Calculating Distance from Earth's Center for 1/10 Gravitational Acceleration

[mosque]

Homework Statement

What is the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Earth is 1/10 its value at the Earth's surface?

Homework Equations

FG= GmM/r^2
g= Gm/r^2

G=6.67x10^-11 (Nm^2)/kg^2
m(earth)=5.97x10^24 kg
g=9.8m/s^2

The Attempt at a Solution

g=1/10g (earth)
g= Gm/r^2

r^2= (Gm/g)1/10
r^2= (Gm/g)10

Answer: r=2.02x10^7 m^2

Using dimensional analysis I somehow got m^2, but I know the units should be in meters, not meters squared.
My other concern is where I change r^2= (Gm/g)1/10 to r^2= (Gm/g)10

I'm not even sure if it was correct to do that. By changing the multiplication from 1/10 to 10 was the only way I got the right answer. I could maybe post an attachment with a picture of my work if it would help you understand what I did.

Any help is greatly appreciated.

 

[TSny]

Hi mosque. Welcome to PF!

The acceleration is g/10. So you have $r^2= \frac{GM}{(g/10)}$

Try simplifying this by multiplying the numerator and denominator by 10.

Note that you had to take a square root to get the answer. What does that do to the units?

 

[mosque]

Wow that's so simple. Now I understand what I was doing wrong. Now that I think about it dividing the gravity by 10 is the same as multiplying it by 10, but instead I worked out Gm/g first and then tried to multiply by 1/10.

Also, once I plugged in the number I did take the square root to get the final answer, but I did the dimensional analysis after so I completely forgot to take the square of the meter.

Thank you so much! :)