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April 1, 2015

April 1, 2015

Posted by **rory** on Friday, December 21, 2007 at 1:04pm.

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?

- physics -
**Damon**, Friday, December 21, 2007 at 4:55pmThe 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.

- physics -
**rory**, Saturday, December 22, 2007 at 1:29ami dont think the conversion will be necessary because it wants the answer to be in scientific notation

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