Finding Weightless Point on Earth-Moon Center Line - 6.67E^-11

[uno]

1. Locate the position of a spaceship on the Earth-Moon center line such that, at that point, the tug of each celestial body exerted on it would cancel and the craft would literally be weightless. Please answer in meters from the Moon



2. The only thing I can think of is that G=6.67E^-11



3. I am not sure how to approach this problem

 

[Rake-MC]

Okay, you know this formula I'm sure: $$F = \frac{GMm}{r^2}$$

But unforunately the force depends on the masses of both bodies, so let's divide by m to find the acceleration of M.

$$a = \frac{GM}{r^2}$$

For the distance (r) at which both have the same gravitational acceleration, you must make two equations equal each other. Be careful with your r, because you are measuring from the moon.

 

[uno]

o.k the mass of Earth is 5.97E24 kg and the moon is 7.36E22kg. First I am looking for acceleration. I am unclear when you suggested that I divide by m to find the acceleration of M.

 

[Rake-MC]

That was just showing you how I derived the acceleration due to gravity formula. You can ignore the first part of that post now.

Focus on this formula:

$$a = \frac{GM}{r^2}$$

This applies to all bodies. Therefore you can have the acceleration due to the Earth's field:

$$a = \frac{GM_e}{r^2}$$ where r is measured from the centre of the earth.

and the acceleration of the moon is:

$$a = \frac{GM_m}{r^2}$$ where r is measured from the centre of the moon.

When these two equations equal each other, you have the point you are looking for.
However! The radii are measured from two different locations. You need to change the form of 'r' in one of the equations.