Example 9 (J89/II/2)

Values for the gravitational potential due to the Earth are given in the table below.

(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 $\frac{d}{}$ ϕ ⅆ r = - - 10.01 - ( 20.00 ) 4 x 1 0 7 - 2 x 1 0 7

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