Acceleration Due to Gravity on earth

AI Thread Summary
To find the altitude where the acceleration due to gravity equals g/3, the equation ag = G Me/(Re + H)^2 is used. The user derived H = √3 Re - Re, leading to a calculated altitude of 17*10^6 meters. However, this result was questioned regarding its correctness and units. The radius of Earth (Re) was confirmed as 6.37*10^6 meters, and the user was prompted to check their calculations for possible errors. Accurate calculations are essential for determining the correct altitude above Earth's surface.
eagles12
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Homework Statement


At what altitude above the Earth's surface is the acceleration due to gravity equal to g/3?



Homework Equations



ag=G Me/(Re+H)^2

The Attempt at a Solution



I solved for H and got H=√3 Re-Re
using this i got 17*10^6
but it is saying this is not correct
 
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eagles12 said:

Homework Statement


At what altitude above the Earth's surface is the acceleration due to gravity equal to g/3?



Homework Equations



ag=G Me/(Re+H)^2

The Attempt at a Solution



I solved for H and got H=√3 Re-Re
using this i got 17*10^6
but it is saying this is not correct
Can you show the details of your calculation? What value did you use for Re?
 
eagles12 said:
I solved for H and got H=√3 Re-Re
using this i got 17*10^6
17*10^6 what? Distance isn't just a number. It has units. What are the units of this number 17*10^6, what units is the answer supposed to be in, and what value did you use for the radius of the Earth?
 
for re i used 6.37*10^6m
the answer is also sopossed to be in meters

i started with ag=g Me/(Re+H)^2
then plugged in 1/3*9.8=G Me/(Re+H)^2
I reduced that to ag=1/(Re+H)^2=1/3(1/Re^2)
from there i got
(Re+H)^2=3Re^2
Re+H=sqrt(3)*Re
H=sqrt(3)Re-Re
 
Your derivation looks okay, so you must be having finger problems with the calculator :smile:
 
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