# Gravity: Point where Earth's Gravity and the Moon's cancel each other out?

1. Nov 27, 2006

### ND3G

The mass of the Moon is 7.35*10^22 Kg. At some point between Earth and the Moon, the force of Earth's gravitational attraction on an object is cancelled by the Moon's force of gravitational attraction. If the distance between Earth and the Moon (centre to centre) is 3.84*10^5 Km (3.84*10^8 m), calculate where this will occur, relative to Earth.

Given:
m Moon = 7.35*10^22 kg
m Earth = 5.98*10^24 kg
r = 3.84*10^8 m
G = 6.67x10^-11 N * m^2/kg^2

Required: r1

Analaysis: Fg = (G*m1*m2) / r^2

Fge = Fgm

Solution:

Let m2 = 1 kg (the mass of an object between the earth and the moon)

For Earth:

Fge = [(6.67*10^-11)(5.98*10^24)(1)] / r1 ^2

For the Moon:

Fgm = Fgm = [(6.67*10^-11)(7.35*10^22)(1)] / (3.84*10^8 - r1)^2

Fge = Fgm

[(5.98*10^24 kg) / r1^2] = [(7.35*10^22 kg) / (3.84*10^8 - r1)^2]

Am I on the right track here, and if so, can someone help me finish off the equation to solve for r1? Thanks in advance.

Last edited: Nov 27, 2006
2. Nov 27, 2006

### Staff: Mentor

Looks good so far. Just square out the denominators, cross-multiply and you're there.

3. Nov 27, 2006

### ND3G

Thanks. It has been a while since I did any real math though.... can you show what you mean by squaring out the denominators?

4. Nov 27, 2006

### Staff: Mentor

It's just basic algebra. Funny that you got a moderately difficult problem (for basic physics) and are rusty on algebra, but whatever. Something like this:

$$\frac{5.98*10^{24} kg}{r1^2} = \frac{7.35*10^{22} kg}{(3.84*10^8 - r1)^2}$$

$$(5.98*10^{24} kg)(3.84*10^8 - r1)^2 = (7.35*10^{22} kg)(r1^2)$$

When you square the lefthand side, you get an r1 term and an r1^2 term. The righthand sinde only has an r1^2 term. So you may need to use the Quadratic Equation to solve for r1. If you don't remember the form of the Quadratic Equation, you can probably find it explained at wikipedia.org.

Hope that helps some.

5. Mar 21, 2009

### nikkor180

Greetings:

I believe the point of interest to be none other than the center of mass which is at

mer / (me + mm) meters from the Earth along the ray extending from Earth through the moon.

Regards,

Rich B.
rmath4u2@aol.com

6. Mar 21, 2009

### D H

Staff Emeritus
We have a policy here at PF against giving out answers on this forum, nikkor180. Giving out answers that are doubly wrong is particularly bad. That point is neither the center of mass nor the answer to the question.

7. Apr 18, 2010

### Nocholas

I am working on the same problem. I had got as far as berkeman's post. However, I cannot seem to go any further without coming to an unreasonable answer. Can anyone provide any more help please?

8. Apr 19, 2010

### Staff: Mentor

Show us your work so far.

9. Apr 19, 2010

### Nocholas

I believe I have finally found the answer.

E= *10^
1. 5.98E24/r^2 = 7.35E22/ (3.84E8)^2 - r^2

2. 5.98E24/r^2 = 7.35E22/1.47E17 - r^2

3. 5.98E24*1.47E17-5.98E24r^2=7.35r^2

4. 8.79E41= (7.35E22+5.98E24)r^2

5. 8.79E41= 6.05E24r^2

6 √(8.79E41/6.05E24) = r

The answer then suggests the point of Fg=0 is a lot closer to the moon than Earth, which makes sense.

Please let me know if I did this correct.

You may edit this if you believe it contains too much information for these forums.

10. Nov 26, 2010

### Plano Physics

Um sorry I'm working on the same problem but my worksheet gives us the answer on what it's suppose to be and I'm suppose to solve towards it yet when I repeat the process above it does not get the answer which is

3.44x10^8(from earth)
3.82x10^7(from moom)

can someone explain?

Thank you

11. Nov 27, 2010

### D H

Staff Emeritus
Which procedure? berkeman practically gave the answer away in post #4. Post #5 and #9 are wrong.

