Dimensional Analysis of Constant Acceleration at 1 g

  • Thread starter Gregor
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  • #1
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Newton's universal formula for calculating Gravity: Fg = G Mm / r²
which states that two objects attract each other with a force equal to the product of their masses
divided by the square of their separation,
times a constant of proportionality (G), where G = 6.6742(10) x 10^-11

makes it possible to calculate the mean acceleration due to gravity (g)
for any planet

in the case of Earth

g = G*Me / Re²

earth's radius (Re] = 6 378 137 m
this radius squared = 40 680 631 590 769
earth's mass (Me) = 5 977 347 368 296 994 158 859 250.7577676 kg
G = 0,0000000000667421 m³ kg^-1 s-²

applying these values to the above formula

5 977 347 368 296 994 158 859 250.7577676
/ 40 680 631 590 769
= 146 933 494 750.68959472357028022792
*0.0000000000667421
= 9.80665

therefore 1 g = 9.80665 m/s²


from this point a dimensional analysis can be made
on the premise that

d = integral of v dt over time interval
since gravity is constant acceleration

v = g*t

d = g*t²/2

as illustrated in following table

exact values from rest to 0.2 seconds - in 0.01 s intervals

Velocity m/s.........Time (s)..........Distance (m)

0 ..............................0 .................... 0
0.0980665 ................. 0.01 ............... 0.0004903325
0.196133 ................... 0.02 ............... 0.00196133
0.2941995 ................. 0.03 ............... 0.0044129925
0.392266 ................... 0.04 ............... 0.00784532
0.4903325 ................. 0.05 ............... 0.0122583125
0.588399 ................... 0.06 ............... 0.01765197
0.6864655 ................. 0.07 ............... 0.0240262925
0.784532 ................... 0.08 ............... 0.03138128
0.8825985 ................. 0.09 ............... 0.0397169325
0.980665 .................. 0.10 ................ 0.04903325

1.0787315 ................. 0.11 ............... 0.0593302325
1.176798 ................... 0.12 ............... 0.07060788
1.2748645 ................. 0.13 ............... 0.0828661925
1.372931 ................... 0.14 ............... 0.09610517
1.4709975 ................. 0.15 ............... 0.1103248125
1.569064 ................... 0.16 ............... 0.12552512
1.6671305 ................. 0.17 ............... 0.1417060925
1.765197.................... 0.18 ............... 0.15886773
1.8632635 ................. 0.19 ............... 0.1770100325
1.96133 .................... 0.20 ............... 0.196133




exact values from rest to 2 seconds - in 0.1 s intervals

Velocity m/s.........Time (s)..........Distance (m)

0 .............................0 ..................... 0
0.980665 .................. 0.1 .................. 0.04903325
1.96133 .................... 0.2 .................. 0.196133
2.941995 .................. 0.3 .................. 0.44129925
3.92266 .................... 0.4 .................. 0.784532
4.903325 .................. 0.5 .................. 1.22583125
5.88399 .................... 0.6 .................. 1.765197
6.864655 .................. 0.7 .................. 2.40262925
7.84532 .................... 0.8 .................. 3.138128
8.825985 .................. 0.9 .................. 3.97169325
9.80665 ................... 1 ..................... 4.903325

10.787315 ................ 1.1 .................. 5.93302325
11.76798 .................. 1.2 .................. 7.060788
12.748645 ................ 1.3 .................. 8.28661925
13.72931 .................. 1.4 .................. 9.610517
14.709975 ................ 1.5 .................. 11.03248125
15.69064 .................. 1.6 .................. 12.552512
16.671305 ................ 1.7 .................. 14.17060925
17.65197 .................. 1.8 .................. 15.886773
18.632635 ................ 1.9 .................. 17.70100325
19.6133 ................... 2 ..................... 19.6133


at 1 g

an object will (from rest) fall 1 m in 0.452 s ---- reaching a velocity of 4.4326058 m/s

an object will (from rest) reach a velocity of 1 m/s in 0.10197162129779282425700927431896 s
and will (during this interval) travel a distance of 0.050985810648896412128504637158521 m
 
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Answers and Replies

  • #2
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anyone have any criticisms of my calculations now? :rolleyes:
 
  • #3
Hootenanny
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And the purpose of these calculations is...?
 
  • #4
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ok, what was the point of moving this topic to Homework & Coursework Questions ?

this is neither Homework nor Coursework
it's a Classical Physics topic

And the purpose of these calculations is...?
a demonstration of my ability to perform Dimensional Analysis :cool:
 
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  • #5
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correct me if I'm wrong (I only just started a topic of motion in a horizontal circle yesterday) but wouldn't the centre directed accelleration due to an objects position on the earth be added to the normal gravitational force?...thus making results correct to 32 s.f. perhaps a bit premature? (i might be wrong and if I am I do apologise)...and even falling from rest for only 1m...I have a vague expectation that drag would also skew those results.
 
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  • #6
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Why don't you go buy a book on physics, you still seem to think [tex] ms^2 = \frac{m}{s^2} [/tex].
 
  • #7
Hootenanny
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GregA said:
correct me if I'm wrong (I only just started a topic of motion in a horizontal circle yesterday) but wouldn't the centre directed accelleration due to an objects position on the earth be added to the normal gravitational force?...thus making results correct to 32 s.f. perhaps a bit premature? (i might be wrong and if I am I do apologise)
Yes, you are quite correct. There are a number of reasons why it is inappropriate to quote to 32 sf. Primarly because the accuracy of some data is only given to 5 sf. Also, this theory assumes that the earth is spherical, which it isn't. Also, the apparent force of gravity is altered by the earth's rotation and significantly more by the movent of other massive bodies (moon and sun etc); in fact, these massive bodies can cause fluctuations of upto [itex]2\mu \;\; m\cdot s^{-1}[/itex] per day.

