Gravitational pull with the earth and moon

[alicia113]

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

The Attempt at a Solution

ok now this is from another thread, and i completely understand how she got everyting but the 81! could someone please explain to me how she got the 81.. thts the only part I am lost in..

We assume an object is at the L1 point (the point where the gravitational fields of the two objects cancels out). We will call this object X

r is the radius between the Earth and the Moon
We will call
r1
the point from the center of the Earth to x and
r2
the point from x to the center of the moon
Therefore
(1)
r=r1+r2Using this assumption of the object X we get..

Fg=Gm1xr21
where
m1
is the mass of the earth, and x is the mass of object X
Fg=Gm2xr22
where
m2
is the mass of the moon, and x is the mass of object X

Since we know that, at the point where the object X is situated, the gravitiational pull from both the Earth and the Moon will be equal, we can equate the above two equations getting the following:
(2)
Gm1xr21=Gm2xr22Now, we obtain a ratio between
m1
and
m2
so we can express
m1
in terms of
m2m1m2=81
(Note: I rounded to 81 just to make typing it up here easier...)
Therefore,
m1=81m2Now, going back to equation (2), G and x will cancel out and we replace
r1
with
r−r2
(from equation (1)) leaving us with:
81m2(r−r2)2=m2r22m2
will cancel out:
81(r−r2)2=1r2281=(r−r2)2r22i take the square root of both sides, and put the denominator on the left side
9r2=r−r210r2=rr2=r/10r2=3.84∗104Since
r1=r−r2r1=3.84∗105−3.84∗104r1=3.456∗105

 

[ehild]

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

The Attempt at a Solution

ok now this is from another thread, and i completely understand how she got everyting but the 81! could someone please explain to me how she got the 81.. thts the only part I am lost in..

We assume an object is at the L1 point (the point where the gravitational fields of the two objects cancels out). We will call this object X

r is the radius between the Earth and the Moon
We will call
r1
the point from the center of the Earth to x and
r2
the point from x to the center of the moon
Therefore
(1)
r=r1+r2Using this assumption of the object X we get..

Fg=Gm1xr21
where
m1
is the mass of the earth, and x is the mass of object X
Fg=Gm2xr22
where
m2
is the mass of the moon, and x is the mass of object X

Since we know that, at the point where the object X is situated, the gravitiational pull from both the Earth and the Moon will be equal, we can equate the above two equations getting the following:
(2)
Gm1xr21=Gm2xr22Now, we obtain a ratio between
m1
and
m2
so we can express
m1
in terms of
m2m1m2=81
(Note: I rounded to 81 just to make typing it up here easier...)
Therefore,
m1=81m2Now, going back to equation (2), G and x will cancel out and we replace
r1
with
r−r2
(from equation (1)) leaving us with:
81m2(r−r2)2=m2r22m2
will cancel out:
81(r−r2)2=1r2281=(r−r2)2r22i take the square root of both sides, and put the denominator on the left side
9r2=r−r210r2=rr2=r/10r2=3.84∗104Since
r1=r−r2r1=3.84∗105−3.84∗104r1=3.456∗105

Alicia,

your post is very hard to read. Use the x² button for a square and x₂ for an index.

So distance between the Earth and the object at point L₁ is r₁, the distance between the Moon and the object is r₂, the force from the Earth is Gm₁x/r₁², the force from the Moon is Gm₂x/r₂², they are equal, so m₁r₂²=m₂r₁². You can find the mas of Earth in any table about Earth. Google "mass of Earth" if you do not have it on your Physics book. It is 5.9736˙10²⁴ kg. You see that the ratio of the masses of Earth and Moon is about 81. ehild

 

[alicia113]

Alicia,

your post is very hard to read. Use the x² button for a square and x₂ for an index.

So distance between the Earth and the object at point L₁ is r₁, the distance between the Moon and the object is r₂, the force from the Earth is Gm₁x/r₁², the force from the Moon is Gm₂x/r₂², they are equal, so m₁r₂²=m₂r₁².


You can find the mas of Earth in any table about Earth. Google "mass of Earth" if you do not have it on your Physics book. It is 5.9736˙10²⁴ kg. You see that the ratio of the masses of Earth and Moon is about 81.


ehild

Thank you very much and sorry for my horrible computer skills! I'll try that next time !