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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 / 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)( ) = (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
7.3.8.4.2 Model
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- http://iwant2study.org/lookangejss/02_newtonianmechanics_7gravity/ejs/ejs_model_GFieldandPotential1Dv7EarthMoon.jar
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- Parent Category: 02 Newtonian Mechanics
- Category: 08 Gravity
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