# The gravitational force between the aircraft and Moon

1. Nov 26, 2004

### Omid

Here is a problem, please let me know whether my answer is right or wrong.

Locate the position of a spaceship on the Earth-Moon center line such that the tug of each celestial body exerts on it would cancel and the craft would literally be weightless.

I found two equations:
A.
(Distance between the aircraft and the Earth) + (Distance between the aircraft and Moon) = (Distance between the Earth and Moon)

B.
(The gravitational force between the aircraft and Moon) = ( The gravitational force between the aircraft and the Earth)

After doing the Algebra:
The distance between the aircraft and the Earth = 350 * 10^6 meters
Distance between the aircraft and Moon = 34 * 10^6 meters

As I expected, the air craft is nearer to Moon. So the less distance will cancel by the more mass of the Earth. But there is a question. If my answer is right, I'll ask it.
Thanks

2. Nov 26, 2004

### arildno

It's rather difficult to answer this, because it requires that the respondent remembers the actual VALUES involved. It would have been much better if you gave your answer in algebraic form, say:
$$\frac{r_{m}}{r_{e}}=\sqrt{\frac{m_{m}}{m_{e}}}$$
Subscripts are for "moon" and "earth" respectively, and the equation then says:
The ratio between the spaceship's distances to the moon and the Earth, equals the square-root of the respective mass ratio.

You're therefor saying that the moon has about a hundredth of the Earth's mass; I would have thought it to be much less.

Note:
I was wrong; I looked it up, and found:
$$m_{m}\approx0.0123m_{e}$$

Last edited: Nov 26, 2004
3. Nov 26, 2004

### jcsd

4. Nov 26, 2004

### jcsd

I looked up the values on wikipedia which is quite a useful source for data about the planets and their satellites.

5. Nov 26, 2004

### arildno

I had to check..I was wrong (as I've edited my post to)..