assignment #028 028. Orbital Dynamics Physics II 12-28-2007
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01:14:17 `q001. Note that this assignment contains 11 questions. The planet Earth has a mass of approximately 6 * 10^24 kg. What force would therefore be experienced by a 3000 kg satellite as it orbits at a distance of 10,000 km from the center of the planet?
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RESPONSE --> F(r1) m * gravitational field at distance r1= mg(r1/R)^2 R=6400km r1=10,000km mass=6*10^24 kg Field= (10000/6400km)^2*g Field=1.5625^2*g Field=2.44g F=2.44*9.8m/s^2*6*10^24kg F=23.93m/s^2 *6*10^24kg F=143.55*10^24 N confidence assessment: 1
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01:20:28 The force would be F = G m1 m2 / r^2, with m1 and m2 the masses of the planet and the satellite and r the distance of the satellite from the center of the planet. Thus we have F = 6.67 * 10^-11 N m^2 / kg^2 * (6 * 10^24 kg) * 3000 kg / (10,000,000 meters) ^ 2 = 12,000 Newtons.
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RESPONSE --> Where did the G come from? is it different from 9.8m/s^2
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01:23:56 `q002. What force would the same satellite experience at the surface of the Earth, about 6400 km from the center.
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RESPONSE --> F= G m1 m2/r^2 F= 6.67*10^-11Nm^2/kg^2 *6.0*10^24kg*3000kg/6,400,000m= confidence assessment: 1
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01:25:54 The force would be F = G m1 m2 / r^2, with m1 and m2 the masses of the planet and the satellite and r the distance of the satellite from the center of the planet. Thus we have F = 6.67 * 10^-11 N m^2 / kg^2 * (6 * 10^24 kg) * 3000 kg / (6,400,000 meters) ^ 2 = 29,000 Newtons. Note that this is within roundoff error of the F = m g = 3000 kg * 9.8 m/s^2 = 29400 N force calculated from the gravitational acceleration experienced at the surface of the Earth.
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RESPONSE --> F=6.67*10^-11Nm^2/kg^2 *6*10^24kg*3000kg/6,400,000m)^2=29,000N F=29,000N self critique assessment: 2
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01:28:09 `q003. What would be the acceleration toward the center of the Earth of the satellite in the previous two questions at the distance 10,000 km from the center of the Earth? We may safely assume that no force except gravity acts on the satellite.
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RESPONSE --> F=ma Force=12000N mass=3000kg 12000N/3000kg=4m/s^2 confidence assessment: 1
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01:28:26 The force at the 10,000 km distance was previously calculated to be 12,000 Newtons, the mass of the satellite being 3000 kg. Since the only force acting on the satellite is that of gravity, the 12,000 Newtons is the net force and the acceleration of the satellite is therefore a = 12,000 N / 3000 kg = 4 m/s^2.
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RESPONSE --> ok self critique assessment: 3
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01:33:56 `q004. The centripetal acceleration of an object moving in a circle of radius r at velocity v is aCent = v^2 / r. What would be the centripetal acceleration of an object moving at 5000 m/s in a circular orbit at the distance of 10,000 km from the center of a planet, and how this this compare to the 4 m/s^2 acceleration net would be experienced by an object at this distance from the center of the Earth?
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RESPONSE --> acent=v^2/r v=5000m/s r=10,000km a= (5000m/s)^2/10000km a=25,000,000m^2/s^2/10,000,000m a=2.5m/s^2 This is less of an acceleration than the 4m/s^2 acceleration that would be experienced by an object at this distance from the center of the Earth confidence assessment: 1
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01:34:08 The centripetal acceleration of the given object would be aCent = (5000 m/s)^2 / (10,000,000 m) = 2.5 m/s^2. This is less than the acceleration of gravity at that distance.
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RESPONSE --> ok self critique assessment: 3
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01:38:15 `q005. What would be the centripetal acceleration of an object moving at 10,000 m/s in a circular orbit at the distance of 10,000 km from the center of a planet, and how does this compare to the 4 m/s^2 acceleration that would be experienced by an object at this distance from the center of the Earth?
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RESPONSE --> aCent=v^2/r v=10,000m/s r=10,000km aCent=(10,000m/s)^2/10000000m=10m/s^2 This is larger than the 4m/s^2 acceleration that would be experienced by an object at this distance from the center of the Earth confidence assessment: 2
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01:38:27 The centripetal acceleration of this object would be aCent = v^2 / r = (10,000 m/s)^2 / (10,000,000 m) = 10 m/s^2, which is greater than the acceleration of gravity at that distance.
