# In which point between Earth and Moon is gravity field zero

1. Nov 29, 2014

1. The problem statement, all variables and given/known data
I already searched for solution and i found similar one but i'm still confused.
-In which point between Earth and Moon gravity field equals 0 if distance between moon and earth is 60.3 * radius Earth and mass of the earth is 81 * mass of the moon.

2. Relevant equations
ɣ*M/(d-x)2 =ɣ*m/x2

3. The attempt at a solution

O----------------------o
M......(d-x).....|........x.......m

consider this a sketch, (d-x) is distance between earth and that point, x is distance between moon and that point. M is mass of the earth, m is mass of the moon.

M/(d-x)2=m/x2
81/(d-x)2=1/x2
81x2=(d-x)2 /root
9x=d-x
10x=d
x=6.03Re

not sure if i made mistake or my book author where he says it has 2 solutions and isn't near as simple as my answer.

2. Nov 29, 2014

### Staff: Mentor

Your answer is correct if you take Earth and Moon as static objects in space. If you account for their rotation around each other, then it gets more complicated and you get 5 solutions of co-rotation (they are called Lagrange points).

3. Nov 29, 2014

### voko

The second equation is not correct. It should be $\pm 9x = d - x$, and that gives you two solutions. You need to be able to choose the correct solution, and explain why it is the correct one.

4. Nov 29, 2014

Thank you both, this is solution from my book and it doesn't match with mine at all.
http://pokit.org/get/img/2a25e10dddd37ab72b3d28fde419fecb.jpg [Broken]

Last edited by a moderator: May 7, 2017
5. Nov 29, 2014

### haruspex

Don't confuse 'net field zero' with Lagrange points. Motion of the bodies does not affect the instantaneous field.

6. Nov 29, 2014

### haruspex

The long equation in the middle of the page is wrong. The multiplying out has been done incorrectly. Your answer is right.

Last edited by a moderator: May 7, 2017
7. Nov 29, 2014

### Staff: Mentor

Did you see the "co-rotation" in my post? The motion does not influence the field in a significant way, but stationary points in the rotating reference frame need a net force.

I did not what the book calculated to get a difference, so I made a guess.

8. Nov 29, 2014

### haruspex

Yes, but I don't read that as meaning a non-inertial frame. Am I wrong?

9. Nov 29, 2014

### Staff: Mentor

Non-inertial frame if you don't want motion, inertial frame if you prefer circular motion (but then it is also important that earth and moon move). Same thing, different ways to express it.

10. Nov 30, 2014

this problem is in part before we learned about inertial and non-inertial systems(frames?) and it isn't mentioned in problem so my solution is correct? it cant be -6.03Re, it doesnt make sense

11. Nov 30, 2014

### voko

The negative value simply means it is on the other side of the Moon. That alone could be sensible. You need some other consideration to rule it out.

12. Nov 30, 2014

### haruspex

If you look through your working, you can see that the equation you started with only says that the two forces have the same magnitude. That's why you get two solutions (but the other solution is not -6.03).

13. Dec 1, 2014

### rude man

The given data look screwy, which to be sure is not a prime concern for the problem as given, but it is well known that the center of mass of the earth-moon system lies INSIDE the earth.

EDIT: OK there are two points where net gravity = 0, and the OP has the right other point. Since the problem stated "between the earth and the moon" I guess the point inside the earth doesn't count.

Still, the c.g. of the earth-moon system is inside the earth, i.e. the two bodies rotate about that point in (essentially) circles.

Last edited: Dec 1, 2014
14. Dec 1, 2014

### Staff: Mentor

So what? We don't look for this point here. This point is not a point of zero net gravity.

There is one point inside earth, true, but at that point you get crushed and boiled so much you probably don't care about gravity.

15. Dec 1, 2014

### rude man

It's not? I beg to differ ...

16. Dec 1, 2014

### haruspex

D = earth-moon distance, M = earth mass, m = moon mass, R = earth radius.
There is a point of zero net gravity at distance x < R from Earth's centre where $\frac{M x}{R^3} = \frac{m}{(D-x)^2}$.
If y is the distance of the CoM from Earth's centre then $My = m(D-y)$.
If we treat Earth as a point mass then there is only one solution for zero net gravity, the one in the OP.
Have I misunderstood?

17. Dec 2, 2014

### rude man

Your formula assumes uniform density within the earth which of course is not the case.
However, thank you for seconding my contention that such a zero-net-gravity point exists somewhere inside the earth.
A third source of support would be Dr. Feynman's Lectures in Physics where he explains why there are two tides per day, which I must admit I never did quite figure out, but is based on the c.g. of the earth-moon system as located inside the earth.

18. Dec 2, 2014

### Staff: Mentor

It is not.
The barycenter of the earth/moon system is roughly at 2/3 the radius of earth (as distance from the center of earth). Even with a uniform mass distribution, you would feel 2/3 g towards the center of earth, with a negligible contribution from the moon (because it is both much more distant and much less massive). With the non-uniform distribution, you actually feel something close to 1g - but certainly not zero g.

This becomes more obvious if we imagine the moon at a distance of 81 times the radius of earth - the barycenter would be at the surface of earth, but we certainly would not float away.

The barycenter takes the linear distance as weight, whereas gravitational force takes inverse squared distance as weight.

The actual point of zero net gravity is very close to the center of earth, less than 1km away (assuming a spherically symmetric mass distribution, which is probably wrong at that level of precision).

No, those two things are completely unrelated.

19. Dec 2, 2014

### rude man

See haruspex's last entry which gives the accurate zero-g point assuming uniform earth density.

20. Dec 2, 2014

### haruspex

I think the argument is because your post #13 gave the impression you were saying that the common mass centre was a point of zero g. I also was not sure why you mentioned the position of the common mass centre.