What is the value of acceleration due to gravity

In summary: 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!
  • #1
parshyaa
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19
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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  • #2
What do you consider a strong enough deviation? 1%?
 
  • #3
DrClaude said:
What do you consider a strong enough deviation? 1%?
I want to know upto what height the value of g remains unchange.
 
  • #4
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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  • #5
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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  • #6
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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  • #7
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.
 
  • #8
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
 
  • #9
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?
 

1. What is the value of acceleration due to gravity?

The value of acceleration due to gravity is approximately 9.8 meters per second squared (m/s^2). This value is commonly denoted by the letter "g" and represents the acceleration of an object in free fall near the Earth's surface.

2. How is acceleration due to gravity calculated?

Acceleration due to gravity can be calculated by dividing the force of gravity acting on an object by its mass. The formula for acceleration due to gravity is g = F/m, where g is the acceleration, F is the force of gravity, and m is the mass.

3. What factors affect the value of acceleration due to gravity?

The value of acceleration due to gravity is affected by the mass and distance between two objects. The greater the mass of the objects, the stronger the force of gravity and the greater the acceleration. The farther apart the objects are, the weaker the force of gravity and the smaller the acceleration.

4. Does acceleration due to gravity vary on different planets?

Yes, acceleration due to gravity can vary on different planets. This is because the mass and distance of each planet's surface differ, resulting in different values of acceleration due to gravity. For example, the acceleration due to gravity on Mars is approximately 3.7 m/s^2, while on Jupiter it is approximately 24.8 m/s^2.

5. How does air resistance affect the value of acceleration due to gravity?

Air resistance can affect the value of acceleration due to gravity by slowing down the acceleration of an object in free fall. This is because air resistance creates a force that opposes the force of gravity, resulting in a lower acceleration. However, this effect is more noticeable for objects with a larger surface area, such as a feather, compared to objects with a smaller surface area, such as a bowling ball.

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