kean (msk955) – Homework 05 – yao – (56725) 1This print-out should have 23 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsThree 8 kg masses are located at points in thexy plane.39 cm53 cmWhat is the magnitude of the resultantforce (caused by the other two masses) onthe mass at the origin? The universal gravita-tional constant is 6 .6726 ×10−11N ·m2/kg2.Correct answer: 3.19 284 × 1 0−8N.Explanation:Let : m = 8 kg ,x = 39 cm = 0.39 m ,y = 53 cm = 0.53 m , andG = 6.6726 × 10−11N · m2/kg2.The force from the mass on the right pointsin the x direction and has magni tudeF1= Gm mx2=G m2x2=(6.6726 × 10−11N · m2/kg2) (8 kg)2(0.39 m)2= 2.80767 × 10−8N .The other force points in the y directionand has magnitudeF2=(6.6726 × 10−11N · m2/kg2) (8 kg)2(0.53 m)2= 1.52028 × 10−8N .F2F1FθThe magnitude of the resultant force isF =qF21+ F22=h(2.80767 ×10−8N)2+ (1.52028 × 10−8N)2i1/2=3.19284 × 10−8N .002 10.0 pointsThe planet Krypton has a mass of5.3 × 1023kg and radius of 4.3 × 106m.What is the acceleration of an object in freefall near the surface of Krypton? The gravita-tional constant is 6.6726 × 10−11N · m2/kg2.Correct answer: 1.9 1264 m/s2.Explanation:Let : M = 5.3 × 1023kg ,R = 4.3 × 106m , andG = 6.6726 × 10−11N · m2/kg2.Near the surface of Krypton, the gravita-tion force on an object of mass m isF = GM mR2,so the acceleration a of a free-fall object isa = gKrypton=Fm= GMR2= (6.6726 ×10−11N · m2/kg2)×5.3 × 1023kg(4.3 × 106m)2=1.91264 m/s2.kean (msk955) – Homework 05 – yao – (56725) 2003 10.0 pointsIn another solar system is planet Driff, whichhas 5 times the mass of the earth and also 5times the r a dius.How does the gravitational acceleration onthe surface of Driff compare to the gravita-tional acceleration on the surface of the earth?1. It’s15th as much. correct2. There is no gravity on Driff because 5times 4000 miles ( the radius of the earth), is20000 miles, far beyond the pull of gravity.3. It’s125th as much.4. It’s 25 times as great.5. It’s 5 times as much.6. It’s the same, 10 m/s2.Explanation:Let : MD= 5 MeandRD= 5 Re.Gravitational force isF = m g = GM mr2g =G Mr2∝Mr2, sogDge=mDr2Dmer2e=MDr2emer2D=(5 me) r2eme(5 re)2=15gD=15ge.004 (part 1 of 2) 10.0 pointsIn order for a satellite to move in a stablecircular orbit of radius 6 659 km at a constantspeed, its centripetal acceleration must beinversely proportional to the square of theradius r of the orbit.What is the speed of t he satellite?The universal gravitati onal constant is6.67259 × 10−11N · m2/kg2and the ma ss ifthe earth is 5.98 × 1024kg.Correct answer: 7740 .93 m/s.Explanation:Let : r = 6659 km = 6.659 × 1 06m ,M = 5.98 × 1024kg , andG = 6.67259 × 10−11N · m2/kg2.The centripetal force is provided by theuniversal force from the earth:GM mr2=m v2r,where M is the mass of the earth. Thusv =rG Mr=s6.67259 × 10−11N · m2/kg26.659 × 106m×p5.98 × 1024kg=7740.93 m/s .005 (part 2 of 2) 10.0 pointsFind the time required to complete one orbit.Correct answer: 1.5 0139 h.Explanation:The time to complete one o rbi t i sT =2 π rv=2 π (6.659 × 106m)7740.93 m/s1 h3600 s=1.50139 h .006 10.0 pointsTwo satellites A and B, where B has twice t hemass of A, orbit the earth in circular orbits.The distance of satelli te B from the earth’skean (msk955) – Homework 05 – yao – (56725) 3center is twice the dist ance of satellite A fromthe earth’s center.What is the ratio of the orbital period ofsatellite B to that of satellite A?1.TBTA= 12.TBTA= 23.TBTA= 1/84.TBTA= 1/25.TBTA=√8 correct6.TBTA= 87.TBTA=p1/88.TBTA=√29.TBTA=p1/210.TBTA= 1/4Explanation:According to Kepler’s thir d law, the squareof the orbital period is proportional to thecub e of the orbital radi us. Therefore,T2AR3A=T2BR3B.Therefore,TBTA=RBRA3/2= 23/2=√8 .007 10.0 pointsAn object of mass m is held at rest a di st anceReabove t he earth’s north pole, where Reisthe radius of t he earth and g is the accelera-tion due t o gravity on the surface.If the object is allowed to fall, with whatspeed does it hit the surface? Ignore the mo-tion o f the earth around the sun, a nd frictionwith the earth’s atmosphere.1. 2pg Re2.rg Re33.12pg Re4.p3 g Re5.pg Recorrect6.13pg Re7.rg Re28.p2 g Re9.rg Re5Explanation:The gravitational potential energy for massm isUg(r) = −G Memr= −m g R2er.The total energy of the o bject is E = K + Ug,soEi= Ef0 −m g R2e2 Re=12m v2−m g R2eRe−12g Re=12v2− g Rev =pg Re.008 10.0 pointsA bla ck hol e is an object so heavy that neithermatter nor even light can escape the influenceof its gravitational field. Since no light canescap e from it, it appears black. Suppose amass approximately the size of the Earth’smass 4.41 × 1024kg is packed into a smalluniform sphere of radius r.Use:The speed of light c = 2.9979 2 × 108m/s .kean (msk955) – Homework 05 – yao – (56725) 4The universal g ravitational constant G =6.67259 × 10−11N m2/kg2.Hint: The escape speed must be the speedof light.Based on Newtonian mechanics, determinethe limiting radius r0when this mass (approx-imately the size of the Earth’s mass) becomesa bla ck hole.Correct answer: 0.00 65482 m.Explanation:Basic Concepts: Energy conservationE = −G m Mr+ K .At minimum escape velocity, E = 0 (the pro-jectile has just enough initial kinetic energyto overcome the gravitational potenti al).Solution: Technically speaking, in a re-gion where gravity is extremely intense, New-ton’s mechanics cannot be used. Rather, oneneeds to apply t he “general theory of relativ-ity” developed by Albert Einstein. Knowingthis is the case, we still would like to seewhat Newtonian mechanics tells us. Settingvesc= c, the limiting radius is given by12m v2esc=12m c2=G m Mr0,orr0=2 G Mc2.It turns out that the theory of relativity givesthe same expression for this li miting radius,referred to as the “Schwarzschild radius”. InNewtonian mechanics, however, the descrip-tion of what happens to objects attracted tothis sphere is often highly inaccurate. But inany case,r0=2 G Mc2= 2 (6.672 59 × 10−11N m2/kg2)×(4.41 × 1024kg)(2.99792 × 108m/s)2=0.0065482 m .009 10.0 pointsThe planet Saturn has a mass 95.2 times thatof the earth and a radius 9.47 times that ofthe earth.Find the escape speed for objects on thesurface of Saturn.Correct answer: 35. 5109 km/s.Explanation:Let : MS= 95.2 MEandRS= 9.47 RE.The escape speed from Saturn