Consider a spaceship located on the Earth-Moon center line (i.e. a line that intersects the centers of both bodies) such that, at that point, the tugs on the spaceship from each celestial body exactly cancel, leaving the craft literally weightless. Take the distance between the centers of the Earth and Moon to be 3.90E+5 km and the Moon-to-Earth mass ratio to be 1.200E-2. What is the spaceship's distance from the center of the Moon?
Bobpursely told me that:
Mm/Me=(d2/d)^2
where mm is mass moon, me mass earth, d2 is distance from craft to moon, and d is the distance from craft to earth.
My only question is, how can I get d in order to solve for d2?
Repeat:
The Law of Universal Gravitation states that each particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Expressed mathematically,
F = GM(m)/r^2
where F is the force with which either of the particles attracts the other, M and m are the masses of two particles separated by a distance r, and G is the Universal Gravitational Constant. The product of G and, lets say, the mass of the earth, is sometimes referred to as GM or mu (the greek letter pronounced meuw as opposed to meow), the earth's gravitational constant. Thus the force of attraction exerted by the earth on any particle within, on the surface of, or above, is F = 1.40766x10^16 ft^3/sec^2(m)/r^2 where m is the mass of the object being attracted and r is the distance from the center of the earth to the mass.
The gravitational constant for the earth, GM(E), is 1.40766x10^16ft^3/sec^2. The gravitational constant for the moon, GM(M), is 1.7313x10^14ft^3/sec^2. Using the average distance between the earth and moon of 239,000 miles, let the distance from the moon, to the point between the earth and moon, where the gravitational pull on a 32,200 lb. satellite is the same, be X, and the distance from the earth to this point be (239,000 - X). Therefore, the gravitational force is F = GMm/r^2 where r = X for the moon distance and r = (239000 - X) for the earth distance, and m is the mass of the satellite. At the point where the forces are equal, 1.40766x10^16(m)/(239000-X)^2 = 1.7313x10^14(m)/X^2. The m's cancel out and you are left with 81.30653X^2 = (239000 - X)^2 which results in 80.30653X^2 + 478000X - 5.7121x10^10 = 0.
From the quadratic equation, you get X = 23,859 miles, roughly one tenth the distance between the two bodies from the moon.
So the spacecraft's distance from the earth is ~215,140 miles. Subtract this from the distance between the earth and moon and you will have your answer.
Checking the gravitational pull on the 32,200 lb. satellite, whose mass m = 1000 lb.sec.^2/ft.^4. The pull of the earth is F = 1.40766x10^16(1000)/(215,140x5280)^2 = 10.91 lb. The pull of the moon is F = 1.7313x10^14(1000)/(23858x5280)^2 = 10.91 lb.
This point is sometimes referred to as L2. There is an L5 Society which supports building a space station at this point between the earth and moon. There are five such points in space, L1 through L5, at which a small body can remain in a stable orbit with two very massive bodies. The points are called Lagrangian Points and are the rare cases where the relative motions of three bodies can be computed exactly. In the case of a body orbiting a much larger body, such as the moon about the earth, the first stable point is L1 and lies on the moon's orbit, diametrically opposite the earth. The L2 and L3 points are both on the moon-earth line, one closer to the earth than the moon and the other farther away. The remaining L4 and L5 points are located on the moon's orbit such that each forms an equilateral triangle with the earth and moon.
The spacecraft's distance from the earth is ~215,140 miles. Subtract this from the distance between the earth and moon and you will have your answer.
I've tried this several times, and I keep coming up with 43765.73 km, which is not correct. ???
215.140 miles = 346,217 km
390,000 - 346,217 = 43,783 km
The actual mean distance between the earth and moon is 238,868 miles or 384,338 km.
Most often the mean distance is quoted as 239,000 miles or 384,551 km.
Then, 384,551 - 346,237 = 38,313 km., or less than using your distance of 3.9x10^5.
What is the answer you are seeking?
To solve for d, you can use the given information about the distance between the centers of the Earth and Moon, and the fact that the spaceship is located on the Earth-Moon center line.
Let's assume that the distance from the spaceship to the center of the Earth (d) is x km. Since the spaceship is located on the Earth-Moon center line, the distance from the spaceship to the center of the Moon (d2) would be the total distance between the Earth and Moon minus the distance between the spaceship and the Earth (3.90E+5 km - x km).
Now we can use the mass ratio equation Mm/Me = (d2/d)^2, as Bobpursely mentioned. Since we know the mass ratio (1.200E-2) and we are solving for d2, we can rearrange the equation to solve for d2.
Mm/Me = (d2/d)^2
Taking the square root of both sides:
sqrt(Mm/Me) = d2/d
Now, substitute the known values:
sqrt(1.200E-2) = (3.90E+5 - x)/x
Solving for x:
1.200E-2 = (3.90E+5 - x)/x
Multiply both sides by x:
1.200E-2 * x = 3.90E+5 - x
Move the -x term to the left side:
1.200E-2 * x + x = 3.90E+5
Combine like terms:
1.200E-2 * x + 1 * x = 3.90E+5
Convert the mass ratio to decimal form:
0.012 * x + 1 * x = 3.90E+5
Combine like terms:
1.012 * x = 3.90E+5
Divide both sides by 1.012 to solve for x:
x = (3.90E+5) / 1.012
By calculating this expression, you will find the value of x, which is the distance from the spaceship to the center of the Earth. Once you have that value, you can substitute it back into the equation (3.90E+5 km - x km) to find the distance from the spaceship to the center of the Moon (d2).