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 / 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 = 0.637x107	-62.72
	10 000 000 = 1x107	-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 rows in the table for 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 = -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 $$ ⅆ ϕ ⅆ 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 ⅆ 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