Question

Show how the Kepler's third planetary motion led Newton to establish the inverse-square rule for gravitation.

Answer 1

@hartnn @Algebraic! @ajprincess @experimentX @experimentX

Answer 2

oops tagged experiment x two times sorry :D

Answer 3

@eyust707

Answer 4

well I think I should do something for u all ..

should I draw a diagram ?

Answer 5

LOL this is easy.

Answer 6

Consider the following image.

Answer 7

ok

Answer 8

https://dl.dropbox.com/u/63664351/Inverse%20Square%20Law.PNG
The lines represent the flux emanating from the source. The total number of flux lines depends on the strength of the source and is constant with increasing distance. A greater density of flux lines (lines per unit area) means a stronger field.

Since the surface area of a sphere (which is 4πr2 ) is proportional to the square of the radius, as the emitted radiation gets farther from the source, it is spread out over an area that is increasing in proportion to the square of the distance from the source. Hence, the intensity of radiation passing through any unit area (directly facing the point source) is inversely proportional to the square of the distance from the point source.

Answer 9

not exactly what I wanted @twitter

do you know the eqn for that?

Answer 10

\[\large{\frac{r^3}{T^2} \textbf{is constant}}\]

Answer 11

indirectly you have to prove that...

Answer 12

Intensity = C/r^2 ; where C is constant and Intensity is the gravitation?

Oops. Looks like I give you the most fundamental one.
To get that equation, you just need to manipulate the algebra.

Answer 13

or simply let the question be :

prove that : \[\large{\frac{r^3}{T^2}\quad \text{is constant}}\]

Answer 14

From Kepler equation? Can you give me that eqn. I forget.

Answer 15

\[\large{F=\frac{mv^2}{R}}\]

or
\[\large{F \propto \frac{v^2}{R}}\]

Answer 16

But seriously the imagine I give you before is really fundamental.
You can even derive electricity, light, electromagnetic radiation and even gravitation F = Gm1m2/R^2 simply from there!

Answer 17

since \(\large {v = \frac{2\pi R}{T}}\)

Answer 18

\[F=\frac{GMm}{r^2} \\ \\ \text{Due \to planets rotating around the sun, }F=\frac{mv^2}{r} \\ \\ \frac{mv^2}{r}=\frac{GMm}{r^2} \\ \\ \frac{\cancel{m}v^2}{\cancel{r}}=\frac{GM\cancel{m}}{r^\cancel{2}} \\ \\ \text{Rotational period, } v=\frac{2 \pi r}{T} \]
\[\frac{4\pi^2 r^2}{T^2}=\frac{GM}{r} \\ \\ r^3=\frac{GM}{4 \pi^2}T^2 \\ \\ \frac{GM}{4\pi^2}is~constant, \\ \\ \text{Therefore, } kT^2=r^3\]

Answer 19

yeah .. this concept is just conversation of gravitational flux
i seem to have forgotten how to arrive at this from kepler's third law.

Answer 20

hence :
\[\large{v \propto \frac{R}{T}}\]
\[\large{v^2\propto \frac{R^3}{T^2} \times \frac{1}{R}}\]

Answer 21

wow 2 proves ...
we can also prove that : \[\large{F \propto \frac{1}{R^2}}\]

Answer 22

Thanks @.Sam. :)

Answer 23

Oops someone done it first

Answer 24

@experimentX you don't arrive to this from Kepler Law.
You derive this to Kepler Law, and other LAWS that have r^2 in it.

Answer 25

i remember it being very simple ... within in 2-3 steps.

Answer 26

Really? I get it using pure logic and calculus.
From the function of circle to get the Area of a sphere.
Voila.

Answer 27

well probably the same way as .sam did ... just manipulation ... ellipses make things uneasy

Answer 28

um. my i know your current education level?
I meant you derive this to get to other fundamental laws in physics. But for high school you don't need to know this.

Answer 29

i'm out of university.

Answer 30

i always wondered how kepler got things without calculus ahead of newton.