Sorry to put a bummer on your work Gregor, but apart from quoting your answers to a inappropriate number of sig figs you have obviously put a fair amount of work in. :smile:
 
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  • #8
Hootenanny
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cyrusabdollahi said:
Why don't you go buy a book on physics, you still seem to think [tex] ms^2 = \frac{m}{s^2} [/tex].
It's probably the fact that I need coffee, but I can't see where he has said that :confused:
 
  • #9
Doc Al
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Gregor said:
a demonstration of my ability to perform Dimensional Analysis
What does all that number crunching have to do with dimensional analysis?
 
  • #10
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Its from another thread. Put your seat belt on, because im sure its BOUND to come up again now that he has proved he can do dimensional analysis

you still seem to think[PLAIN]https://www.physicsforums.com/latex_images/93/938964-0.png[/QUOTE] [Broken]

it was you who posted the expression https://www.physicsforums.com/latex_images/93/938964-0.png [Broken]

which is meaningless

you need serious help


the dimensional analysis is represented in the table i calculated

if you take the time to read the table
you will find that it's absolutely correct

the point has been made, so from here on i will just sit back and watch you make a fool of yourself
 
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  • #11
Integral
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Gregor said:
ok, what was the point of moving this topic to Homework & Coursework Questions ?

this is neither Homework nor Coursework
it's a Classical Physics topic



a demonstration of my ability to perform Dimensional Analysis :cool:
I guess you still do not understand the meaning of dimensional analysis.

Dimensional analysis is MUCH simpler and is used to verify calculations without the use of ANY numbers.

Here is dimensional analysis for your first equation

[tex] g = \frac {G M_e} {R_e^2} [/tex]

[tex] g = \frac {m^3} {kg s^2} * \frac {kg} {m^2} [/tex]

now do basic algebra on the units to get:
[tex] g = \frac m {s^2}[/tex]

this is the correct units for acceleration, now that you know the basic relationships are correct you can proceed with your calculation using valid numbers for your quantities. Here is where you have thoroughly demonstrated a total lack of understanding of the concept of significant digits.
 
  • #12
Doc Al
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Gregor said:
the dimensional analysis is represented in the table i calculated
Open up one of your many physics books and look up what "dimensional analysis" means.

if you take the time to read the table
you will find that it's absolutely correct
What's your point? You just plugged numbers into an equation. Uh... why?

"earth's mass (Me) = 5 977 347 368 296 994 158 859 250.7577676 kg" :rofl:


the point has been made, so from here on i will just sit back and watch you make a fool of yourself
You've made no point.

Please don't start another of these silly threads.
 
  • #13
Hootenanny
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I cannot see any purpose to what you have done gregor. It looks like you've gone the long way round to nowhere. As Doc Al said, dimensional analysis isnt plugging numbers into equations, it looking at the dimensions of the equations.
 
  • #14
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i know what dimensional analysis is

the table was an illustration giving numerical proof


and what the hell do you find so funny about earth's mass?


are all mentor's in this forum narrowminded?
 
  • #15
chroot
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If you're actually measuring the Earth's mass down to the microgram, I'm afraid that you're ignorant.

The earth is constantly losing some mass (e.g. hydrogen) from the top of the atmosphere, while constantly gaining other mass (e.g. dust from meteors, etc.). The earth's mass varies on the order of several tons a day. So it's absolutely laughable to try to quote the Earth's mass to a precision of micrograms.

And it's very apparent, Gregor, that you actually do not know what dimensional analysis is. As has been said, once you start plugging in numerical values, you're by definition no longer doing dimensional analysis.

Please grow up and recognize the limits of your own understanding.

- Warren
 
  • #16
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If you're actually measuring the Earth's mass down to the microgram, I'm afraid that you're ignorant.
the precise value of earth's mass (for those nearsighted thwackheads out there in physics lalaland) may be derived by Newton's Gravitational Formula

Fg = G Mm / r²

since the value of Re is precisely known to be 6 378 137 m
which squared = 40 680 631 590 769
and the value of G is a precise value : 0,0000000000667421
as well as g : 9.80665 m/s²
therefore Me is precisely defined by default as 5 977 347 368 296 994 158 859 250.7577676 kg

More insults deleted
Integral
 
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  • #17
chroot
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Gregor said:
the fact that you couldn't think this out on your own
proves that you couldn't calculate your way out of a public toilet :rolleyes:
:rofl: Where to begin? Where to begin?

The value of G is only known to about six significant figures (you posted it, count them yourself!). This means your calculation of the mass of the earth is also only accurate to about six significant figures. This means your assertion of the mass of the earth in terms of micrograms is entirely meaningless. It's effectively nothing more than "measurement noise."

The most precise value for Me that you could quote, using your own numbers, is 5.97734 x 1024 kg. :rofl:

Are you just a troll, or what?

- Warren
 
  • #18
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the end result must equal the known, measured value for g

this is the only way we can measure earth's mass

Edit:Enough with the insults

Integral

tell that to your Admin....

Chroot - Are you just a troll, or what?
and these are grown up people?
 
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  • #19
chroot
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You've had this explained to you numerous times Gregor. If you still wish to discuss the topic, take it to another website.

- Warren
 

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