Acceleration Due to Gravity on earth

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Homework Help Overview

The discussion revolves around determining the altitude above the Earth's surface where the acceleration due to gravity equals one-third of its value at sea level. The subject area includes gravitational physics and mathematical reasoning related to the equations governing gravitational acceleration.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the gravitational acceleration formula and the calculations involved in solving for altitude. Questions arise regarding the values used for the radius of the Earth and the units of the final answer.

Discussion Status

Some participants have provided calculations and attempted to clarify the steps taken. There is an acknowledgment of potential errors in calculation, and guidance has been offered to ensure proper unit usage and clarity in the derivation process. Multiple interpretations of the problem and its setup are being explored.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or receive. There is a focus on ensuring that all calculations are correctly presented with appropriate units.

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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