12. Nov 27, 2010

### D H

Staff Emeritus
I prefer to work things out symbolically and insert the numbers as late as possible. This often makes the math shorter and often makes it easier to detect stupid mistakes. We all make stupid mistakes (at least I do). Any techniques that militates against stupid math errors is a good thing in my mind. With that in mind, let
• Me denote the mass of the Earth;
• Mm denote the mass of the Moon;
• mr denote the Moon-Earth mass ratio, mr=Mm/Me;
• R denote the distance between the Earth and the Moon;
• r denote the distance between the Moon and this null gravity point; and
• f denote the location of the null gravity point as a fraction of R, f=r/R.

Note that the distance between the null gravity point and the Earth is R-r=(1-f)R. Applying Newton's law of gravity yields the equation for the location of the point in question,

$$\frac{GM_m}{r^2} = \frac {G M_e}{(R-r)^2}$$

Obviously the gravitational constant G cancels out. Expressing r as f*R enables canceling the factor of 1/R2 on both sides. Dividing both sides by the Earth's mass yields the unitless expression

$$\frac{m_r}{f^2} = \frac 1{(1-f)^2}$$

This should be easy to solve.

One final note: Why use the Moon-Earth ratio instead of the masses of each? Three reasons:
• The ratio is unitless. Unitless parameters are often quite useful in physics.
• There is only one number to worry about rather than two.
• I don't know the mass of the Earth or the Moon off the top of my head, by I do know the Moon-Earth mass ratio. This ratio turns out to have an easy-to-memorize value: 0.0123. Better yet, this value very accurate. To six places the value is 0.0123000. You have to go to seven places to get the more random looking value of 0.01229998.

13. Nov 27, 2010

Alternatively,let the zero gravity point be distance x from the earth and distance y from the moon:

GME/x squared=GMm/y squared from which:

x/y=root(Me/Mm).........................................(x+y=r)

These methods give quadratics and therefore two answers and this is because the equation as used does not recognise whether the gravitational force is attractive or repulsive.One of the answers is between the earth and moon(much closer to the moon) and is the correct answer,the gravitational force being attractive from both earth and moon.The answer would be the same if both forces were repulsive.The other answer is on the opposite side of the moon and would be correct if the force from the earth were repulsive(like you) and the force from the moon attractive(like me).If the force from the moon were repulsive and the force from the earth attractive there would be no point where the forces cancel.

Last edited: Nov 27, 2010
14. Mar 3, 2011

### 5hassay

I would like to confirm about the required use of the quadratic equation to solve for the required variable. I would also like to add the following mathematics concept/law, for I noticed the poster stated their difficulty with such, and I had difficulty with it as well: a variable x with a coefficient y subtracted by another like variable x will result with a variable x with a coefficient of the two coefficients difference, or, yx-x=(y-1)x. This should clear stuff up on how to format the quadratic equation if it was not clear before.

Additionally, I was confused on which answer to use that are produced from the quadratic equation. I thought the larger of the two would be the one relative to the Earth, because the only way for an object between the Earth and the Moon to have this cancellation would be that the object was closer to the Moon.

Much appreciation to this thread, for I would have been quite lost with this question without it.

15. Mar 3, 2011

### Telemachus

I wanted to ask if this is the gravity center for the moon-earth system.

16. Mar 3, 2011

### D H

Staff Emeritus
If by that you mean the center of mass, no, it is not.

If you mean something else by that, you need to be explicit on what you mean by "gravity center".

17. Mar 3, 2011

### Telemachus

No, I mean the gravity center, but I think its not, because its usually the same point than the mass center (in absence of other forces). Its the point where the gravity field of a systems seems to be concentrated. But it should be closer to the earth than to the moon, as the mass center.

18. Mar 3, 2011

### D H

Staff Emeritus
What, exactly, do you mean by "gravity center", Telemachus?

19. Mar 3, 2011

### Telemachus

Well, the earth and the moon, both have a gravity field, right? if we now consider the system moon-earth going around the sun, the gravity center is the point where the system behaves as one object with the gravity field composed by the earth and the moon. The thing is that some books distinguish between mass center and gravity center, because they are actually different concepts (and its not the same point actually). But it has nothing to do with this topic, so I'm sorry for desalinating it.

20. Mar 3, 2011

### D H

Staff Emeritus
The center of mass of some set of objects (in English we say "center of mass" rather than "mass center") has nothing to do with the gravity field in which the objects. The center of gravity does. In this case (and in many cases), the center of gravity is very close to the center of mass.

However, while the Earth, Earth-Moon center of mass, and the Moon are collinear, the center of gravity usually is not collinear with the Earth and Moon.

The Earth-Moon center of mass (and their center of gravity with respect to the Sun) are very close to one another, and both are much closer to the center of the Earth than the center of the Moon. The point where gravitational attraction toward the Earth and Moon cancel one another is much closer to the center of the Moon than the center of the Earth.