7.3.8 Relationship between F and U; between g and φ JavaScript HTML5 Applet Simulation Model
 Details
 Parent Category: 02 Newtonian Mechanics
 Category: 08 Gravity
 Created: Thursday, 20 August 2015 15:00
 Last Updated: Thursday, 25 May 2017 15:35
 Published: Friday, 08 April 2016 15:00
 Hits: 5014
About
7.3.8 Relationship between F and U; between g and ϕ
To understand how g is related to ϕ:
 Similarly, compare $g=\genfrac{}{}{0.1ex}{}{GM}{{r}^{2}}$ and $\text{\varphi}=\frac{\text{GM}}{r}$ in the above table.
 If we differentiate $\text{\varphi}=\frac{\text{GM}}{r}$ with respect to r, we will get $\genfrac{}{}{0.1ex}{}{d\varphi}{dr}=\genfrac{}{}{0.1ex}{}{GM}{({r}^{2})}$, which has the same expression as g.
 Hence, mathematically $\genfrac{}{}{0.1ex}{}{d\varphi}{dr}=\genfrac{}{}{0.1ex}{}{GM}{{r}^{2}}=g$
 To understand the meaning of $g=\genfrac{}{}{0.1ex}{}{d\varphi}{dr}$ observe the two graphs carefully, on the right side where r is positive, the gradient of ϕ vs r graph is positive but the value of g will be negative. And on the left side where r is negative, the gradient of ϕ vs r graph is negative but the value of g is positive. Thus, $g=\genfrac{}{}{0.1ex}{}{d\varphi}{dr}$
Similarly, it can be concluded that by multiplying both sides by test mass m.
$mg=\genfrac{}{}{0.1ex}{}{dm\varphi}{dr}$
thus
$F=\genfrac{}{}{0.1ex}{}{dm\varphi}{dr}$
7.3.8.1 Activity To do
ICT inquiry worksheet 1 (E), as well as the "G field and potential" EJS here. The HTML5 version is below, the Java worksheet customization to HTML5 is work in progress.7.3.8.2 Summary
symbol  $g=\genfrac{}{}{0.1ex}{}{GM}{{r}^{2}}$  $\text{\varphi}=\frac{\text{GM}}{r}$ 
name  Field strength  Potential 
units  N kg^{1} or m s^{2}  J kg^{1} 
meaning  Gravitational force per unit mass  Gravitational potential energy per unit mass 
quantity  vector  scalar 
equation  $\mathrm{g}=\genfrac{}{}{0.1ex}{}{GM}{{r}^{2}}$
towards the centre of the source mass 
$\text{\varphi}=\frac{\text{GM}}{r}$ 
relationship to mass  Force, $F=\frac{G{M}_{1}{M}_{2}}{{r}^{2}}$ = mg  Potential energy, $U=m\genfrac{}{}{0.1ex}{}{GM}{r}$ = mϕ 
graph  
computer model if M = 500.  6.67*500/(abs(r)*r)  6.67*500/abs(r) 
relationship between g and ϕ 
$g=\genfrac{}{}{0.1ex}{}{d\varphi}{dr}$


relationship between F and U  $F=\genfrac{}{}{0.1ex}{}{dU}{dr}$ 
7.3.8.3 Model
Translations
Code  Language  Translator  Run  

Software Requirements
Android  iOS  Windows  MacOS  
with best with  Chrome  Chrome  Chrome  Chrome 
support fullscreen?  Yes. Chrome/Opera No. Firefox/ Samsung Internet  Not yet  Yes  Yes 
cannot work on  some mobile browser that don't understand JavaScript such as.....  cannot work on Internet Explorer 9 and below 
Credits
This email address is being protected from spambots. You need JavaScript enabled to view it.; Anne Cox; Wolfgang Christian; Francisco Esquembre
end faq
Other Resources
 http://iwant2study.org/lookangejss/02_newtonianmechanics_7gravity/ejs/ejs_model_GField_and_Potential_1D_v8wee.jar
end faq
Facebook Social Comments