What is the value of acceleration due to gravity

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SUMMARY

The maximum height at which the acceleration due to gravity (g) remains approximately 9.8 m/s² is around 32 kilometers, as gravity decreases by about 1% for every 200 kilometers of Earth's radius. The discussion emphasizes the need for a defined range of acceptable deviation when considering the constancy of g, with participants suggesting variations of 1% to 10% as thresholds. The calculation of g at various heights can be approximated using the formula g' = g(1 - 2h/r), where r is the Earth's radius. The conversation highlights the importance of specifying acceptable accuracy in gravitational calculations.

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parshyaa
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Upto what maximum height, value of g remains equal to 9.8 or close to it.
  • g = GM/r^2
  • And g' = GM/(r+h)^2 = g/(1+ h/r)^2
  • Therefore when h ≥ r , g' decreases
  • Again if h < r , g' agin decreases
  • BUT I want to know what is the maximum height at which g' ≈ 9.8
  • From my calculation maximum height ≈ 19.3 km, I think I am wrong, so what is the corrrct value of maximum height.
 
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What do you consider a strong enough deviation? 1%?
 
DrClaude said:
What do you consider a strong enough deviation? 1%?
I want to know upto what height the value of g remains unchange.
 
parshyaa said:
I want to know upto what height the value of g remains unchange.

it doesn't remain unchanged, it's constantly changing with height
hence @DrClaude 's question which you didn't answer
 
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It might be useful to do a linear approximation to the function ## g'(h)=g \cdot (1+\frac {h}{r})^{-2} ##
Then you would get ##g'(h) \approx g'(0) + \frac {d(g'(0))}{dh} \cdot h ## Sorry for the convoluted notation.
this works out to be
## g' \approx g(1-\frac{2h}{r}) ## Or in other words gravity decreases about 1 percent every ## \frac {1}{200}## of Earth's radius (which is roughly 32 kilometers).
EDIT: The approximation diverges pretty quickly but it gives at least some idea about the change in gravity with height.
 
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Tazerfish said:
It might be useful to do a linear approximation to the function ## g'(h)=g \cdot (1+\frac {h}{r})^{-2} ##
Then you would get ##g'(h) \approx g'(0) + \frac {d(g'(0))}{dh} \cdot h ## Sorry for the convoluted notation.
this works out to be
## g' \approx g(1-\frac{2h}{r}) ## Or in other words gravity decreases about 1 percent every ## \frac {1}{200}## of Earth's radius (which is roughly 32 kilometers).
EDIT: The approximation diverges pretty quickly but it gives at least some idea about the change in gravity with height.
Thanks tazerfish
 
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parshyaa said:
Thanks tazerfish

This is puzzling. How did that answer your question, when you were insistent on finding the height where "... the value of g remains unchange (sic)..."? tazerfish answer was exactly what DrClaude was asking you about, and you never did fully address.

Or are you even aware of what just happened here?

Zz.
 
Actually I didn't understand what Dr.cloude asked. my question is simple , upto which height their is not much deviation in value of g , it remains between 9.8 to 10 . So what is the height
 
Value of g remains in between 9.8 and 10
parshyaa said:
Actually I didn't understand what Dr.cloude asked. my question is simple , upto which height their is not much deviation in value of g , it remains between 9.8 to 10 . So what is the height
 
  • #10
parshyaa said:
Actually I didn't understand what Dr.cloude asked. my question is simple , upto which height their is not much deviation in value of g , it remains between 9.8 to 10 . So what is the height

Well then me explain. YOU need to specify the RANGE of values where you consider that it has not changed! Some people think a variation of 10% is sufficient. Some 1%. And if you are doing LIGO experiment, the variation can be orders of magnitude smaller!

Can't you even see what's going on here? You want something within some accuracy, but YOU never specified the level of accuracy that you are looking for!

Why is this so difficult to comprehend?

Zz.
 
  • #11
Just now I told you that the value should be inbetween 9.8 and 10

ZapperZ said:
Well then me explain. YOU need to specify the RANGE of values where you consider that it has not changed! Some people think a variation of 10% is sufficient. Some 1%. And if you are doing LIGO experiment, the variation can be orders of magnitude smaller!

Can't you even see what's going on here? You want something within some accuracy, but YOU never specified the level of accuracy that you are looking for!

Why is this so difficult to comprehend?

Zz.
toldto
 
  • #12
parshyaa said:
Just now I told you that the value should be inbetween 9.8 and 10 toldto

That makes even LESS sense. As you go higher in height, the value of "g" drops! So if you are insisting that the value stays betweeen 9.8 and 10 (btw, my students get penalized for not writing down units) m/s^2, then you'll never find such a value, will you?

Why can't you just specify a percentage drop of when you consider to be within your acceptable range?

Zz.
 
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  • #13
Suppose I am at the top of burj khalifa , it is at a height of 830 m , the value of g acting on me is approximately equals 9.7679 ≈ 9.8 , I want to know upto what such heights, the value of g remains closer to 9.8, suppose some budy asked a question that what is the force of attraction of Earth on a object which is at a height of 200 km or 300 km or etc etc but not much larger from the Earth surface.Then upto what value of h , without using the Newtons gravitational formula I can directly take g ≈ 10. And find the force by multiplying it to the mass of a object.
 
  • #14
parshyaa said:
Suppose I am at the top of burj khalifa , it is at a height of 830 m , the value of g acting on me is approximately equals 9.7679 ≈ 9.8 , I want to know upto what such heights, the value of g remains closer to 9.8, suppose some budy asked a question that what is the force of attraction of Earth on a object which is at a height of 200 km or 300 km or etc etc but not much larger from the Earth surface.Then upto what value of h , without using the Newtons gravitational formula I can directly take g ≈ 10. And find the force by multiplying it to the mass of a object.
Just use the equation from Newton's theory of gravitation:

https://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation

It's the equation part-way down the page under "Modern" version. The distance r that you use is measured from the center of the Earth, so be sure to add that distance to the distance your object is above the surface of the Earth.
 
  • #15
parshyaa said:
Suppose I am at the top of burj khalifa , it is at a height of 830 m , the value of g acting on me is approximately equals 9.7679 ≈ 9.8 , I want to know upto what such heights, the value of g remains closer to 9.8, suppose some budy asked a question that what is the force of attraction of Earth on a object which is at a height of 200 km or 300 km or etc etc but not much larger from the Earth surface.Then upto what value of h , without using the Newtons gravitational formula I can directly take g ≈ 10. And find the force by multiplying it to the mass of a object.
So 9.7679 is acceptably close to 9.8 for your purposes.
Would 9.7402 be unacceptably low for your purposes?
 
  • #16
Why don't you substitute the value which you deem to be the first not acceptable answer for gravity, and work it backwards to get your distance?
 
  • #17
jbriggs444 said:
So 9.7679 is acceptably close to 9.8 for your purposes.
Would 9.7402 be unacceptably low for your purposes?
Yes value between 9.6 to10 will be accepetable
 
  • #18
parshyaa said:
Yes value between 9.6 to10 will be accepetable

you didn't read ZZ's post #12 did you ?
 
  • #19
How long is a piece of string?

Can a mentor please put this thread out of its misery?
 
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  • #20
Vanadium 50 said:
Can a mentor please put this thread out of its misery?
Yes.

@parshyaa, please reread this thread carefully and think about it. Why was your request incomplete?
 

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