- 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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22.214.171.124 Example 9 (J89/II/2)
126.96.36.199.1 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 / ms-2|
|0||6 370 000 = 6.37x106= 0.637x107||-62.72|
|10 000 000 = 1.0x107||-40.10|
(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)( ) = (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
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