Earth and moon where does gravity cancel out

In summary, the gravity between Earth and moon cancels out to some degree. However, if you are located between the two, you would not experience a force.
  • #1
lozzajp
14
0

Homework Statement


where does the gravity cancel out between the Earth and moon?

Homework Equations


s=3.84405x10^8m
earth m = 5.98x10^24kg
moon m =7.35x10^22kg
G = 6.67x10^-11N m^2/kg^2
mass of object between = 1kg for simplicity
F = G x m / r^2
r = ?

F Earth = F moon
G x Earth / r^2 = G x moon / (s-r)^2

The Attempt at a Solution


I get so far with that and I've ended up with:
earth = (moon x r^2) / (s-r)^2

not sure if its right and how do i isolate r?
 
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  • #2
Let's call the distance away from Earth r, and the total distance between the Earth and moon R.

A mass in between the Earth and moon will experience a gravitational force from both.

Fearth = GMem/r2

Fmoon = GMmm/(R-r)^2

For the net force to be zero the two above forces must equal each other.
 
  • #3
i understand that but i don't know how to solve for r (R = s in my part)
 
  • #4
You need to use simple algebra to isolate r.

start with multiplying both sides by the rights side denominator. then you need to multiply out Earth and divide everything by r2. By doing this you should be able to get a single r2 term. then just isolate r.
 
  • #5
lozzajp said:
i understand that but i don't know how to solve for r (R = s in my part)

I worked it out, and it's definitely doable but the algebra is a little more intense than it probably needs to be. I thought of an easier way to do it. Let R be the total distance between the moon and the earth.

The distance away you are from the Earth must be some percentage of total R. So if you're halfway of the total distance then .5R , if you're a quarter then .25R, etc. So let's say you're distance from Earth is aR where a is some number <1

The distance from the moon must then be 1-a

So we can say this.

GM/aR = Gm/(1-a)R here I called M the mass of Earth and m the mass of the moon.

Now you can just solve for a, that is significantly easier. Multiply a by the total distance and that's how far from Earth you need to be (well the center of the earth)
 
  • #6
thats a good way to do it! thanks for that.

do post the intense algebra? :)
 
  • #7
lozzajp said:
thats a good way to do it! thanks for that.

do post the intense algebra? :)

You cancel G and cross multiply to get

(R-r)2/r2 = m/M

Foil out R-r

(R2-2Rr+r2)/r2 = m/M

You cross multiply the m/M to bring it to the left and r2 to bring it to the right.

Distribute the new M/m term on the left. Now subtract r2 so it's back on the left. Give r2 a common denominator wit the rest of the terms. Now you can factor out r2

The end result will be a constant plus r times some constant plus r2 times some constant = 0

So you can plug that into the quadratic formula.
 
  • #8
Oops.
 
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  • #9
haruspex said:
I guess that's what the question is after, but it isn't strictly correct. If you were at that place you would not find that the two gravities cancelled. That's because once inside the Earth there's no net force from from the parts of the Earth that are further from the centre than you are. But I hope that's not to be taken into account here.


I'm not sure what you mean? What does being inside the Earth have to do with this? You're in space located in between the moon and the earth.
 
  • #10
aftershock said:
I'm not sure what you mean? What does being inside the Earth have to do with this? You're in space located in between the moon and the earth.
Oops, was thinking of a different question entirely. Sorry for the noise.
 
  • #11
aftershock said:
You cancel G and cross multiply to get

(R-r)2/r2 = m/M

Foil out R-r

(R2-2Rr+r2)/r2 = m/M

You cross multiply the m/M to bring it to the left and r2 to bring it to the right.

Distribute the new M/m term on the left. Now subtract r2 so it's back on the left. Give r2 a common denominator wit the rest of the terms. Now you can factor out r2

The end result will be a constant plus r times some constant plus r2 times some constant = 0

So you can plug that into the quadratic formula.

So it is impossible without quadratic formula.. guess I have some reading to do thanks for all your help!
 

Related to Earth and moon where does gravity cancel out

1. What is the concept of gravity cancellation between the Earth and the Moon?

The concept of gravity cancellation between the Earth and the Moon refers to the point in space where the forces of gravity between the two objects are equal and opposite, resulting in a nullification of their gravitational pull on each other.

2. Where does gravity cancel out between the Earth and the Moon?

The point where gravity cancels out between the Earth and the Moon is known as the Lagrange point or L1 point. It is located about 238,855 miles (384,400 kilometers) from the Earth towards the Moon.

3. Why is the L1 point significant?

The L1 point is significant because it allows for a stable orbit between the Earth and the Moon. Objects placed at this point will maintain their relative position to both the Earth and the Moon, making it an ideal location for space exploration and satellite missions.

4. How does the Earth-Moon distance affect the L1 point?

The L1 point is directly influenced by the distance between the Earth and the Moon. As the distance between the two objects changes due to their orbital paths, the L1 point also shifts accordingly. This means that the L1 point is not a fixed location, but rather a constantly moving point in space.

5. Can gravity cancel out between other celestial bodies?

Yes, the concept of gravity cancellation can be applied to other celestial bodies as well. For example, the Lagrange points also exist between the Sun and the Earth, and between the Earth and other planets in our solar system. These points have been utilized in space missions for various purposes, such as studying the Sun or observing other planets.

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