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Worked solutions Chapter 3 Gravity and satellites - PEGSnet

**Worked** **solutions****Chapter** 3 **Gravity** **and** **satellites****Chapter** review1 F = GMm/R 2then R 2 = GMm/F= (6.67 × 10 –11 × 5.69 × 10 26 × 1.05 × 10 21 )/(2.79 × 10 20 )**and** R = 3.78 × 10 8 m2 At a height of 2 Earth radii above the Earth’s surface, a person is a distance of 3 Earth radiifrom the centre of the Earth.Then F = 900/3 2 = 900/9 = 100 N so D.3 a ∑F = F N +F gF N = ∑F – F g = 2400↑ – 780↓ = 3200N ↑ so Db The normal force is about 4 times greater than usual, so the astronaut will feel about 4times heavier, so B.c Weight F g = mg = 80 × 9.8 = 780 N, so C.d During orbit, the astronaut travels with an acceleration of 8.2 m s –2 **and** so is in free-fall.F N = 0, so A.e Weight F g = mg = 80 × 8.2 = 660 N, so D.4 M/(0.8R) 2 = m/(0.2R) 2**and** M/m = 1/4 2 = 1 : 16 = 0.0635 a g = GM/R 2= (6.67 × 10 –11 × 1.02 × 10 26 )/(2.48 × 10 7 ) 2= 11.1 N kg –1b C, it will accelerate at a rate given by the gravitational field strength, i.e. at 11.1 m s –26 The orbital speed of the Moon around the Earth is given by:v = √(GM/R)= [(6.67 × 10 –11 × 5.98 × 10 24 )/(3.84 × 10 8 )] ½= 1.02 × 10 3 m s –1The period of the Moon around the Earth T = 2πR/v= 2π × 3.84 × 10 8 /(1.02 × 10 3 )= 2.367 × 10 6 s= 2.367 × 10 6 / (60 × 60 × 24) = 27.4 days7 a v = √(GM/R)= [(6.67 × 10 –11 × 1.90 × 10 27 )/1.10 × 10 10 ] ½ = 3.39 × 10 3 m s –1b a = v 2 /R= (3.39 × 10 3 ) 2 /(1.10 × 10 10 )= 1.05 × 10 –3 m s –2 towards Jupiterc T = 2πR/v= 2π × 1.10 × 10 10 /3.39 × 10 3= 2.048 × 10 7 s = 2.048 × 10 7 / (60 × 60 × 24) = 235 days8 a C, a satellite is said to be in a geosynchronous orbit if its orbital period is the same as thatof the Earth’s rotation (i.e. 24 hours or 86 400 s).b This satellite will always be directly above the same point on the Earth’s surface, **and** cantherefore transmit radio signals to any point that can see it.c R 3 = GMT 2 /4π 2 ,Heinemann Physics 12(3e)Copyright © Pearson Education Australia 2008Teacher’s Resource **and** Assessment Disk(a division of Pearson Australia Group Pty Ltd) ISBN 978144250126310

**Worked** **solutions****Chapter** 3 **Gravity** **and** **satellites**then R 3 = (6.67 × 10 –11 × 6.0 × 10 24 × 86400) 2 /4π 2**and** R = 4.2 × 10 7 m9 a R 3 = GMT 2 /4π 2= (6.67 × 10 –11 × 3.30 × 10 23 × (5.07 × 10 6 ) 2 /4π 2**and** R = 2.43 × 10 8 mb v = 2πR/T= 2π × 2.43 × 10 8 /(5.07 × 10 6 ) = 301 m s –1c a = v 2 /r= (301) 2 /(2.429 × 10 8 ) = 3.73 × 10 –4 m s –2 towards Mercury10 a The increase in the kinetic energy of the rock= area under the graph from R = 2.5 × 10 6 m to R = 3.0 × 10 6 m= 5.6 squares × 10 × 0.5 × 10 6=2.8 × 10 7 Jb The initial E k = ½ mv 2= 0.5 × 20 × (1.0 × 10 3 ) 2 = 1.0 × 10 7 JThe final E k = 2.8 × 10 7 J + 1.0 × 10 7 J = 3.8 × 10 7 Jc 3.8 × 10 7 J = 0.5 × 20v 2**and** v = 1.9 × 10 3 m s –1d At r = 2.5 × 10 6 m, F g = 70 N = mgg = 70/20 = 3.5 N kg –111 a (T 1 /T 2 ) 2 = (R 1 /R 2 ) 3**and** T 1 /T 2 = (R 1 /2R) 3/2 = 1/8 so Db v 1 /v 2 = (2πR 1 /T 1 )/(2πR 2 /T 2 ) = (T 2 /T 1 )(R 1 /R 2 ) = 8 /2 = 2 , so B.c a 1 /a 2 = (v 2 1 /R 1 )/(v 2 2 /R 2 ) = 2 × 2 = 4, so D.12 Transposing GMm/R 2 = 4π 2 Rm/T 2 to make M the subject gives: M = 4π 2 R 3 m/GT 2 .Substituting into this using R = 1.5 × 10 11 m **and** T = 365.25 × 24 × 60 × 60 gives M =2.0×10 30 kg.13 From graph, when distance = 300 km = 3.0 × 10 5 m, g = 9.0 N kg –114 D is correct. Area units are: N kg –1 × m = N m kg –1 = J kg –115 C is the correct answer. The area under the graph gives the energy change per kilogram. In thisexample, the satellite is falling closer to Earth, so it is losing gravitational potential energy.16 This can be determined by finding the area under the graph between R = 2 × 10 5 m **and**R = 6 × 10 5 m.Counting squares give area = 35 × 1.0 × 1.0 × 10 5 = 3.5 × 10 6 J kg-1Energy = mass × area = 1000 × 3.5 × 10 6 = 3.5 × 10 9 J17 No, air resistance would mean that a significant amount of kinetic energy would be transformedinto heat energy.18 D, the mass of the moons would need to be known before the forces can be compared.19 Charon: T 2 /R 3 = 6.4 2 /1960 3 = 5.0 × 10 –9Nix: T 2 /R 3 = 5.0 × 10 –9Heinemann Physics 12(3e)Copyright © Pearson Education Australia 2008Teacher’s Resource **and** Assessment Disk(a division of Pearson Australia Group Pty Ltd) ISBN 978144250126311

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