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January 29, 2015

January 29, 2015

Posted by **Rory still need help damon and others** on Wednesday, December 26, 2007 at 12:09am.

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.

- PHYSICSS!!! -
**Rory still need help damon and others**, Wednesday, December 26, 2007 at 12:11amyeah 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?

- PHYSICSS!!! -
**Damon**, Wednesday, December 26, 2007 at 6:13amWell, 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)

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