Gravitational attraction

hydrocarbon
The mass of the moon is 7.35*10^22Kg. At some point between Earth and the moon, the force of Earth's gravitational attraction on an object is canceled 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, calculate where this will occur, relative to earth.

required: r1

analysis: Fge=Fgm

therefore: (6.67*10^-11)(5.98*10^24)/(6.38*10^6)^2 = (6.67*10^-11)(7.35*10^22)/r^2

(5.98*10^24)(3.84*10^8)/(6.38*10^6)^2(7.35*10^22)= r^2

I got 282m which is clearly wrong can i please get some assistance with the steps? thank you very much

Homework Helper
Look at the post for tascja. Are you in the same class? You posted the same question.

hydrocarbon
yes i see it. i posted before him, but it's ok. i still don't understand how to do the steps,the explanation on the posts are still too vague. i understand that Fge=Fgm as well as the universal formula. my problem is how to do the steps fully. the answers i keep getting are too small.

Homework Helper
The mass of the moon is 7.35*10^22Kg. At some point between Earth and the moon, the force of Earth's gravitational attraction on an object is canceled 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, calculate where this will occur, relative to earth.

required: r1

analysis: Fge=Fgm

therefore: (6.67*10^-11)(5.98*10^24)/(6.38*10^6)^2 = (6.67*10^-11)(7.35*10^22)/r^2

(5.98*10^24)(3.84*10^8)/(6.38*10^6)^2(7.35*10^22)= r^2

I got 282m which is clearly wrong can i please get some assistance with the steps? thank you very much
Here's a tip: say what you are doing: I can guess that the numerator your first line is GM for Earth and moon respectively but what distances are you using in the denominator?
If "r" be the distance from the object to the center of the Earth and "R" the distance between moon and earth, then the distance from the object to the center of the moon is R-r. Letting Me and Mm be the masses of the Earth and moon respectively, The two gravitational forces are
$$\frac{GM_em}{r^2}= {GM_mm}{(R- r)^2}$$

hydrocarbon
$$\frac{GM_em}{r^2}= {GM_mm}{(R- r)^2}$$
do i plug in the value for r^2 on the left hand side which would be the distance of the earth? and leave r on the right alone?

hydrocarbon
can anyone actually help me with the calculations?

Homework Helper
$$\frac{GM_em}{r^2}= {GM_mm}{(R- r)^2}$$
do i plug in the value for r^2 on the left hand side which would be the distance of the earth? and leave r on the right alone?

It should be pretty obvious Hall's missed a '/' on the right hand side. Now just multiply both sides by r^2*(R-r)^2 to clear out the fractions. Then you have a quadratic equation. Solve it for r. r is the distance to the earth, R is the total distance Earth to moon so (R-r) is distance to moon.

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hydrocarbon
this what i got.

GMem/r^2=GMmm/(R-r)^2 therefore r^2GMmm=GMem(R-r)^2

(R-r)=(6.38*10^6)^2(5.98*10^24)/7.35*10^22

(R-r) = 2.43*10^38/7.35*10^22

(R-r) = square root 3.31*10^15 = 5.75*10^7

-r = -3.84*10^8 + 5.75*10^7 = -3.27*10^8

r = -3.27*10^8/-1

r = 3.27*10^8

please let me know if that's the distance that the Earth and moon's gravities cancel each other out. Thanks

Homework Helper
The first line is good, but now you have a quadratic equation. Mm*r^2=Me*(R-r)^2. Then I have no idea what you are doing but it doesn't look like legal algebra. You want to solve for r. Expand it out. You should get an equation like ar^2+br+c=0. What are the a, b and c in terms of Mm, Me and R?

hydrocarbon
if i was expected to use quadratic formula for this question then it should have been mentioned in the lesson. there should be another way to do this

Homework Helper
if i was expected to use quadratic formula for this question then it should have been mentioned in the lesson. there should be another way to do this

There is if you dislike the quadratic equation. Take the square root of both sides of Mm*r^2=Me*(R-r)^2.

hydrocarbon
can you help me with the first step on squaring both sides? I'm a bit confused with that. if it was just one side then i would know what to do. thanks

Homework Helper
Don't square, square root. As in, if you have 4*9=36, then it's correct to say sqrt(4*9)=sqrt(36). sqrt(4*9)=sqrt(4)*sqrt(9)=2*3. sqrt(36)=6. 2*3=6. Do the same thing with your expression. sqrt(r^2)=r, e.g.

hydrocarbon
ok so. sqrt(r^2)=r , sqrt(R-r)^2=R-r? then Mm*r=Me*R-r?

Mm*r/Me then subtract R =-r then then divide by negative 1? then I'm left with the answer for r? is this right?

Homework Helper
ok so. sqrt(r^2)=r , sqrt(R-r)^2=R-r? then Mm*r=Me*R-r?

Mm*r/Me then subtract R =-r then then divide by negative 1? then I'm left with the answer for r? is this right?

No offense, but your algebra skills don't seem to be up for this. Are you in the same class as tascja? Because neither of you seem very prepared for this question. sqrt(Mm)*r=sqrt(Me)*(R-r).

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hydrocarbon
i can do math it's just that i can get confused when things don't get properly explained to me. but not everyone is good at what they do.

oh and offense taken.

Staff Emeritus
Homework Helper
ok so. sqrt(r^2)=r , sqrt(R-r)^2=R-r? then Mm*r=Me*R-r?

No.

You need to take the square root of all terms on each side of the equation:
Mm*r^2=Me*(R-r)^2
Not just the (r^2) and (R-r)^2 terms.