Friday, May 8, 2015

EJSS gravity 07_1

Example 9 (J89/II/2)

Values for the gravitational potential due to the Earth are given in the table below.
Distance from Earth’s surface / mDistance from Earth’s centre / mGravitational potential / MJ kg-1Gravitational Field Strength g / ms-2rate of change of potential with distance  ⅆ ϕ  r/ ms-2
06 370 000 = 0.637x107-62.72


10 000 000 = 1x107-40.10

1.363x1072.0x107-20.00


3.0x107-13.34-0.440.44

4.0x107-10.01


5.0x107-8.01-0.160.16

Infinity0

(i) given that radius of Earth = 6370 km, fill in the missing rows in the table for Distance from Earth’s surface / m.
(ii)    Calculate the change in potential if an object travels from r = 5.0x10m to r = 2.0x107 m


Answer: Δφ = φfinal - φinitial = -19.99x106 - (-8.01x106) = -11.95x106 J/kg
(ii) Hence or otherwise, calculate the change in potential energy if the object-satellite has a mass 0.01x1024 kg.
Answer:  ΔPE = ΔU = m(Δφ) = 0.01x1024(-11.95x106) = -1.196x1029 J = -0.20x1030 J
(iii) Determine the potential energy lost by object-satellite of 0.07x1024 kg from a height of 13 630 000 to the Earth's surface.
Answer: ΔPE = m(Δφ) = 0.07x1024 (-6.67x10-11)(6.0x1024)( 1 ( 6.37 x 1 0 6 ) - 1 ( 13630000 + 6.37 x 1 0 6 ) ) = (0.07x1024 )(-4.28x107 ) = 2.996x1030 = 3.00x1030 J


iv) 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.
v) Hence, suggest a relationship between g and   ϕ  r
vi) Explain why the accuracy of determining g at r = 3x107 is poor when using the following values at r = 2x107 and 4x107 m where ⅆ ϕ  r = - - 10.01 - ( 20.00 ) 4 x 1 0 7 - 2 x 1 0 7

Java Model

http://iwant2study.org/lookangejss/02_newtonianmechanics_7gravity/ejs/ejs_model_GFieldandPotential1Dv7EarthMoon.jar

Model

https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_model_gravity07_1/gravity07_1_Simulation.xhtml