Finding the point where Earth and Moon's gravity balance

[Alameen Damer]

Homework Statement

Homework Equations

fg=Gm1m2/r^2

The Attempt at a Solution

Let FgE be the force of gravity of Earth, and FgM be force of gravity of the moon.

We need a net gravitational force of 0 N. So:

FgE-FgM=0
FgE=FgM

Can someone attempt to solve this and see if the answer matches the book answer which is 3.8 x 10^7.

I have tried this with many different methods, and the closest I got was 4.2 x 10^7. So if someone gets the same answer as me or the book, can you please reply.

 

[ehild]

Show your work in detail, please. And what is 4.2 x 10^7? mass? distance? force? In what unit?

 

[Alameen Damer]

Distance/radius so its in m, and I got the answer eventually having to use the quadratic formula.

 

[Alameen Damer]

What I did was:

Let mE be mass of Earth
Let mM be mass of the moon
Let mR be mass of the rocket

GmEmR/r^2=GmMmR/(3.8x10^8-r)^2

(3.8x10^8-r)^2(GmEmR)=r^2(GmMmR)

G and MR cancel

(3.8x10^8-r)^2(mE)=r^2(mM)

(5.926x10^24)r^2-4.56x10^34r+8.664x10^41=0

I then used the quadratic formula to solve for r, getting 342017105.1 m. This is the radius from the Earth to the rocket to distance from the moon to the rocket would be:
3.8x10^8m-342017105.1 m=3.8x10^7m.

HOWEVER, i am wondering is there an easier approach to solving this?

 

[ehild]

What I did was:

Let mE be mass of Earth
Let mM be mass of the moon
Let mR be mass of the rocket

GmEmR/r^2=GmMmR/(3.8x10^8-r)^2

(3.8x10^8-r)^2(GmEmR)=r^2(GmMmR)

G and MR cancel

(3.8x10^8-r)^2(mE)=r^2(mM)

You can write the equation in the form
(3.8x10^8-r)^2=r^2(m_M/m_E).
Take the square root of both sides, you get a first order equation.