Gravitational Attraction Question

[wilson_chem90]

1. 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 canceled by the Moon's force of gravitational attraction. If the distance between Earth and the Moon (centre to centre) is 3.84x10^5 km, calculate where this will occur, relative to Earth.



2. Homework Equations
Fg = Gm1m2/r2


3. The Attempt at a Solution
I don't know where to begin.

 

[Doc Al]

Imagine some mass (call it 'm') at a distance X from the center of the Earth. What force does the Earth exert on m? The Moon? What's the net force on m from both?

 

[wilson_chem90]

So i have to find the gravitational attraction exerted on Earth and the moon first? and find the net force from those 2 forces?

 

[Doc Al]

Yes. Do it symbolically first, before doing any arithmetic.

 

[wilson_chem90]

Alright so what i did was i found the mass of the object first.

I used Fg= Gm1m2/r2 and rearranged it to m1 = Fgr2/Gm1

Then i put the information in and got

m1 = (1.74x10^6m)2(1.67 N)/(6.67x10^-11 Nm2/kg2)(7.35x10^22 kg)
m1 = 5.57x10^23 kg

Now should i put that new mass into the Fg for the moon and the earth?

 

[Doc Al]

Alright so what i did was i found the mass of the object first.

The mass of the object is arbitrary. Don't try to "figure it out", just call it 'm'.

I used Fg= Gm1m2/r2 and rearranged it to m1 = Fgr2/Gm1

Then i put the information in and got

m1 = (1.74x10^6m)2(1.67 N)/(6.67x10^-11 Nm2/kg2)(7.35x10^22 kg)
m1 = 5.57x10^23 kg

In order to "figure out" the mass of the object, you'd have to know the gravitational force and its distance from the Earth (or moon). You don't know any of that.

Instead, just write expressions for the force on the object from the Earth and the force on it from the Moon. Call its mass 'm' and its distance from the center of the Earth 'X'.

 

[wilson_chem90]

Alright hopefully this is the right step

Gravitational Attraction of Earth

Fg = Gm1m2/r2
= (6.67x10^-11 Nm2/kg2)(m)(5.98x10^24 kg) / (6.38x10^6 m)2
= 9.80 N

Gravitational Attraction of the Moon

Fg = Gm1m2/r2
= (6.67x10^-11 Nm2/kg2)(m)(7.35x10^22 kg) / (1.74x10^6 m)2
= 1.62 N

Fnet = Fg of Earth - Fg of the Moon
= 9.80 N - 1.62 N
= 8.18 N

now is this correct? lol

 

[Doc Al]

Nope, not correct.

In general, if someone asks you to find the force on an unspecified mass "m", do not respond with a number! You don't have the information to get a numerical answer.

Instead, I just want you to write the formulas for the forces. The formulas will be in terms of 'm' and 'X'. Later, you'll solve for X.

 

[wilson_chem90]

lol alright, this is harder than i thought...

F = ma
Fs = usFn
Fk = ukFn
F = mg

Fg = Gm1m2 / r2
F = mv2/r

i think that's all of them

 

[Doc Al]

I meant for you to write the force formula that you need, not every one that you can think of!

Fg = Gm1m2 / r2

That's the only one you need. What's m1, m2, and r for each case?

 

[wilson_chem90]

ahah i knew that.. ah m1 is the mass of Earth and m2 is the mass of the moon, and then r should be the distance between the centre of Earth to the centre of the moon

 

[Doc Al]

ahah i knew that.. ah m1 is the mass of Earth and m2 is the mass of the moon, and then r should be the distance between the centre of Earth to the centre of the moon

Earth's force on object:
m1 = mass of earth
m2 = m = mass of object
r = distance of object to center of earth = X

Moon's force on object:
m1 = mass of moon
m2 = m = mass of object
r = distance of object to center of moon = D - X (where D is earth-moon distance)

 

[wilson_chem90]

i think i finally got it...

