Register to reply 
Gravity: Point where Earth's Gravity and the Moon's cancel each other out? 
Share this thread: 
#1
Nov2706, 03:29 PM

P: 83

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. 


#2
Nov2706, 03:56 PM

Mentor
P: 41,106

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



#3
Nov2706, 05:46 PM

P: 83




#4
Nov2706, 06:38 PM

Mentor
P: 41,106

Gravity: Point where Earth's Gravity and the Moon's cancel each other out?
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:
[tex]\frac{5.98*10^{24} kg}{r1^2} = \frac{7.35*10^{22} kg}{(3.84*10^8  r1)^2}[/tex] [tex](5.98*10^{24} kg)(3.84*10^8  r1)^2 = (7.35*10^{22} kg)(r1^2)[/tex] 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
Mar2109, 02:36 AM

P: 5

Greetings:
I believe the point of interest to be none other than the center of mass which is at m_{e}r / (m_{e} + m_{m}) meters from the Earth along the ray extending from Earth through the moon. Regards, Rich B. rmath4u2@aol.com 


#6
Mar2109, 06:19 AM

Mentor
P: 15,170

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
Apr1810, 10:47 PM

P: 3

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?



#9
Apr1910, 02:24 PM

P: 3

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*1.47E175.98E24r^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
Nov2610, 11:55 PM

P: 2

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
Nov2710, 03:55 AM

Mentor
P: 15,170

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



#12
Nov2710, 05:07 AM

Mentor
P: 15,170

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
Note that the distance between the null gravity point and the Earth is Rr=(1f)R. Applying Newton's law of gravity yields the equation for the location of the point in question, [tex]\frac{GM_m}{r^2} = \frac {G M_e}{(Rr)^2}[/tex] Obviously the gravitational constant G cancels out. Expressing r as f*R enables canceling the factor of 1/R^{2} on both sides. Dividing both sides by the Earth's mass yields the unitless expression [tex]\frac{m_r}{f^2} = \frac 1{(1f)^2}[/tex] This should be easy to solve. One final note: Why use the MoonEarth ratio instead of the masses of each? Three reasons:



#13
Nov2710, 06:18 AM

PF Gold
P: 2,022

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. 


#14
Mar311, 09:53 AM

P: 82

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, yxx=(y1)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
Mar311, 01:18 PM

P: 552

I wanted to ask if this is the gravity center for the moonearth system.



#16
Mar311, 01:24 PM

Mentor
P: 15,170

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


#17
Mar311, 02:00 PM

P: 552

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
Mar311, 02:10 PM

Mentor
P: 15,170

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



Register to reply 
Related Discussions  
Why does gravity cancel out for all points inside a sphere?  General Physics  38  
Earth's density and its gravity  Introductory Physics Homework  1  
Earth's Gravity  General Physics  11  
Gravity on mass on earth's surface vs sunearth gravity  Introductory Physics Homework  2  
Correcting for the Moon's gravity  Introductory Physics Homework  5 