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Gravitational Attraction

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1. Homework Statement
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 Earth and the Mon (centre to centre) is 3.84x10^5 km, calculate where this will occur, relative to Earth.

2. Homework Equations
Fg = (G)(m1)(m2) / r^2


3. The Attempt at a Solution
FgEARTH = (G)(m1)(mE) / r^2
FgMOON = (G)(m1)(mM) / r^2
dM + dE = 3.84x10^5 km = 3.84x10^8 m


(G)(m1)(mE) / dE^2 = (G)(m1)(mM) / dM^2

(5.98x10^24) / dE^2 = (7.35x10^22) / (3.84x10^8 - dE)^2

(7.35x10^22) / (5.98x10^24) = [(3.84x10^8 - dE)^2](dE^2)

0.01229 = (1.5x10^17 - 3.84x10^8dE + dE^2)(dE^2)

I dont know where to go from here? can i move the 0.01229 to the other side and at the same time move the dE^2 to the other side (so that it gets cancelled out)??
 

Answers and Replies

Dick
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I don't know what you are doing. If Fearth=Fmoon then G*m1*mE/rE^2=G*m1*mM/rM^2 and rM+rE=r, where rE is the distance from earth, rM is the distance from the moon and r is the total distance=3.84x10^5 km. That's two equations in two unknowns, rM and rE.
 
87
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thats exactly what i wrote in mine its dE instead of rE. and then you isolate for rE:
dE = 3.84x10^8 m - dM

and then you can solve by substitution and get what i have at the top, and now i still have the same question
 
87
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okay so i cross multiplied the ratios and then used the Quadratic Formula to solve for dE and i get a number of 5x10^43. Does that sound right?
 
Redbelly98
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(5.98x10^24) / dE^2 = (7.35x10^22) / (3.84x10^8 - dE)^2

(7.35x10^22) / (5.98x10^24) = [(3.84x10^8 - dE)^2](dE^2)
Can you fix the error that occurs between these two steps?
 
87
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yea i multiplied wrong:
so i cross multiplied like this instead:

(5.98x10^24) / dE^2 = (7.35x10^22) / (3.84x10^8 - dE)^2

(7.35x10^22)(dE^2) = (5.98x10^24)(1.5x10^17 - 3.84x10^8dE + dE^2)

0 = 5.9x10^24dE^2 - 2.3x10^34dE + 8.97x10^42

**using quadratic formula i then got an answer of: 5x10^43 m
 
Redbelly98
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You are on the right track, but here are more hints:

The quadratic formula gives two solutions, not just one.
You might double-check carefully the numbers you just posted in those equations.
Perhaps it would help first to divide through by 10^22
 
87
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yea the other answer for the quadratic formula gave me a negative number, and since im looking for distance, negative numbers are meaningless, but ill try your suggestion of dividing by 10^22
 
87
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okay so i realized that some of my answers were off, but now i keep getting a negative under the square root sign??
 
454
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yea i multiplied wrong:
so i cross multiplied like this instead:

(5.98x10^24) / dE^2 = (7.35x10^22) / (3.84x10^8 - dE)^2

(7.35x10^22)(dE^2) = (5.98x10^24)(1.5x10^17 - 3.84x10^8dE + dE^2)

0 = 5.9x10^24dE^2 - 2.3x10^34dE + 8.97x10^42

**using quadratic formula i then got an answer of: 5x10^43 m
you made a mistake in multiplying out (3.84x10^8 - dE)^2
I recommend finding the general formula for dE as a function of m_e m_m and R.
you made additional errors in getting 2.3x10^34dE and 8.97x10^42 as well.

you use the quadratic formula here with b^2 very close to 4*a*c so any mistake will result in
an outcome that is many orders of magnitude off.
 
Last edited:
87
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okay so i restarted the question over (was making way too may mistakes lol) and i got as final answers dE = 7.78x10^10 and 1.7x10^8. Now because it is looking for the distance from the Earth do i just use 7.78x10^10m ??
 
454
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It's still not right. The point where the forces cancel must be between the earth and the moon, so 7.78x10^10m can't be right. the other point is closer to the earth than to the moon. This is really a case where it's better to find the general formula before you start pushing all those powers of 10 around.
 
87
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o okay i see what you mean, sorry i thought you meant there was a specific formula that i didnt know... so for what your saying then:

(mE)(dM^2) = (mM)(dE^2)

and then plug in? for:

(5.98x10^24)(3.84x10^8 - dE)^2 = (7.35x10^22)(dE^2)

and then id foil and multiple the left side and just distribute on the right side?
 