D = 3.84x10^5 km = 3.84x10^8 m

Ge = Gm1m2/X
Gm = Gm1m2/ D-X

Ge = (6.67x10^-11 Nm2/kg2)(5.98x10^24 kg)m2/X
= 3.99x10^14 Nm2/kg(m2)/ X

Gm = (6.67x10^-11 Nm2/kg2)(7.35x10^22 kg) / (3.84x10^8m - X)2
= 4.90x10^12 Nm2/kg m2 / (3.84x10^8 - X)2

Now the X and m2's cancel and i am now left with

(3.99x10^14 Nm2/kg) X (4.90x10^12 Nm2/kg) / (3.84x10^8 m)2
= 1.33x10^10 m2

 

[Redbelly98]

But X does not cancel, only m2 cancels.

I have 3 suggestions for you:

• Don't plug in numbers yet, wait until you have an equation in terms of X before doing that. Keep things in terms of symbols for now.
  
• Let the Earth's mass be mE, and the moon's mass be mM, so that you're not using the same symbol (m1) for two different things.
  
• If something is squared in your equations, write that as ^2. I.e., X^2 for "X squared". That way, you're not using m2 to mean both "mass 2" and "meters squared".

Ah, one more thing. The expression for Ge should have /X^2 at the end of it.

So, what equation results from setting Ge and Gm (using symbols, not numbers) equal to each other?

 

[wilson_chem90]

GMe/X^2 = GMm/(X-D)^2

 

[Doc Al]

GMe/X^2 = GMm/(X-D)^2

Good. But since D > X, I'd use D-X (instead of X-D) to keep things positive. (That will matter when you take the square root.)

Now cancel what you can, rearrange, and solve for X.

 

[wilson_chem90]

alrght so i got square root(Me-D/X^2Mm) = X

and if i rearrange the X^2 i get square root(Me-D/Mm) = X(X^2)

 

[Doc Al]

Starting with this equation:

GMe/X^2 = GMm/(X-D)^2

Or better, this one:
GMe/X^2 = GMm/(D-X)^2

Cancel the G, invert, then take the square root of both sides.

 

[wilson_chem90]

alright so then i get

square root X/Me = square root(D-X)/Mm

 

[Doc Al]

alright so then i get

square root X/Me = square root(D-X)/Mm

Almost. What you get is:

square root {X^2/Me} = square root {(D-X)^2/Mm}

What's the square root of X^2?

 

[wilson_chem90]

isn't the square root of X^2 just {X}?

 

[Doc Al]

isn't the square root of X^2 just {X}?

Right!

 

[wilson_chem90]

so that means the square root ({X}/Me) = square root ({D-X}/Mm)

 

[Doc Al]

so that means the square root ({X}/Me) = square root ({D-X}/Mm)

Nope. √(A²B) = A√B ≠ √(AB)

 

[wilson_chem90]

sorry, but I'm kind of lost at the moment.. does that mean only X and D-X is square rooted?

 

[Doc Al]

$$\sqrt{\frac{X^2}{M_e}} = \frac{\sqrt{X^2}}{\sqrt{M_e}} = \frac{X}{\sqrt{M_e}}$$

 

[wilson_chem90]

oh okay yeah, that's actually what i got, i just wasn't completely sure

 

[wilson_chem90]

alright so the new equation should be X/{Me} = D-X/{Mm}

 

[Redbelly98]

Is X/{Me} the same as $$\frac{X}{\sqrt{M_e}} \$$ ?

 

[wilson_chem90]

Yeah it is

 

[wilson_chem90]

the { } is square root

 

[Redbelly98]

the { } is square root

Ah, okay. Then yes, this is correct:

alright so the new equation should be X/{Me} = D-X/{Mm}

Next is to solve that equation for X.

 

[wilson_chem90]

i got to X = D-X{Me} / {Mm}

 

[Doc Al]

i got to X = D-X{Me} / {Mm}

OK, write that as:
X = (D-X) {Me/Mm} (using {} for square root)

Now solve for X. (Get all the X terms on the same side.)

 

[wilson_chem90]

so would it just be X+X = D {Me/Mm}