keywords 012 part 1 of 1 10 points Given G 6 N m 2 kg 2 A

# Keywords 012 part 1 of 1 10 points given g 6 n m 2 kg

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keywords:012(part 1 of 1) 10 pointsGiven:G= 6.67259×10-11N m2/kg2.A projectile is fired vertically upward fromthe surface of planet Moorun of mass 3×1024kg and radius 9×106m.If this projectile is to rise to a maximumheight above the surface of Moorun equal to6×106m, what must be the initial speed ofthe projectile?Correct answer: 4.21824 km/s.Explanation:Basic Concept:Total energy (Kineticand Gravitational Potential) isE=12m v2-G M mr.Solution:Denote the mass of the projectilebym.The initial total energy (before take-off) isEi=12m v2-G M mR.The final total energy (when the projectile isat the maximum height) isEf= 0-G M mR+h,since the projectile has zero velocity at itshighest point.Energy conservation impliesEi=Ef(neglecting air friction), so12m v2-G M mR=-G M mR+h.Solving forv, we havev=sG2M hR(R+h)(1)=(6.67259×10-11N m2/kg2)×2 (3×1024kg) (6×106m)h(9×106m) (1.5×107m)12= 4218.24 m/s= 4.21824 km/s, Alvarado, Patrick – Homework 10 – Due: Nov 16 2006, 10:00 am – Inst: Andrei Sirenko5whereR+h= (9×106m) + (6×106m)= 1.5×107m.Comment:Noticethat if we had appliedordinary kinematics, we would have utilizedthe familiar formulav21-v20= 2a(x1-x0),which in our problem, is0-v2= 2 (-g)(h-0),so that the initial velocity to rise to heighthwould bev=p2g h(2)=q2 (9.8 m2/s2) (6×106m)= 10.8444 km/s,which isNOTthe right answer.This isbecause the attraction of gravity decreases aswe go further away from the planet, requiringless initial velocity than if the attraction hadbeen constant (which our simple kinematicsformula assumes).It is, however, a goodapproximation when we are close to a planet’ssurface.Try picking a smaller heighthand see whenthe simpler description (2) presents a goodapproximation to the velocity of formula (1) !keywords:013(part 1 of 2) 10 pointsGiven:G= 6.67259×10-11N m2/kg2Two hypothetical planets of masses 3.9×1023kg and 7.8×1023kg and radii 3.5×106mand 5.4×106m, respectively, are at rest whenthey are an infinite distance apart.Becauseof their gravitational attraction, they headtoward each other on a collision course.When their center-to-center separation is3×108m, find their relative velocity.Correct answer: 721.001 m/s.Explanation:At infinite separation the potential energyUis zero, and at rest the kinetic energyKiszero. Since energy is conserved we have0 =12m1v21+12m2v22-G m1m2d.The initial momentum is zero and momentumis conserved, so0 =m1v1-m2v2.Combine these two equations to findv1=m2s2Gd(m1+m2)= 7.8×1023kg×s2 (6.67259×10-11N m2/kg2)(3×108m) (1.17×1024kg)= 480.667 m/sv2=m1s2Gd(m1+m2)= 3.9×1023kg×s2 (6.67259×10-11N m2/kg2)(3×108m) (1.17×1024kg)= 240.334 m/s.  #### You've reached the end of your free preview.

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