Dick
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o okay i see what you mean, sorry i thought you meant there was a specific formula that i didnt know... so for what your saying then:

(mE)(dM^2) = (mM)(dE^2)

and then plug in? for:

(5.98x10^24)(3.84x10^8 - dE)^2 = (7.35x10^22)(dE^2)

and then id foil and multiple the left side and just distribute on the right side?
Don't be in such a hurry to plug in. Replace dM with R-dE, where R is the total distance from earth to moon. Now you have a quadratic equation in dE. Try solving it without putting numbers in, like kamerling suggests. It's easy to make mistakes with all of the numbers floating around. They all look alike.
 
87
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okay: no numbers at all would be:

(mE)(r^2) - 2(mE)(R)(dE) + [(mE)(dE^2) - (mM)(dE^2)] = 0

then from here there really is no way to solve except by quadratic formula is there?
 
454
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okay: no numbers at all would be:

(mE)(r^2) - 2(mE)(R)(dE) + [(mE)(dE^2) - (mM)(dE^2)] = 0

then from here there really is no way to solve except by quadratic formula is there?
This is right if r is the same as R. There is quite a lot you can cancel after using the
quadratic formula.
 
87
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yes r is the same as R i just forgot to capitalize...

and i dont really see what can cancel:
** let the numbers inside {} be like under the square root sign
= -b +/- {b^2 -4ac} / 2a

= -[-2(mE)(R)(dE)] +/- { [-2(mE)(R)(dE)]^2 - 4[(mE)(dE^2) - (mM)(dE^2)](mE)(r^2)} / 2[(mE)(dE^2) - (mM)(dE^2)]
 
454
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and i dont really see what can cancel:
** let the numbers inside {} be like under the square root sign
= -b +/- {b^2 -4ac} / 2a

= -[-2(mE)(R)(dE)] +/- { [-2(mE)(R)(dE)]^2 - 4[(mE)(dE^2) - (mM)(dE^2)](mE)(r^2)} / 2[(mE)(dE^2) - (mM)(dE^2)]
you're solving for dE so you can't get that in the result of the quadratic formula.

what are a, b and c in [tex]\frac { -b \pm \sqrt {b^2 - 4 a c}} { 2 a } [/tex]
 
Redbelly98
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yes r is the same as R i just forgot to capitalize...

and i dont really see what can cancel:
** let the numbers inside {} be like under the square root sign
= -b +/- {b^2 -4ac} / 2a

= -[-2(mE)(R)(dE)] +/- { [-2(mE)(R)(dE)]^2 - 4[(mE)(dE^2) - (mM)(dE^2)](mE)(r^2)} / 2[(mE)(dE^2) - (mM)(dE^2)]
Work on the terms inside the square root sign next. Something cancels.
 
Kurdt
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I'd suggest going through some of your notes or a text book on algebra again. Most of the problems you're having will disappear and it will set you up for the rest of your studies.
 
87
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okay so simplifying under the square root sign will give:

3(mE^2)(R^2)(dE^2) + (mE)(mM)(R^2)(dE^2)
 
Redbelly98
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You're dropping the "4" that appears in the expression.

{ [-2(mE)(R)(dE)]^2 - 4[(mE)(dE^2) - (mM)(dE^2)](mE)(r^2)}
 
87
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i see so under the square root its actually only:

4(mE)(mM)(R^2)(dE^2)
 
Redbelly98
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Almost, but as Kammerling said dE should not appear, since dE is the variable you're trying to solve for. Sorry, I missed that point myself up to now.
So it's really 4(mE)(mM)(R^2) inside the square root.

From you're post #17 in this thread, we now "sort of" have
= -[-2(mE)(R)(dE)] +/- { 4(mE)(mM)(R^2) } / 2[(mE)(dE^2) - (mM)(dE^2)]

And I say "sort of" because, as mentioned before, the dE's should simply be omitted.
 
87
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yea i kinda realized that too so i rewrote the equation and its:

= 2(mE)(R) +/- {4(mE)(mM)(R^2)} / 2mE - 2mM

and i plugged everythign in and i still get a funny number: 3.8x10^8 which if you look at the original question that is the distance between Earth and the Moon (centre to centre) so it really doesnt make sense as an answer??
 

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