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RESPONSE --> ok self critique assessment: 3
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01:41:58 `q006. An object will move in a circular orbit about a planet without the expenditure of significant energy provided that the object is well outside the atmosphere of the planet, and provided its centripetal acceleration matches the acceleration of gravity at the position of the object in its orbit. For the satellite of the preceding examples, orbiting at 10,000 km from the center of the Earth, we have seen that the acceleration of gravity at that distance is approximately 4 m/s^2. What must be the velocity of the satellite so that this acceleration from gravity matches its centripetal acceleration?
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RESPONSE --> a=4m/s^2 aCent=v^2/r 4m/s^2=v^2/10,000,000m 40,000,000m^2/s^2=v^2 v=+-6324.6m/s confidence assessment: 2
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01:42:52 The velocity must be such that aCent = v^2 / r matches the 4 m/s^2. Solving aCent = v^2 / r for v we obtain v = `sqrt( aCent * r ), so if aCent is 4 m/s^2, v = `sqrt( 4 m/s^2 * 10,000,000 m ) = `sqrt( 40,000,000 m) = 6.3 * 10^3 m/s.
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RESPONSE --> aCent=v^2/r v=`sqrt(4m/s^2*10,000,000m)=6.3*10^3m/s self critique assessment: 3
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`q007. The orbital velocity of a satellite in a circular orbit is that velocity for which the centripetal acceleration of the satellite is equal to its gravitational acceleration. The satellite in the previous series of examples had a mass of 3000 kg and orbited at a distance of 10,000 km from the center of the Earth. What would be the acceleration due to Earth's gravity of a 5-kg hunk of space junk at this orbital distance?
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RESPONSE --> aCent=v^2/r r=10,000,000m gravitational acceleration=9.8m/s^2 9.8m/s^2=v^2/10,000,000m v^2=98000000m^2/s^2 v=9899.5m/s a=9.8m/s^2 confidence assessment: 1
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18:46:38 The force of gravity on the junk hunk is easily found from Newton's Law of Universal Gravitation. Using F = G m1 m2 / r^2 we see that the force of gravity must be Fgrav = (6.67 * 10^-11 kg) * (6 * 10^24 kg) * (5 kg) / (10,000,000 m)^2 = 20 Newtons, approx.. Its acceleration due to gravity is thus a = Fgrav / m = 20 Newtons / 5 kg = 4 m/s^2. We note that this is the same gravitational acceleration experienced by the 3000 kg mass, and conjecture that any mass will experience the same gravitational acceleration at this distance from the center of the planet.
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RESPONSE --> F=Gm1m2/r^2 Fgravity=(6.67*10^-11kg)*(6*10^24kg)(5kg)/(10,000,000m)^2=20 Newtons acceleration=Fgrav/m=20N/5kg=4m/s^2 self critique assessment: 2
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18:48:58 `q008. What therefore will be the orbital velocity of the 5-kg piece of junk?
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RESPONSE --> aCent=v^2/r a=4m/s^2 4m/s^2=v^2/10,000,000m 40,000,000m^2/s^2=v^2 v=6324.6m/s confidence assessment: 2
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18:49:55 Orbital velocity is calculated from distance and gravitational acceleration by solving a = v^2 / r for v, where a is the centripetal acceleration, which is the same as the gravitational acceleration. We get v = `sqrt( a * r), just as before, and v = `sqrt( 4 m/s^2 * 10,000,000 m) = 6.3 * 10^3 m/s, the same velocity as for the 3000 kg satellite.
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RESPONSE --> The 5kg hunk had the same velocity as the 3000 kg satellite. self critique assessment: 2
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18:54:01 `q009. Is it true that the gravitational acceleration of any object at a distance of 10,000,000 meters from the center of the Earth must be the same as for the 3000-kg satellite and the 5-kg hunk of space junk? (Hint: We have to find the acceleration for any mass, so we're probably going to have to let the mass of the object be represented by symbol. Use mObject as a symbol for the mass of the object. While dealing in symbols, you might as well leave G and r in symbols and let mEarth stand for the mass of the Earth. Find an expression for the force, then using this expression and Newton's Second Law find an expression for the acceleration of the object).
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RESPONSE --> F= G mObject*mEarth/r^2 a=F/mObject confidence assessment: 1
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18:57:44 We know that the gravitational force on the object is Fgrav = G * mEarth * mObject / r^2, where G is the universal gravitational constant, r the distance from the center of the Earth, mEarth the mass of the Earth and mObject the mass of the object. The acceleration of the object is a = Fgrav / mObject, by Newton's Second Law. Substituting the expression G * mEarth * mObject / r^2 for Fgrav we see that a = [ G * mEarth * mObject / r^2 ] / mObject = G * mEarth / r^2. We note that this expression depends only upon the following: G, which we take to be univerally constant, the effectively unchanging quantity mEarth and the distance r separating the center of the Earth from the center of mass of the object. Thus for all objects at a distance of 10,000 km from the center of the Earth the acceleration due to the gravitational force must be the same.
