DHS 07 Gravitational Field (Tutorial Solutions)

2024 H2 Physics (9749) Gravitation Field_Tutorial Solution 9749 Physics (2024) Topic 7: Gravitational Field 6 – Tutorial Solution - 1 For Internal Use Only LO Sub-topic Tut Qn a Show an understanding of the concept of a gravitational field as an example of field of force and define gravitational field strength at a point as the gravitational force exerted per unit mass placed at that point. b Recognise the analogy between certain qualitative and quantitative aspects of gravitational and electric fields. c Recall and use Newton's law of gravitation in the form F = Gm1m2 / r2. S1,S2 d Derive, from Newton's law of gravitation and the definition of gravitational field strength, the equation g = GM / r2 for the gravitational field strength of a point mass. e Recall and apply the equation g = GM / r2 for the gravitational field strength of a point mass to new situations or to solve related problems. S3,S4,S7, S8,D1,D2 f Show an understandi ng that near the surface of the Earth g is approximately constant and equal to the acceleration of free fall. g Define potential at a point as the work done per unit mass in bringing a small test mass from infinity to the point. h Solve problems using the equation φ = −GM / r for the gravitational potential in the field of a point mass. S5,S6,S9, D3,D4,D5, D6,D10, D11, D12, D13, D14 i Analyse circular orbits in inverse square law fields by relating the gravitational force to the centripetal acceleration it causes. S10,D7,D9 j Show an understanding of geostationary orbits and their application. D8 Dunman High School (Senior High Physics) Gravitational Field Tutorial Solution

2024 H2 Physics (9749) Gravitation Field_Tutorial Solution 9749 Physics (2024) Topic 7: Gravitational Field 6 – Tutorial Solution - 2 For Internal Use Only No Solution Mark S1 F = 2 (3)(2)G r F1 = 2 (3)(2) (2 ) G r = 2 1 (3)(2) 4   G r = 4 F 1 1 S2 Ans: C On the Earth’s surface, F = 32(5760 10 )× EsGM m - - - - (1) At a height of 144 km, F1 = 32[(5760 144) 10 ]+× EsGM m = 32(5904 10 )× EsGM m - - - - (2) (2) (1) : 1F F = 2 2 5760 5904 = 0.952 ≈ 95%  i.e. less by 5% S3 Ans: C On the surface of the planet at distance R from its centre, gravitational field strength due to the planet = 2 GM R S4 Ans: A g = 2 GM r , is a vector. The g ravitational field strength near Earth is directed towards Earth, which is to the left, represented by a negative value. The gravitational field strength near Moon is directed towards Moon, which is to the right, represented by a positive value. S5 Ans: D If gravitational potential energy is zero at infinity , it is always negative anywhere else, i.e. U =− GMm r . Q is a point further away from the Earth’s centre, hence U at point Q is greater than that at point P. i.e. UQ is less negative than UP. S6 Ans: A From φ =− GM r : at φ = –60 MJ kg−1, r is smaller, i.e. closer to the Earth. ∆U = m(φf – φi) = (50)[(–60) – (–20)] = –2000 MJ (negative indicates a loss) S7(a) (i) 1 M1 M2 Y X d g1

2024 H2 Physics (9749) Gravitation Field_Tutorial Solution 9749 Physics (2024) Topic 7: Gravitational Field 6 – Tutorial Solution - 3 For Internal Use Only (ii) g1 = 1 2 GM d 1 (iii) F = (g1)(M2) = 12 2 GM M d 1 (b) force on M3 by M1 = force on M3 by M2 13 2 GM M x = 23 2() − GM M dx  2 2 () −dx x = 2 1 M M  −dx x = 2 1 M M Cross-multiply: x 2M = d 1M − x 1M  x( )12MM+ = d 1M  x = d 1 12   + M MM 1 1 Note: • Since d > x, negative 2 1 M M is rejected. S8 Gravitational field strength at X due to the Earth: gE = − ( ) 282.0 10× EGM (directed towards Earth, taken as negative) Gravitational field strength at X due to the Moon: gM = ( ) 281.8 10× MGM (directed towards Moon, taken as positive) Resultant g = gM + gE = 6.67 × 10−11 ( ) ( ) ( ) ( ) 22 24 2288 7.4 10 6.0 10 1.8 10 2.0 10  ×× − ×× = −9.85 × 10−3 N kg−1 (towards Earth) 1 1 S9 Ans: C F =− pdE dr S10(a) (i) v = rω = r 2π T = (1.5 × 1011) π  ××× 2 365 24 60 60 = 2.99 × 104 m s−1 1 1 (ii) a = 2v r = 42 11 (2.99 10 ) 1.5 10 × × = 5.96 × 10−3 m s−2 1 1 (iii) F = ma = (6.0 × 1024)(5.96 × 10−3) = 3.58 × 1022 N 1 1 (iv) g = F m = 5.96 × 10−3 N kg−1 1

