Where is the Point of Zero Gravitational Force Between Earth and the Moon?

[Plutonium88]

Homework Statement

At a ertain point between Earth and the moon the total gravitation force exted on an object by both planets is 0. The Earth - moon distance is 3.84 x 10^5 and the moon has 1.2% of the mass of earth. Where is this point located.

Homework Equations

Fg=GmM/R^2

The Attempt at a Solution

So i drew a diagram and put the distance R between the Earth and the moon and the object somewhere closer to the moon. I then wrote that the distance from the object from the Earth is (R-X). I then said that the distance from the moon to the object was X.

Sine it says the Gravitational force exerted on the object at a point is the same... i get

Fgm=Fge
(GmMm)/(X^2) = (GmMe)/(R-X)^2

x<R

i simplify this too
X^2 = (R-X)^2
0 = 0.988x^2 - 0.024RX + (0.012)*R^2

if i put this into the quadratic formula i get values

x= 0.5

and x = 9327.44m


Now if my answer are correct, does this mean that both are answers? so i would say?

this point is located 9327.44 meters from the moon and 374.672.56 meters from earth.

and this point is also located

0.5 meters from the moon and 383999.5 meters from the earth?


but at this second point it couldn't match my first point cause the force of gravity from the Earth would decrease while the force of gravity on the moon would increase so how could the force exerted be the same? Anyway i would appreciate any help i can get.

 

[voko]

Taking Me ti be the mass of the Earth, and Mm the mass of the Moon, (GmMm)/(X^2) = (GmMe)/(R-X)^2 is correct. The next step is not: X^2 = (R-X)^2 is incorrect. Me and Mm are different so cannot both disappear.

 

[Bashyboy]

Sine it says the Gravitational force exerted on the object at a point is the same... i get

Fgm=Fge
(GmMm)/(X^2) = (GmMe)/(R-X)^2

If the two forces were equivalent, that would imply the two forces pull with the same magnitude and in the SAME DIRECTION. If the two forces pulled in the same direction, and were of the same magnitude, the net force wouldn't be zero. The key idea is that the net force is in fact zero; hence, the two forces have to balance each other out (or cancel).

I hope this helps. If anyone finds fault, please correct me.

 

[SteamKing]

You also need to get your units straight. Is the distance from the Earth to the moon in feet, meters, miles, furlongs, etc? What about the units of G? What units is the mass of the Earth measured in?

 

[Plutonium88]

Taking Me ti be the mass of the Earth, and Mm the mass of the Moon, (GmMm)/(X^2) = (GmMe)/(R-X)^2 is correct. The next step is not: X^2 = (R-X)^2 is incorrect. Me and Mm are different so cannot both disappear.

Yea you're right it hsould have een X^2 =(0.012)*(R-X)^2 cause Mm = 0.012*Me but it was included in that step in calculation of the the next step

 

[Plutonium88]

If the two forces were equivalent, that would imply the two forces pull with the same magnitude and in the SAME DIRECTION. If the two forces pulled in the same direction, and were of the same magnitude, the net force wouldn't be zero. The key idea is that the net force is in fact zero; hence, the two forces have to balance each other out (or cancel).

I hope this helps. If anyone finds fault, please correct me.

So If i choose my + direction toward earth.

i would get Fnet = Fge - Fgm = 0
Fnet = (GmMe)/(r-x)^2 - (Gm(0.012)*Me)/x^2 = 0
Fge = Fgm ?

and solve for this?

 

[Plutonium88]

You also need to get your units straight. Is the distance from the Earth to the moon in feet, meters, miles, furlongs, etc? What about the units of G? What units is the mass of the Earth measured in?

yea sorry i don't know why i didn't include it in the givens. Everything is in meters for distances. The radius R and X as well in meters. The mass of the Earth is in kilograms.

G is measured by (N*M^2)/Kg^2

Sorry about not including that in my given statement.

Me=5.98x10^24kg
Re=6.38 x 10^6 meters
G= 6.67x10^-11 (N*M^2)/kg^2

 

[voko]

Here is what happens. $$ \frac {M_m} {X^2} = \frac {M_e} {(R - X)^2}
\\
\frac {(R - X)^2} {X^2} = \frac {M_e} {M_m}
\\
(\frac {R - X} {X})^2 = \frac {M_e} {M_m}
\\
\frac R X - 1 = \pm \sqrt {\frac {M_e} {M_m}}
$$

You need to choose the right sign. Which one is right?

 

[Plutonium88]

Here is what happens. $$ \frac {M_m} {X^2} = \frac {M_e} {(R - X)^2}
\\
\frac {(R - X)^2} {X^2} = \frac {M_e} {M_m}
\\
(\frac {R - X} {X})^2 = \frac {M_e} {M_m}
\\
\frac R X - 1 = \pm \sqrt {\frac {M_e} {M_m}}
$$

You need to choose the right sign. Which one is right?

The positive sign cause a distance can't be negative? So since this question asks for the Point at which the force of gravitational force exerted on the object is equal, R/X -1 is that point? Or do i solve for that X value, and express the point as a distance from the moon to the object and from the Earth to the object?

x=R/[√(1/0.012) +1]

therefore the moon is "R/[√(1/0.012) +1]" from the object and the Earth is "R - R/[√(1/0.012) +1]" from the object?

I'm just having a hard time understanding how you consider the exact point.

 

[haruspex]

So since this question asks for the Point at which the force of gravitational force exerted on the object is equal, R/X -1 is that point? Or do i solve for that X value, and express the point as a distance from the moon to the object and from the Earth to the object?

Voko set X as the distance of the point from the moon. Did you understand voko's equations?

 

[voko]

Recall what R and X are - you defined them, so you should know!

For R/X - 1 to be negative, R has to be less than X, or one of them has to be negative - is that physically sensible?

 

[Plutonium88]

So since this question asks for the Point at which the force of gravitational force exerted on the object is equal, R/X -1 is that point? Or do i solve for that X value, and express the point as a distance from the moon to the object and from the Earth to the object?
/QUOTE]
Voko set X as the distance of the point from the moon. Did you understand voko's equations?

Yes his equations are essentially the same as mine, except he was smart and took the two square pieces and combined them, where i turned it into a quadratic.

GmMe/(R-X)^2 = GmMm/X^2
He combined the X^2 + (R-X^2)

where i said

Mm(R-X)^2 = Me*X^2
0.012*Me(R-X)^2 = Me*X^2
(0.012)(R^2 - 2RX + X^2) = X^2** FIX **
0 = -0.988x^2 - 0.024RX + (0.012)R^2
So both his equation and my equation should yield the same answers FOR X no?

 

[Plutonium88]

Recall what R and X are - you defined them, so you should know!

For R/X - 1 to be negative, R has to be less than X, or one of them has to be negative - is that physically sensible?

R cannot be less than X because i defined X as the smaller portion, and no neither of them would be negative cause they are lengths

Also that's a cool way of thinking about it, If R is LESS* than x the fraction is less than 1 so it will be negative... cooolio

 

[haruspex]

0 = 0.988x^2 - 0.024RX + (0.012)*R^2

I believe you have a sign wrong there.

 

[Plutonium88]

I believe you have a sign wrong there.

yessir i put a fix up top when i saw that.

 

[Plutonium88]

Also the Distance R is measured in kilometers... I don't know why i said meters, i must have missread the question.

So when i solve the quadratic i get

X1 = 37912.03 KM

or x2 = -47239.97 KM

 

[voko]

The negative answer, as you have discovered, is meaningless physically.