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Posted (edited)

Sir Isaac Newton's law of gravity deduces or measures acceleration due to gravity constant as g=9.8 m/sec square which we apply universally in other equations..

How this was measured or deduced?

I mean why not 10 m/sec square Or 9.5 m/sec square or any other value ?.  

 

Thanks & Regards,

Prashant S Akerkar

 

Edited by prashantakerkar
Content
Posted (edited)

Thanks.

From how much distance : 1st floor, 2nd floor, 3rd floor ...etc in meters or feet ?

What's the reference point i.e. Height / Altitude?

Thanks & Regards,

Prashant S Akerkar

Edited by prashantakerkar
Content
Posted (edited)
1 hour ago, prashantakerkar said:

Sir Isaac Newton's law of gravity deduces or measures acceleration due to gravity constant as g=9.8 m/sec square which we apply universally in other equations..

g=9.81 m/s^2 is not real constant. It's variable. Depends on distance from the source of gravitation of this planet, the Earth center. The further from the center, the smaller gravitation.

So, altitude does matter. On high mountain you have different gravitation, than on sea level beach, and different when you're flying in airplane 10 km above the ground..

[math]F(r)=\frac{GMm}{r^2}[/math]

[math]F=m a[/math]

[math]a(r)=\frac{GM}{r^2}[/math]

There are also local variations of gravitation caused by presence of e.g. heavy metal ores in the ground, and therefore they can be detected.

Read article:

https://en.wikipedia.org/wiki/Gravity_of_Earth

"Effective gravity on the Earth's surface varies by around 0.7%, from 9.7639 m/s2 on the Nevado Huascarán mountain in Peru to 9.8337 m/s2 at the surface of the Arctic Ocean.[4] In large cities, it ranges from 9.7760[5] in Kuala Lumpur, Mexico City, and Singapore to 9.825 in Oslo and Helsinki."

1 hour ago, prashantakerkar said:

which we apply universally in other equations..

Only in very basic calculations.

 

Edited by Sensei
Posted (edited)
47 minutes ago, prashantakerkar said:

From how much distance : 1st floor, 2nd floor, 3rd floor ...etc in meters or feet ?

What's the reference point i.e. Height / Altitude?

Distance: something you can measure accurately. Obviously in meter if you use m/s^2

Reference point you determine yourself. Usually where the fall starts.

I have used Tracker to have students calculate g with nothing but a smartphone camera and a ruler to surprising accuracy. To avoid blurry images, you can also roll a marble down a sloped rail and apply some trigonometry to find g.

Edited by Bender
Posted
4 minutes ago, prashantakerkar said:

That means, We can calculate g for all eight planets ?

Yes, we can calculate it- which means you can calculate it.

You can also google it.

Posted
1 hour ago, prashantakerkar said:

What's the reference point i.e. Height / Altitude?

I am going to give you +1 for thinking for yourself.

This discussion shows much improvement on your earlier ones.

:)

 

 

Sensei has told you some important things about the variation of gravity,

One very important conclusion about this it that it doesn't only matter what value (magnitude) you measure for gravity, that value is worthless without stating where it was measured.
This means you also have to have sufficient positioning capability to make use of the variation.

Over Earth the variation is very small so when you look at gravity survey maps they are shown in 'milligals' or even 'microgals' where 1 gal is about 1 millionth of standard g.

There is also a slight variation over the Earth in the direction  that the acceleration points in, compared to the perpendicular to the 'Earth's surface at that point.

In other words g is not perfectly perpendicular and surveyors have to take this variation of the vertical into account.

Posted
1 hour ago, prashantakerkar said:

What's the reference point i.e. Height / Altitude?

The simple answer to that is "sea level"

https://en.wikipedia.org/wiki/Sea_level

But even at the tops of the highest mountains, it doesn't change much.

But it does mean that sensitive analytical balances in laboratories  need recalibrating if you move then up and down stairs.

Posted
4 hours ago, prashantakerkar said:

Thanks.

From how much distance : 1st floor, 2nd floor, 3rd floor ...etc in meters or feet ?

What's the reference point i.e. Height / Altitude?

Thanks & Regards,

Prashant S Akerkar

If you want g in m/s^2, you'd better be using meters.

The reference points are whatever you choose in your experiment. 

3 hours ago, studiot said:

Over Earth the variation is very small so when you look at gravity survey maps they are shown in 'milligals' or even 'microgals' where 1 gal is about 1 millionth of standard g.

1 gal is 1 cm/s^2, meaning it is approximately a thousandth of a "g"

Posted
24 minutes ago, swansont said:
3 hours ago, studiot said:

Over Earth the variation is very small so when you look at gravity survey maps they are shown in 'milligals' or even 'microgals' where 1 gal is about 1 millionth of standard g.

1 gal is 1 cm/s^2, meaning it is approximately a thousandth of a "g"

Oops, how did that slip in

You are dead right on thousandth of standard g

Posted
2 hours ago, studiot said:

Oops, how did that slip in

You are dead right on thousandth of standard g

What's three orders of magnitude between friends?

Posted (edited)

Of course, the specific value depends upon the units you are using.  In the "mks", "meter-kilogram-second", system, the acceleration due to gravity, g, is, on the surface of the earth, approximately 9.8 meters per second squared. In "cgs", "centimeter, gram, second, it is 980 centimeters per second squared, and in the "English" system, it is 32.2 feet per second squared.  That number is only  an average- even on the surface of the earth it varies from place to place, largely changing with altitude but there are places inside the earth that are denser than others so even at the same altitude g may vary slightly.  And, of course, on different planets, with different masses and radii, g is different.

You may be confusing "g" with "Newton's universal gravitational constant, "G", used in his formula, [math]F= \frac{GMm}{r^2}[/math],  for the gravitational force between two objects of masses M and m with distance r between their centers.   If a mass m is on a planet with mass M and distance r from the center of the planet, then it feels gravitational force $F= \frac{GMm}{r^2}$ and, since "F= ma", would fall with acceleration [tex]g= a= \frac{GM}{r^2}[/tex].  The mass of the earth and the average radius of the earth is such that that works out to g= 9.8 meters per second squared.

Edited by Country Boy

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