2024 H2 Physics (9749) Gravitation Field_Tutorial Solution 9749 Physics (2024) Topic 7: Gravitational Field 6 – Tutorial Solution - 4 For Internal Use Only (b) Using T2 = 24 π   sGM r3  T2 = kr3 (Kepler’s 3rd law, Ms = mass of Sun) For Earth, 12 = k(1.5 × 1011)3 - - - - (1) For Mars, T2 = k(2.3 × 1011)3 - - - - (2) (2) (1) : 2 21 T = 3 3 2.3 1.5  T = 1.9 years 1 1 1 D1 Ans: C M = ρV = ρ 34 3  π r g = 2 GM r = 3 2 4 3ρπ  Gr r = 4 3 πρrG D2 Ans: A Using g = 2 GM r , when r = 1, g = 64  64 = 2(1) GM  GM = 64. So, when r = 2, g2 = 2(2) GM = 2 64 (2) = 16 When r = 4, g4 = 2 64 (4) = 4 Note: • Using Desmos and the three reference points in ( A), if a graph of g against 2 1 r is plotted and extrapolated, a straight line graph passing the origin will be obtained. D3 Ans: C A: True. A mass will move from position Y of less negative gravitational potential to X of more negative potential when released, hence the work done by an external force is negative. B: True. P oints that lie on the same equipotential line have the same potential. C: False. A mass will move from a position of less negative gravitational potential to one of more negative potential when released, hence the work done by the gravitational field to move a mass further away from Earth is negative. D: True. φ ∝ 1− r D4 From the graph, g = φ− d dr , i.e. resultant g is zero at φ = −1.3 MJ kg−1 Minimum energy required (to move the object to zero resultant g from Earth’s surface) = m(φf – φi) = (2.0)[ (–1.3) – (−62.3)] = 122 MJ 1 1

2024 H2 Physics (9749) Gravitation Field_Tutorial Solution 9749 Physics (2024) Topic 7: Gravitational Field 6 – Tutorial Solution - 5 For Internal Use Only (Upon reaching the point of zero resultant g, if the object has a velocity towards the Moon, it will move towards the M oon pass the point of zero resultant g. Subsequently, the Moon’s gravitational pull would predominate and a resultant force accelerates it to the Moon’s surface.) D5 By conservation of energy, Loss in U = Gain in KE  Ui – Uf = KEf – KEi 3 − GMm R −− GMm R = 1 2 mv2 – 0 2 3 GMm R = 1 2 mv2  v = 1 24 3   GM R 1 1 D6(a) The gravitational potential at a point is defined as the work done per unit mass in bringing a small test mass from infinity to that point. 1 (b) Potential at infinity is taken to be zero. Due to the attractive nature of the gravitational force, work done by an external force to bring any mass from infinity to that point is always negative, thus a decrease in potential (from zero). Hence the potential at any point must always be negative. 1 1 D6(c) (i) ∆φ = φf − φI = GM 11  − −−     firr = (6.67 × 10−11)(6.0 × 1024) 67 6 11 (6.4 10 1.3 10 ) 6.4 10  − −− ×+× ×  = 4.19 × 107 J kg−1 1 1 (ii) By conservation of energy, KE loss = U gain 1 2 mv2 – 0 = m∆φ v = 2 φ∆ = 72(4.19 10 )× = 9.15 × 103 m s−1 1 1 (iii) The escape velocity v = 2 Earth GM R = 11 24 6 2(6.67 10 )(6.0 10 ) 6.4 10 −×× × = 1.12 × 104 m s−1 OR v = 2 EarthgR = 62(9.81)(6.4 10 )× = 1.12 × 104 m s−1 1 1 (d) The equation is not appropriate because the acceleration is not uniform. 1 D7(a) Gravitational force provides the centripetal force for the satellite in orbit: 1

2024 H2 Physics (974