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RESPONSE --> a=Fgrav/mObject G*mEarth*mObject/r^2=Fgrav a(G*mEarth*mObject/r^2)/mObject=G*mEarth/r^2 The mObject cancels out. For all objects at a distace of 10.000km from the center of the Earth, the acceleraton due to the gravitational force is the same. self critique assessment:
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18:59:02 We know that the gravitational force on the object is Fgrav = G * mEarth * mObject / r^2, where G is the universal gravitational constant, r the distance from the center of the Earth, mEarth the mass of the Earth and mObject the mass of the object. The acceleration of the object is a = Fgrav / mObject, by Newton's Second Law. Substituting the expression G * mEarth * mObject / r^2 for Fgrav we see that a = [ G * mEarth * mObject / r^2 ] / mObject = G * mEarth / r^2. We note that this expression depends only upon the following: G, which we take to be univerally constant, the effectively unchanging quantity mEarth and the distance r separating the center of the Earth from the center of mass of the object. Thus for all objects at a distance of 10,000 km from the center of the Earth the acceleration due to the gravitational force must be the same.
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RESPONSE --> self critique assessment: 2
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19:04:54 `q010. How much work would have to be done against gravity to move the 3000 kg satellite from a circular orbit at a distance of 10,000 km to a circular orbit at a distance of 10,002 km?
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RESPONSE --> Work=Force *distance F=G m1m2/r^2 Force= 6.67*10^-11N m^2/kg^2 * 6*10^24kg * 3000kg/ (10,000,000m)^2 distance=2km= 2000m confidence assessment: 0
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19:07:55 As found previously the object experiences a force of approximately 12,000 N at a distance of 10,000 km. At a distance of 10,002 km, the force of gravity will be slightly less than at 10,000 km, but only by about 5 Newtons or .0004 of the force. That is, over the 2 km distance the force of gravity doesn't change by very much. Therefore to move 2 km = 2000 m further from the center of the planet would require the application of a force very close to 12,000 N in the direction away from the center. The work done by this force is therefore `dW = 12,000 Newtons * 2000 m = 24,000,000 Joules.STUDENT QUESTION: I understand this mathmatically, I'm not sure I understand practically. How do you gain KE if one object was intially stationary? It would seem that the first object would lose and the second object would gain what was lost but not more than what was lost... INSTRUCTOR RESPONSE: As stated it isn't possible for total KE to increase unless there is some other source of energy involved. For example if there is a coiled spring on one object it could uncoil on collision and add extra KE. Momentum conservation does not say anything about energy. Momentum and energy are two completely independent quantities. **
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RESPONSE --> `dW=12000N*2000m=24,000,000J Required 12000N at 10,000km at 10,002km= 5 Newtons less self critique assessment: 2
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19:09:51 `q011. Does it therefore follow that the work done to move a 3000 kg satellite from the distance of 10,000 km to a distance of 10,002 km from the center of the Earth must be 24,000,000 Joules?
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RESPONSE --> Work =Force *distance `dW = 12,000 Newtons * 2000 m = 24,000,000 Joules confidence assessment: 1
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19:13:44 It might seem so, but this is not the case. The net force does work, but when the radius of the orbit changes the velocity and hence the kinetic energy of the satellite also changes. The work done by the net force is equal to the sum of the changes in the KE and the gravitational PE of the satellite. The change in gravitational PE is the 24,000,000 J we just calculated, and if there is no KE change this will be equal to the work done by the net force. However if KE increases the net force must do more than 24,000,000 J of work, and if KE decreases the net force must do less than 24,000,000 J of work. In this case, as we move further away the KE decreases so the net force must do less than 24,000,000 J of work. (See also Conservation of Energy in Orbit under Q&A)
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RESPONSE --> When the radius of the orbit changes, the velocity and thus the kinetic energy of the satellite changes. Work done by the net force= changes in KE plus the changes in the gravitational PE of the satellite change in gravitational PE= 24,000,000J IF no KE change= work done by the net force IF KE increases, the net force must do more than 24,000,000J of work If KE decreases, the net force must do less than 24,000,000J of work As we move further away from Earth, the KE decreases so the net force must do less than 24,000,000J of work self critique assessment: 2
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