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The height at which the gravity acceleration is 5x smaller

  • Thread starter Amar
  • Start date
12
0
1. 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

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

3. 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.
 

NascentOxygen

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Welcome to Physics Forums.

Your approach sounds right. Can you show the steps in your calculations so we can look for errors?
 
12
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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

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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

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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

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9,215
1,056
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.
 
12
0
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
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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.
 
12
0
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 :)
 

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