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May 29, 2020, 10:08:53 pm

"There is nothing either good or bad, but thinking makes it so." ..."Shakespeare (154-1616, English dramatist and poet) "

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 Author Topic: Newton's law of gravitation/board:26-3-  (Read 10971 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
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 « Embed this message on: September 29, 2009, 05:01:34 pm »

Can I request for simulation that show the objective of theoretical predictions from Newton's law of gravitation ( recall and use Newton's law of gravitation in the form F = G(m1m2)/r2), please ? Thank you
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Fu-Kwun Hwang
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 « Embed this message Reply #1 on: September 29, 2009, 05:22:14 pm » posted from:Taipei,T'ai-pei,Taiwan

Do you want it to be a 2D or 3D simulation?
What parameters you want to be able to be changed in the simulation?
m1,m2 and r ?

The gravitational constant $G=6.674 \times 10^{-11} N (m/kg)^2$ is a very small number.
Do you want the force be calculated with the above formula?
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 « Embed this message Reply #2 on: September 29, 2009, 05:41:08 pm »

I want it to be in 3D. Can it be all m1, m2 & r? Yes, I want the force to be calculated using the formula. Thank you
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 « Embed this message Reply #3 on: September 29, 2009, 08:21:11 pm » posted from:Taipei,T\'ai-pei,Taiwan

What is the unit for the length? for example: a simulation with 640x480 pixels were created.
What is the distance from left to right? 640km? 640 m? 640mm?
If the mass is in unit of kg, then the value of the force will be very small.
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 « Embed this message Reply #4 on: September 29, 2009, 11:57:39 pm » posted from:Taipei,T\'ai-pei,Taiwan

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i want the length to be in km & mass to be in g.

As you wish, the value has been calculated. However, the force is very small for your cases!
You can use mouse to drag charge particles or use sliders to change their location!

Actually, you can calculate it with a calculator easily by yourself,too.
Just calculate the distance between two charge particles, then calculate the force between them.

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Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!
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 « Embed this message Reply #5 on: October 02, 2009, 10:48:14 pm » posted from:Dallas,Texas,United States

Can this simulation used to analyse circular orbits in inverse square law fields by relating the gravitational force to the centripetal acceleration it causes? Thank you
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 « Embed this message Reply #6 on: October 03, 2009, 08:51:21 am » posted from:Taipei,T\'ai-pei,Taiwan

The condition for circular motion is $F=m\frac{v^2}{r}$ where $\vec{v}$ is the tangential velocity.
And if the force is provided from gravitation force $F=\frac{GmM}{r^2}$,
it impliy that $v=\sqrt{\frac{GM}{r}}$
Or $m\frac{v^2}{r}=m\frac{(2\pi r/T)^2}{r}=m \frac{4\pi^2 r}{T^2}=\frac{GmM}{r^2}$
so $\frac{r^3}{T^2}=\frac{GM}{4\pi}$ which is a constant.

The trajectory of planet motion could be circular only if the above condition is satisfied, otherwise you might find an ellipse if the total energy is less than zero (closed orbit).

Please check out the following related simulations at this web site:
planet motion for two stars
Circular motion: acceleration always perpendicular to velocity
How to determined the trajectory of a planetary motion?
motion of Moon, Earth relative to the Sun
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"There is nothing either good or bad, but thinking makes it so." ..."Shakespeare (154-1616, English dramatist and poet) "