still cant get this one?
so damon i know you wanna help!
or anyone else im open for suggestions
haha
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 3.72E+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?
The distance from the moon to this Lagrange point we can call x
Then the distance from the earth to that point is (3.72*10^8 - x) meters (note conversion to meters)
the gravitational attraction of the earth on our ship is
G * mass earth * mass ship /(3.72*10^8 -x)^2
the gravitational attraction of the moon on our ship is
G *1.200^10^-2 mass earth * mass ship /x^2
Set those attractions equal and you have your point. Notice that the gravitational constant G cancels as does the mass of our spaceship.
Solve for x, convert it to kilometers from meters.
yeah i get the whole equation you're telling me to set up but its gets confusing because there ends up being like X^4 because its X^2 on one side and on the other its (3.72*10^8-X)^2 so either you lose all the X's or you have X^4
can you help with the algebra?
Well, what I see is
1/[3.72*10^8 - x]^2 = 1.2*10^-2/x^2
so
x^2 =1.2*10^-2 [3.72*10^8 - x]*2
x*2 = 16.6*10^14 - 8.93*10^6 x + 1.2*10^-2 x^2
or
.988 x^2 + 8.93*10^6 x - 16.6*10^14 = 0
solve that quadratic and use the positive answer (after checking my arithmetic carefully)
To find the spaceship's distance from the center of the Moon, we need to set the gravitational attractions from the Earth and Moon equal to each other.
Let's consider the distance from the Moon to the Lagrange point as "x". Then, the distance from the Earth to the Lagrange point would be (3.72 * 10^8 km - x) (note the conversion to meters).
The gravitational attraction of the Earth on the spaceship can be represented as:
(G * mass of Earth * mass of spaceship) / (3.72 * 10^8 km - x)^2
The gravitational attraction of the Moon on the spaceship can be represented as:
(G * (1.200 * 10^-2) * mass of Earth * mass of spaceship) / x^2
Setting these two attractions equal to each other, we have:
(G * mass of Earth * mass of spaceship) / (3.72 * 10^8 km - x)^2 = (G * (1.200 * 10^-2) * mass of Earth * mass of spaceship) / x^2
Now, we can simplify the equation by canceling out the gravitational constant "G" and the mass of the spaceship:
1 / (3.72 * 10^8 km - x)^2 = (1.200 * 10^-2) / x^2
To solve for x, we can take the reciprocal of both sides:
(3.72 * 10^8 km - x)^2 / x^2 = 1 / (1.200 * 10^-2)
Next, we can cross multiply:
(3.72 * 10^8 km - x)^2 * (1.200 * 10^-2) = x^2
Expanding the left side of the equation, we get:
(3.72 * 10^8 km)^2 - 2 * (3.72 * 10^8 km) * x + x^2 * (1.200 * 10^-2) = x^2
Simplifying further, we have:
(3.72 * 10^8 km)^2 - 2 * (3.72 * 10^8 km) * x + (1.200 * 10^-2) * x^2 = x^2
Rearranging the terms, we get a quadratic equation in terms of x:
x^2 - [(3.72 * 10^8 km)^2 - 2 * (3.72 * 10^8 km) * x] + [(1.200 * 10^-2) * x^2] = 0
Now, we can solve this quadratic equation for x using any preferred method such as factoring, completing the square, or using the quadratic formula. Once you find the value of x, convert it from meters to kilometers.
Finally, the obtained value of x will be the spaceship's distance from the center of the Moon.