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course Phy 241
7/20/2013 8:00 pm
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Question: `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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Your solution:
F = G m1 m2 / r^2
F = 6.67 * 10^-11 N m^2 / kg^2 * (6 * 10^24 kg) * 3000 kg / (10,000,000 meters) ^ 2
F = 12,000Newtons
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Question: `q002. What force would the same satellite experience at the surface of the Earth, about 6400 km from the center.
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Your solution:
F = G m1 m2 / r^2
F = 6.67 * 10^-11 N m^2 / kg^2 * (6 * 10^24 kg) * 3000 kg / (6,400,000 meters) ^ 2
F = 29,000 Newtons
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Question: `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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Your solution:
a = Fnet / m
a = 12,000 N / 3000 kg
a = 4 m/s^2
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Question: `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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Your solution:
aCent = (5000 m/s)^2 / (10,000,000 m)
aCent = (2.5*10^7 m^2/s^2) / (10^7 m)
aCent = 2.5 m/s^2
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Question: `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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Your solution:
aCent = v^2 / r = (10,000 m/s)^2 / (10,000,000 m)
aCent = (10^8 m^2/s^2) / (10^7 m)
aCent = 100 m/sec^2
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Question: `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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Your solution:
v = sqrt( aCent * r )
v = sqrt( 4 m/s^2 * 10,000,000 m )
v = sqrt( 40,000,000 m)
v = 6.3 * 10^3 m/s
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Question: `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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Your solution:
F = G m1 m2 / r^2
Fgrav = (6.67 * 10^-11 N m^2/kg^2) * (6 * 10^24 kg) * (5 kg) / (10,000,000 m)^2
Fgrav = 20 Newtons
a = Fgrav / m
a = 20 Newtons / 5 kg
a = 4 m/s^2
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Question: `q008. What therefore will be the orbital velocity of the 5-kg piece of junk?
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Your solution:
v = `sqrt( a * r)
v = `sqrt( 4 m/s^2 * 10,000,000 m)
v = sqrt (4 * 10^7 m^2/s^2)
v = 6.3 * 10^3 m/s
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Question: `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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Your solution:
Fgrav = G * mEarth * mObject / r^2
a = Fgrav / mObject
a = (G * mEarth * mObject / r^2) / mObject
a = G * mEarth / r^2
From this equation, the acceleration of all objects at a distance of 10,000 km from the center of the Earth comes from the fact that the gravitational force must be the same.
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Question: `q010. How much work would have to be done against gravity to move the 3000 kg satellite from a distance of 10,000 km to a distance of 10,002 km?
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Your solution:
2 km = 2000 m
dW = F * `ds
dW = 12,000 Newtons * 2000 m
dW= 24,000,000 J
dW = 2.4 * 10^7 J
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Question: `q011. Does it therefore follow that the work done to move a 3000 kg satellite from a circular orbit at distance 10,000 km to a circular orbit at distance 10,002 km from the center of the Earth must be 24,000,000 Joules?
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Your solution:
No, because it violates the requirements for the conservation of energy theory. dKE of the center of the Earth is -24,000,000 J and according to the rule of conservative energy (dW_nc_ON = `dPE + `dKE), the dW_nc_ON will be less than the dPE of the center of the Earth, instead of being greater.
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If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
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Question: `q012. What would be the gravitational attraction between the Earth and a 1200 kg satellite, if the satellite was in a circular orbit 8000 km from the center of the Earth?
What orbital velocity would be required so that the centripetal force on the satellite was equal to the gravitational attraction?
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Your Solution:
v = sqrt((6.67 *10^-11N m^2/kg^2) (6 *10^24 kg) / (8,000,000 m)^2)
v = sqrt((4.002 * 10^14N m^2/kg) / 6.4 * 10^13 m^2)
v = sqrt(6.253N / kg)
v = 2.50 m/s
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Question: `q013. By how much would the PE of the satellite in the preceding question change if its orbit was changed to a circular orbit 7900 km from the center of the Earth?
By how much would the KE of the satellite change?
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Your Solution:
PE = - G M m / r
PE = - (6.67 * 10^-11 N m^2 / kg^2)(6 * 10^24 kg)(1200 kg)/(7,900,000 m)
PE = - (4.002 * 10^14 N m^2/kg)(1200 kg)/(7,900,000 m)
PE = - (4.802 * 10^17 N m^2)/(7,900,000 m)
PE = - 6.07 * 10^10 J
KE = G M m/ r *1/2
KE = 6.07 * 10^10 J / 2
KE = 3.04 * 10^10 J
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