# Newton's law of universal gravitation!

1. Mar 31, 2012

### wow22

1. The problem statement, all variables and given/known data

is there a point between the earth and the moon for which the net gravitational force on an object is zero? Where is this point located? Note that the mass of the earth is 5.98x10^24 kg, the mass of the moon is 7.35x10^22kg, and the distance between the centres of earth and moon is 3.84x10^8m.

2. Relevant equations

Fge=Fgm

so GMeMp/x^2 = GMmMp/(r-x)^2
p being the point, x being the radius of earth

3. The attempt at a solution

i tried to rearrange...to find r. and got MeMm=x^2*(r-x)^2 ???
so that means 5.98*10^24(7.35*10^22)= x^2(3.84*10^8-x)^2 ?!?!

Can someone tell me how to rearrange it, cause it's supposed to be rearraned to Mm/Me(x^2)=R^2-2r+x^2 ... Thanks!
1. The problem statement, all variables and given/known data

2. Mar 31, 2012

### SammyS

Staff Emeritus
How did you ever get MeMm=x2*(r-x)2

from GMeMp/x2 = GMmMp/(r-x)2

???

3. Mar 31, 2012

### wow22

GMp crosses out both sides.. to make Me/X^2 = Mm/(R-x)^2
so cross multiply, MeMm=X^2*(r-x)^2

..why? how was i supposed to do it?

4. Mar 31, 2012

### SammyS

Staff Emeritus
That's not what you get from cross multiplying !

Cross multiply this:
$\displaystyle\frac{M_e}{x^2}=\frac{M_m}{(r-x)^2}$​
Mm and Me should end up on opposite sides of the equation from each other.

5. Mar 31, 2012

### wow22

Oh wow.. How did I not realize that ...
haha so its Me(r-x)^2 = Mm(x^2)
Thankkss!

6. Mar 31, 2012

### Cecilia48

http://www.infoocean.info/avatar2.jpg [Broken]GMp crosses out both sides..

Last edited by a moderator: May 5, 2017