gravity07

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 $\frac{d ϕ}{d r}$ / 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 = 5.0x10⁷m 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)( $\frac{1}{(6.37 x 1 0^{6})} - \frac{1}{(13630000 + 6.37 x 1 0^{6})}$ ) = (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 $\frac{d ϕ}{d r} = - \frac{- 10.01 - (20.00)}{4 x 1 0^{7} - 2 x 1 0^{7}}$

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.

7.3.8.4.2 Model