatively short distance whereas in the system escape scenario we are integrating all the way out to infinity. This more than compensates for the slightly greater gravitational force due to Earth in the inward launch scenario.

VI REFERENCES

Modern University Physics: Richards, Sears, Wehr, and Zermansky, 1960, Addison-Wesley

http://pauldunn.dynip.com/solarsystem/earth.html#Satellites