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gravity07_1

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   d ϕ d r / ms-2
0 6 370 000 = 6.37x106= 0.637x107 -62.72


10 000 000 = 1.0x107 -40.10

1.363x107
2.0x107 -20.00


3.0x107 -13.34 -0.44
0.44

4.0x107 -10.01


5.0x107 -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 = 5.0x107 m to r = 2.0x107 m

Answer: Δϕ = ϕfinal - ϕinitial = -20.00x106 - (-8.01x106) = -11.95x106 J/kg

(iii) 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

(iv) 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 )(-6.27x107 -(-20.00x107))= (0.07x1024 )(-4.28x107 ) = 2.996x1030 = 3.00x1030 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 = 3x107 is poor when using the following values at r = 2x107 and 4x107 m where  d ϕ d r = - - 10.01 - ( 20.00 ) 4 x 1 0 7 - 2 x 1 0 7

7.3.8.4.2 Model

  1. Run Sim
  2. http://iwant2study.org/ospsg/index.php/60
 

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

 

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