7.3.8.4 Gravitational Field and Potential Earth JavaScript HTML5 Applet Simulation Model
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 Parent Category: 02 Newtonian Mechanics
 Category: 08 Gravity
 Created: Thursday, 20 August 2015 15:00
 Last Updated: Thursday, 25 May 2017 15:37
 Published: Thursday, 07 April 2016 15:00
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7.3.8.4 Example 9 (J89/II/2)
7.3.8.4.1 Values for the gravitational potential due to the Earth are given in the table below.
Distance from Earth’s surface / m  Distance from Earth’s centre / m  Gravitational potential / MJ kg^{1}  Gravitational Field Strength g / ms^{2}  Rate of change of potential with distance $\genfrac{}{}{0.1ex}{}{d\varphi}{dr}$/ ms^{2} 
0  6 370 000 = 6.37x10^{6}= 0.637x10^{7}  62.72  
10 000 000 = 1.0x10^{7}  40.10  
1.363x10^{7} 
2.0x10^{7}  20.00  
3.0x10^{7}  13.34  0.44 
0.44 

4.0x10^{7}  10.01  
5.0x10^{7}  8.01  0.16  0.16  
Infinity  0 
(i) given that radius of Earth = 6370 km, fill in the missing values in the column of Distance from Earth’s surface / m.
(ii) Calculate the change in potential if an object travels from r = 5.0x10^{7 }m to r = 2.0x10^{7} m
Answer: Δϕ = ϕfinal  ϕinitial = 20.00x10^{6}  (8.01x10^{6}) = 11.95x10^{6} J/kg
(iii) Hence or otherwise, calculate the change in potential energy if the objectsatellite has a mass 0.01x10^{24} kg.
Answer: ΔPE = ΔU = m(Δϕ) = 0.01x10^{24}(11.95x10^{6}) = 1.196x10^{29} J = 0.20x10^{30} J
(iv) Determine the potential energy lost by objectsatellite of 0.07x10^{24} kg from a height of 13 630 000 to the Earth's surface.
Answer: ΔPE = m(Δϕ) = 0.07x10^{24 }(6.67x1011)(6.0x10^{24})($\frac{1}{\left(6.37x1{0}^{6}\right)}\frac{1}{(13630000+6.37x1{0}^{6})}$ ) = (0.07x10^{24 })(6.27x10^{7} (20.00x10^{7}))= (0.07x10^{24 })(4.28x10^{7} ) = 2.996x10^{30} = 3.00x10^{30} J
v) by means of using the simulation, move the mass and record down and
fill in the last 2 missing columns for Gravitational Field Strength and
rate of change of potential with distance.
vi) Hence, suggest a relationship between g, ϕ and r.
vii) Explain why the accuracy of determining g at r = 3x10^{7} is poor when using the following values at r = 2x10^{7} and 4x10^{7} m where $\genfrac{}{}{0.1ex}{}{d\varphi}{dr}=\frac{10.01\left(20.00\right)}{4x1{0}^{7}2x1{0}^{7}}$
7.3.8.4.2 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 
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end faq
Versions
 http://iwant2study.org/lookangejss/02_newtonianmechanics_7gravity/ejs/ejs_model_GFieldandPotential1Dv7EarthMoon.jar