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Title: Newton's law of gravitation/board:26-3-Post by: Rare 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 :D
Title: Re: Newton's law of gravitation/board:26-3-Post by: Fu-Kwun Hwang on September 29, 2009, 05:22:14 pm
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? Title: Re: Newton's law of gravitation/board:26-3-Post by: Rare 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
Title: Re: Newton's law of gravitation/board:26-3-Post by: Fu-Kwun Hwang on September 29, 2009, 08:21:11 pm
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. Title: Re: Newton's law of gravitation/board:26-3-Post by: Fu-Kwun Hwang on September 29, 2009, 11:57:39 pm
Quote 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. Title: Re: Newton's law of gravitation/board:26-3-Post by: Rare on October 02, 2009, 10:48:14 pm
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
Title: Re: Newton's law of gravitation/board:26-3-Post by: Fu-Kwun Hwang on October 03, 2009, 08:51:21 am
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 (http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=766.0) Circular motion: acceleration always perpendicular to velocity (http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=693.0) How to determined the trajectory of a planetary motion? (http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=629.0) motion of Moon, Earth relative to the Sun (http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=547.0) |