The gravitational force between the aircraft and Moon

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Homework Help Overview

The discussion revolves around determining the position of a spaceship along the Earth-Moon center line where the gravitational forces from both celestial bodies would cancel each other, resulting in a weightless condition for the craft. The subject area includes gravitational forces and celestial mechanics.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to derive the position using gravitational force equations and algebraic manipulation. Some participants suggest expressing the solution in algebraic form to clarify the relationship between distances and masses of the Earth and Moon. Others question the mass ratio assumptions and the implications for the calculated distances.

Discussion Status

Participants are actively engaging with the problem, with some expressing uncertainty about the mass values involved and their impact on the solution. There is a mix of validation and questioning regarding the original poster's calculations, indicating a productive exploration of the topic without a clear consensus on the correctness of the answer.

Contextual Notes

There is mention of the need for specific mass values and ratios, which some participants find challenging to recall. The discussion also reflects on the importance of using reliable sources for such data.

Omid
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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
 
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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:
[tex]\frac{r_{m}}{r_{e}}=\sqrt{\frac{m_{m}}{m_{e}}}[/tex]
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:
[tex]m_{m}\approx0.0123m_{e}[/tex]
that is, your answer is probably right..
 
Last edited:
Your answer appears correct.
 
arildno said:
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:
[tex]\frac{r_{m}}{r_{e}}=\sqrt{\frac{m_{m}}{m_{e}}}[/tex]
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.

I looked up the values on wikipedia which is quite a useful source for data about the planets and their satellites.
 
jcsd said:
I looked up the values on wikipedia which is quite a useful source for data about the planets and their satellites.
I had to check..I was wrong (as I've edited my post to)..:redface:
 

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