Deviation of Newton's Gravity Equations

In summary, Daniel explains that the time equals 2 times the distance over the acceleration of gravity on Earth. The distance is multiplied by 2 because that is the integral of a constant acceleration. The equation is only valid when the initial velocity is zero.
  • #1
eNathan
352
2
[tex]T = \sqrt { \frac{2d} {g} }[/tex]
Therefore (and I had to come up with this on my own because I could not find it on the internet :smile:)
[tex]D = \frac{t^2 * g} {2}[/tex]

But I cannot find the logic in why the time equals 2 times the distance over the acceleration of gravity on earth. Now, I do understand the units. The Time is in seconds, and the Distance is in meters, because the acceleration is in meters per second (any other units will work just fine).

But why is the distance multiplied by 2?
And why is the square root function used?
 
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  • #3
hehe, I only wish I did. I am researching it in my spare time, and I plan to take classes on it but in shorts no I don't know calc.

But it won't hurt if you try to explained it to me anyway if that was your plan :)
 
  • #4
To integrate Newton's second law in scalar form

[tex] \frac{d^{2}y(t)}{dt^{2}}=g [/tex]

subject to the initial conditions

[tex] y\left(t_{0}\right)=y_{0} [/tex]

[tex]\frac{dy}{dt}\left(t_{0})=:v\left(t_{0}\right)=v_{0} [/tex]


Daniel.
 
  • #5
More simply, I'm sure you know the position equation for an object in constant acceleration:

[tex] x = x_0 + v_0t + \frac{gt^2}{2} [/tex]

Notice if [itex] x_0 = v_0 = 0 [/tex] then solving for t:

[tex] t = \sqrt{\frac{2x}{g}} [/tex]

Your equation is only valid when the initial velocity is zero.
 
  • #6
whozum said:
More simply, I'm sure you know the position equation for an object in constant acceleration:

[tex] x = x_0 + v_0t + \frac{gt^2}{2} [/tex]

Notice if [itex] x_0 = v_0 = 0 [/tex] then solving for t:

[tex] t = \sqrt{\frac{2x}{g}} [/tex]

Your equation is only valid when the initial velocity is zero.

[tex] x = x_0 + v_0t + \frac{gt^2}{2} [/tex]

Hmn, I think I get your point, but why do you square numbers all the time anyway? I mean, I see it everywhere. e=mc^2 ^2 ^2 blah every equation in physics involves [tex]x^2[/tex] :smile: Maybe if you explain why everything need be squared, I can understand it :smile:

Thx
 
  • #7
eNathan said:
[tex]T = \sqrt { \frac{2d} {g} }[/tex]
Therefore (and I had to come up with this on my own because I could not find it on the internet :smile:)
[tex]D = \frac{t^2 * g} {2}[/tex]

But I cannot find the logic in why the time equals 2 times the distance over the acceleration of gravity on earth. Now, I do understand the units. The Time is in seconds, and the Distance is in meters, because the acceleration is in meters per second (any other units will work just fine).
Hope it is a type, and you use meters per second squared for the acceleration. Also, this thread should be titled "Deduction (or derivation) of Galilean law for free falling bodies"; Newton has no role here except to confirm us that acceleration g is approximately constant at Earth surface.

Indeed the original proof of the time square formula can be found in internet :biggrin: here:
http://www.mpiwg-berlin.mpg.de/Galileo_Prototype/DHTML/D202.HTM
 
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  • #8
arivero said:
Indeed the original proof of the time square formula can be found in internet ::biggrin:: here:
http://www.mpiwg-berlin.mpg.de/Galileo_Prototype/DHTML/D202.HTM
This requires a bit of history. In order to get students, Galileo did not publish the parabolic trajectory until Cavalieri did (rightly attributing it to master Galileo). Then he become furious for a moment but finally he thought better and he finished his http://oll.libertyfund.org/ToC/0416.php in a notebook of Galileo, mss 72, probably used during the composition of the book.
 
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  • #9
eNathan said:
[tex] x = x_0 + v_0t + \frac{gt^2}{2} [/tex]

Hmn, I think I get your point, but why do you square numbers all the time anyway? I mean, I see it everywhere. e=mc^2 ^2 ^2 blah every equation in physics involves [tex]x^2[/tex] :smile: Maybe if you explain why everything need be squared, I can understand it :smile:

Thx

The reason is calculus, and definitions of acceleration, velocity, and position.

Acceleration is the rate of change of velocity:

[itex] a = \frac{dv}{dt} [/tex] so then [itex] \int{a}{dt} = \int{dv} [/tex]

Integral of a constant acceleration "a" is just 'a' times the dependant variable, t. v = at+v_0. v_0 is the integration constant.

[itex] v = at+v_0 [/itex]

Velocity is the rate of change of position:

[itex] v = at+v_0 = \frac{dx}{dt} [/itex] so then [itex] x = \int{(at+v_0)}{dt} [/tex]

Integral of a linear function of time 'at' is 'at^2/2':

[tex] x = \int{at+v_0}{dt} = v_0t+\frac{at^2}{2} + x_0 [/tex]
 

Related to Deviation of Newton's Gravity Equations

1. What is Newton's law of gravity?

Newton's law of gravity is a fundamental principle in physics that describes the force of attraction between two masses. It states that the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

2. What is the deviation of Newton's gravity equations?

The deviation of Newton's gravity equations refers to discrepancies or variations from the predicted behavior of gravity based on Newton's law. These deviations can occur when dealing with very large masses, very small distances, or extremely high speeds, and are better explained by Einstein's theory of general relativity.

3. How does Einstein's theory of general relativity explain the deviation of Newton's gravity equations?

Einstein's theory of general relativity proposes that gravity is not a force between masses, but rather a curvature of spacetime caused by the presence of mass. This explains the deviations observed in Newton's law, as it accounts for the effects of extremely large masses and high speeds on the curvature of spacetime.

4. Are there any experimental evidences of the deviation of Newton's gravity equations?

Yes, there have been numerous experimental observations that support the deviation of Newton's gravity equations. One of the most famous examples is the precession of the orbit of Mercury, which could not be explained by Newton's law but was accurately predicted by Einstein's theory of general relativity.

5. How does the deviation of Newton's gravity equations impact our understanding of the universe?

The deviation of Newton's gravity equations highlights the limitations of Newton's law and the need for a more comprehensive theory of gravity. It has led to the development of Einstein's theory of general relativity, which has greatly improved our understanding of the universe and has been crucial in many important discoveries and innovations in modern physics.

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