# The height at which the gravity acceleration is 5x smaller

## Homework Statement

The problem states to calculate the height at which the gravity acceleration is 5x smaller than on the surface of the earth. So
g=9,81 m/s^2
R=6370km
M=6x10^24
gamma(don't know how to put it :P)=6,67x10^-11

## Homework Equations

g=gamma x M/(R+h)^2

## The Attempt at a Solution

I was pretty confident in this h = sqrt (g x R)/(gamma x M) but the result turned out bad.

The book I use states that the result is 7874 km. [/B]

NascentOxygen
Staff Emeritus
Welcome to Physics Forums.

Your approach sounds right. Can you show the steps in your calculations so we can look for errors?

Thank you :)

Firstly I calculated the 5x smaller gravity ( g/5 ) which resulted in 1,962 m/s^2
Then I converted the 6370km to meters ( 6 370 000 m )
Then, using my approach sqrt( 1,962 x 6370000 ) / ( 6,67 x 10^-11 x 6x10^24 ) i ended up with sqrt 18383940 / 40,02 x 10^23 which ended up as sqrt 0.0000000459368816 and I found it useless to continue as the result is obviously wrong

gneill
Mentor
Feel free to write G for the gravitational constant.

You might find it simpler to first find the radial distance where the acceleration is 1/5 g. Then subtract one Earth radius to yield the height. Work with symbols at first: stuff will cancel out and you can save fingers and calculator batteries. Use ratios!

RUber
Homework Helper
g_1=gamma x M/(R+0)^2
g_2 = gamma x M/(R+h)^2
g_1/g_2 = 5.
Lots of stuff should cancel out.

NascentOxygen
Staff Emeritus
Firstly I calculated the 5x smaller gravity ( g/5 ) which resulted in 1,962 m/s^2
Then I converted the 6370km to meters ( 6 370 000 m )
Then, using my approach sqrt( 1,962 x 6370000 ) / ( 6,67 x 10^-11 x 6x10^24 ) i ended up with sqrt 18383940 / 40,02 x 10^23 which ended up as sqrt 0.0000000459368816 and I found it useless to continue as the result is obviously wrong
You seem to have this upside-down, or something.

haruspex
Homework Helper
Gold Member
2020 Award
g=gamma x M/(R+h)^2
Right.
h = sqrt (g x R)/(gamma x M)
Wrong. Have another go at that algebra. If you get the same wrong result, please post all your steps.

Right.

Wrong. Have another go at that algebra. If you get the same wrong result, please post all your steps.

After a rather long period of trying different variations I figured it out. It resulted in h = [ sqrt (G x M) / g ] - R

You seem to have this upside-down, or something.

This really helped out since it was kinda upside-down...

So thank you everyone for leading me to the right solution ( and I'm really grateful for not getting a straight answer because I feel I've improved my algebra lots ).

One more question : RUber and gneill mentioned the method of canceling out, I did try that but I got R^2=(R-h)^2 and I was stuck there. I'm pretty sure I messed up somewhere in the process but could you guys explain this one to me since I want to understand this method too ( mainly because it sounds easier ).

Once again thank you very much ! :D

RUber
Homework Helper
RUber and gneill mentioned the method of canceling out, I did try that but I got R^2=(R-h)^2 and I was stuck there.
You should have gotten that 5 R^2 = (R+h)^2, since you are looking for the gravity to be 1/5 of what it is on earth.

g_1=gamma x M/(R+0)^2
g_2 = gamma x M/(R+h)^2
g_1/g_2 = 5.
Lots of stuff should cancel out.
##g_1/g_2 =\frac{ \frac{ \gamma M }{R^2}}{\frac{ \gamma M }{(R+h)^2}}=\frac{(R+h)^2}{R^2}=5##
This will give you a quadratic equation in h to solve.

You should have gotten that 5 R^2 = (R+h)^2, since you are looking for the gravity to be 1/5 of what it is on earth.

##g_1/g_2 =\frac{ \frac{ \gamma M }{R^2}}{\frac{ \gamma M }{(R+h)^2}}=\frac{(R+h)^2}{R^2}=5##
This will give you a quadratic equation in h to solve.

Ahh I see now. I forgot about the 5 and didn't quite understand that concept. Thank you very much :)