Gravitational Equilibrium Between Earth and Moon: Location and Other Points

[thomasrules]

Hi for some reason I can't get this easy/hard question. I've tried different things but they are wrong. Please help, here's the question:

At a certain point between the Earth and the moon the total gravitational force exerted on an object by the Earth and the moon is zero. If the Earth-moon distance is 3.84x10^5 km and the moon has 1.2% of the mass of the Earth, where is this point located? Are there any other such points?

Thanks in advance

 

[DB]

Do you mean Lagrangian points?
http://en.wikipedia.org/wiki/Lagrangian_point

 

[dextercioby]

He meant L_{1}.To the OP:apply Newton's laws...



Daniel.

 

[thomasrules]

so can you help me with that part? applying the laws

 

[dextercioby]

Yeah,though i assume u know how to do the simple math involved
F_{attr}=G\frac{m_{1}m_{2}}{d^{2}}

That's Newton's gravity law.
Write Newton's second law of dynamics for the body and put the condition of equilibrium:
m\vec{a}=\sum_{k} \vec{F}_{k}
,where \vec{a} is the resulting acceleration of the body,and \vec{F}_{k} is the force labaled by "k" which acts on the body.


Daniel.

 

[thomasrules]

I'm sorry but your going to have to explain it, sorry I'm really stupid...lol
I got the Universal gravity equation already, but don't know what to do with it. I haven't seen the second equation yet, and don't know what the "E" is.

 

[dextercioby]

I'm sorry but your going to have to explain it, sorry I'm really stupid...lol
I got the Universal gravity equation already, but don't know what to do with it. I haven't seen the second equation yet, and don't know what the "E" is.

You're not stupid...And this isn't funny...What "E" are u talking about?
Where did u get the problem from? The second equation I've poted is the second law of dynamics by Isaac Newton...Have u seen it before??

Daniel.

 

[DB]

I think he means the summation sigma.
here, I had the same question not long back.
https://www.physicsforums.com/showthread.php?t=57580

 

[thomasrules]

yes the sigma...i don't know what that is...but are you sure there is no other way because my teacher didn't teach us that...

 

[dextercioby]

Okay,have it your way... Is that any more clear?
m\vec{a}=\vec{F}_{attr.Moon}+\vec{F}_{attr.Earth}

Can u take it from here?

Daniel.

 

[thomasrules]

alright so that definitely puts things easier. I don't want to waste ur time or anything its just that I'm not exactly sure what to do with that..
Um...
so is ma=0 in this case?

 

[dextercioby]

Yes,that's the condition for equilibrium.So u'll be getting a system of equations which will not be hard to solve.

Post your work...


Daniel.

 

[BobG]

The point you're looking for has to be between the Earth and Moon. In other words, they are pulling in opposite directions - one has a positive force; the other has a negative force.

In dextercioby's post he gave you the formula for the force of gravity:

F_{attr}=G\frac{m_{1}m_{2}}{d^{2}}
That's the general equation for gravitational force.
m_2 is your object
m_1 is either the mass of the Earth or the mass of the Moon (each has it's own force of gravity. We don't really want any more variables and the question told you how much mass the moon has relative to the Earth. So it's best to let m_1 be the mass of the Earth and .012m_1 be the mass of the moon.
G is the universal gravity constant.
d^2 is the square of the distance between the Earth and Moon. You need to know the principle of this. In actuality, you're going to stick something else in there.

The distance you need is some distance, x, where (d-x) is the distance between your object and the Moon and x is the distance of your object from the Earth.

At equilibrium, the net force on your object is zero. That's what dextercioby's last equation (plus post) describes:
m\vec{a}=\vec{F}_{attr.Moon}+\vec{F}_{attr.Earth}
The left side of the equation is zero. Since one of the forces from either the Moon or the Earth is negative, you can move it to the left side, making the force from the Moon equal to the force from the Earth.

You have a G on both sides and a m_2 on both sides. That means you can cancel those and don't even have to know their value.

That leaves you with:
\frac{.012m_1}{(d-x)^2}=\frac{m_1}{x^2}
And that's the problem you have to solve. You know what d is from the question. You also still have m_1 on both sides. The only unknown left is going to be x.

 

[thomasrules]

Thanks very much Bob that was excellenty explained. I understand everything thank you...