# Homework Help: How to find the L1 (First Langrangian Point) between the Earth and the Moon?

1. Jun 5, 2006

### ootz0rz

Hi everyone :) I'm new to the forums, need some help with a homework question I just can't seem to be able to get...

The question as described in my book is as follows:

The mass of the moon is 7.35x10^22 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 the earth and the moon (center to center) is 3.84x10^5 km, calculate where this will occur, relative to the Earth.

Earlier in the book, I am given the mass of the Earth as 5.98x10^24 kg and I am supposed to use that.

What I've got so far (not sure if any of this is correct at all... ):

$$F_g = \frac{GMm}{r^2}$$

Where...
$$G = 6.67x10^{-11} \frac{Nm^2}{kg^2}$$
M = The larger mass
m = the smaller mass

Problem is, we are not given an object being at the L1 point (which I will refer to as X), so it can't be used for that. It can be used however to find the force of gravity from earth that is acting on the moon at that radius. But anyways, I needed a way to remove the m variable, so I replaced Fg with an equivalent equation:

$$\frac{mv^2}{r} = \frac{GMm}{r^2}$$

m and one r cancels out and we get

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

this however introduces a new unknown variable into, v - being the velocity
However, v can be expressed as

$$v = \frac{2 \pi r}{T}$$
therefore $$v^2 \Rightarrow v^2 = \frac{4 \pi^2 r^2}{T^2}$$

T is the period it takes to complete one revolution, yet another fun unknown...

at this point we have
$$\frac{GM}{r} = \frac{4 \pi^2 r^2}{T^2}$$

Then if we bring r^2 to the left side...

$$\frac{GM}{r^3} = \frac{4 \pi^2}{T^2}$$

I could further juggle the equation so it would solve for r...however...I'm not sure what to do with the T variable

That's about as far as I was able to get :/ Any help would be greatly appreciated.
Thank you :)

Last edited: Jun 5, 2006
2. Jun 5, 2006

### Andrew Mason

Welcome to PF!

You just have to find the point at which the force of gravity on an object from the moon is equal and opposite to the force of gravity on the earth. Take the distance between the earth and moon as d. So $r_e = d - r_m$

AM

3. Jun 5, 2006

### ootz0rz

I just tried doing this another way, as follows (and was able to successfully solve for "r" using this method). Can someone please look it over? :)

We assume an object is at the L1 point (the point where the gravitational fields of the two objects cancels out). We will call this object X

r is the radius between the Earth and the Moon
We will call $$r_1$$ the point from the center of the earth to x and $$r_2$$ the point from x to the center of the moon
Therefore
(1)$$r = r_1 + r_2$$

Using this assumption of the object X we get..

$$F_g = \frac{G m_1 x}{r^2_1}$$ where $$m_1$$ is the mass of the earth, and x is the mass of object X
$$F_g = \frac{G m_2 x}{r^2_2}$$ where $$m_2$$ is the mass of the moon, and x is the mass of object X

Since we know that, at the point where the object X is situated, the gravitiational pull from both the Earth and the Moon will be equal, we can equate the above two equations getting the following:
(2)$$\frac{G m_1 x}{r_1^2} = \frac{G m_2 x}{r_2^2}$$

Now, we obtain a ratio between $$m_1$$ and $$m_2$$ so we can express $$m_1$$ in terms of $$m_2$$

$$\frac{m_1}{m_2} = 81$$ (Note: I rounded to 81 just to make typing it up here easier...)
Therefore, $$m_1 = 81m_2$$

Now, going back to equation (2), G and x will cancel out and we replace $$r_1$$ with $$r-r_2$$ (from equation (1)) leaving us with:
$$\frac{81m_2}{(r-r_2)^2} = \frac{m_2}{r_2^2}$$

$$m_2$$ will cancel out:
$$\frac{81}{(r - r_2)^2} = \frac{1}{r_2^2}$$

$$81 = \frac{(r - r_2)^2}{r_2^2}$$

i take the square root of both sides, and put the denominator on the left side
$$9 r_2 = r - r_2$$

$$10 r_2 = r$$

$$r_2 = r / 10$$

$$r_2 = 3.84*10^{4}$$

Since $$r_1 = r - r_2$$

$$r_1 = 3.84*10^{5} - 3.84*10^{4}$$

$$r_1 = 3.456*10^{5}$$

my actual answer is a little bit different since i rounded differently for posting on here...but i'd like to know if the method is correct?