course Phy201
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Given Solution: 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. SLIGHT ERROR IN STUDENT SOLUTION: F = 6.67e^-11Nm^2/kg^2 * 6E^24 *3000kg /10000km^2 F = 1.2 E10N INSTRUCTOR'S CORRECTION: You have m^2 in the numerator, km^2 in the denominator. These units don't divide out. To agree with the units of the universal gravitational constant G, you need to expres r in meters. That's the only discrepancy between your solution and the correct solution-- everything else is just as it should be. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok I forgo th the constant = 6.67*10^-11N M^2/kg^2 *6*10^24kg * 3000kg/(10,000,000)^2 =12,000N I understand this ------------------------------------------------ Self-critique rating #$&*ent:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q002. What force would the same satellite experience at the surface of the Earth, about 6400 km from the center. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: g= 6.67*10^-11N m^2/kg^2 * (6*10^24 kg * 3000kg)/(6,400,000km)^2= 29,311.52N confidence rating #$&*2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT SOLUTION USING PROPORTION (correct except proportion is reversed): 10000/6400 = 1.56 = 1.6 F = 12000 * (1/1.6^2) F = 4687.5N INSTRUCTOR CORRECTION: You made excellent use of ratios. Your only error was that the ratio was 'upside down'. The force varies as 1 / r^2. If r1 and r2 are two radii, then the force ratio would be F2 / F1 = (1 / r2)^2 / (1 / r1)^2 = r1^2 / r2^2 = (r1 / r2)^2. So if F2 is the force at the 6400 km distance, we would have F2 / F1 = (r1 / r2)^2 = (10 000 km / (6400 km) )^2 = 1.6^2 = 2.6, approx, and the force would be about 2.6 * 12000 N = 30,000 N. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok I understand the f= g* M1 * m2 /r^2 way better ------------------------------------------------ Self-critique rating #$&*ent:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q003. What would be the acceleration toward the center of the Earth of the satellite in the previous two questions at the distance 10,000 km from the center of the Earth? We may safely assume that no force except gravity acts on the satellite. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F= 12,000N M= 3000kg a= 12000N/3000kg = 4m/s^2 confidence rating #$&*2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok I got it ------------------------------------------------ Self-critique rating #$&*ent:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q004. The centripetal acceleration of an object moving in a circle of radius r at velocity v is aCent = v^2 / r. What would be the centripetal acceleration of an object moving at 5000 m/s in a circular orbit at the distance of 10,000 km from the center of a planet, and how this this compare to the 4 m/s^2 acceleration net would be experienced by an object at this distance from the center of the Earth? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v= 5000m/s r= 10,000km aCent= v^2/r aCent= (5000m/s)^2/10,000,000m =2.5 m/s^2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ????????????????????????????Ok, so does this mean that as the object orbit around the centripetal acceleration is less than the acceleration if the object were to be falling toward the center of the earth?????????????????????????????????????????????????? I guess that would make sense. ------------------------------------------------ Self-critique rating #$&*ent:3 ********************************************* Question: `q005. What would be the centripetal acceleration of an object moving at 10,000 m/s in a circular orbit at the distance of 10,000 km from the center of a planet, and how does this compare to the 4 m/s^2 acceleration that would be experienced by an object at this distance from the center of the Earth? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: aCent= v^2/r v= 10,000 r= 10,000km aCent= (10,000)^2/ 10,000,000m=10m/s^2 it is much higher, so as the velocity doubled the centripetal acceleration almost doubled. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok, I understand this ------------------------------------------------ Self-critique rating #$&*ent:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q006. An object will move in a circular orbit about a planet without the expenditure of significant energy provided that the object is well outside the atmosphere of the planet, and provided its centripetal acceleration matches the acceleration of gravity at the position of the object in its orbit. For the satellite of the preceding examples, orbiting at 10,000 km from the center of the Earth, we have seen that the acceleration of gravity at that distance is approximately 4 m/s^2. What must be the velocity of the satellite so that this acceleration from gravity matches its centripetal acceleration? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: aCent=v^2/r 4m/s^2= v^2/10,000,000m 40,000,000m^2/s^2= V^2 v= 6324.56m/s confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok i got that ------------------------------------------------ Self-critique rating #$&*ent:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q007. The orbital velocity of a satellite in a circular orbit is that velocity for which the centripetal acceleration of the satellite is equal to its gravitational acceleration. The satellite in the previous series of examples had a mass of 3000 kg and orbited at a distance of 10,000 km from the center of the Earth. What would be the acceleration due to Earth's gravity of a 5-kg hunk of space junk at this orbital distance? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F= 6.67 *10^-11 N m^2/kg^2 * (6*10^24kg * 5kg)/(10,000,000m)^2= 20.01N F= M*a a=20.01N/5kg= 4.0m/s^2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok I understand this ------------------------------------------------ Self-critique rating #$&*ent:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q008. What therefore will be the orbital velocity of the 5-kg piece of junk? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Acent= v^2/r 4.0m/s^2= v^2/10,000,000m v=6324.56m/s confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand this ------------------------------------------------ Self-critique rating #$&*ent:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q009. Is it true that the gravitational acceleration of any object at a distance of 10,000,000 meters from the center of the Earth must be the same as for the 3000-kg satellite and the 5-kg hunk of space junk? (Hint: We have to find the acceleration for any mass, so we're probably going to have to let the mass of the object be represented by symbol. Use mObject as a symbol for the mass of the object. While dealing in symbols, you might as well leave G and r in symbols and let mEarth stand for the mass of the Earth. Find an expression for the force, then using this expression and Newton's Second Law find an expression for the acceleration of the object). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes it is true that the gravitational acclereration of any object at at distance of 10,000,000 meters from the centr of the earth will be the same whether it is 5kg or 3000kg. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok, I understand this. ------------------------------------------------ Self-critique rating #$&*ent:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q010. How much work would have to be done against gravity to move the 3000 kg satellite from a distance of 10,000 km to a distance of 10,002 km? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f= 12000N dw= fnet * 'ds 'dw= 12000N * 2000m 'dw= 24,000,000J confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 mathematically, 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. ** STUDENT QUESTION They ask what the work is to move from 10,000 km to 10,002 km. The force in between here is 12006 - 12002 = 4 N. Why do you use the whole 12000N in the work equation? Shouldn't it be W = 4* 2000? If it was asking for the work to move from 0 to 10,000 km, then I would understand the use of 12000N as the force, but the net force is only 4 N. INSTRUCTOR RESPONSE Work is average force * displacement, not change in force * displacement. In symbols `dW = F * `ds, not `dF * `ds. If you drive up the road for 4 hours, gradually increasing your speed from 60 mph to 65 mph, you travel between 240 and 260 miles, not (65 mph - 60 mph) * 4 hours = 20 miles. This situation is completely analogous. You have to exert a force which is never under 12 000 N, and do so for 2000 meters. Your approximation will use the average force 12 004 N. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok, Iunderstand this. ------------------------------------------------ Self-critique rating #$&*ent:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: `q011. Does it therefore follow that the work done to move a 3000 kg satellite from a circular orbit at distance 10,000 km to a circular orbit at distance 10,002 km from the center of the Earth must be 24,000,000 Joules? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes it does follow that confidence rating #$&*2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: It might seem so, but this is not the case. Gravity does -24 000 000 J of work (force exerted by gravity is in the direction opposite the displacement from 10 000 km to 10 002 km so work by gravity is negative) on the satellite, so its PE increases by 24 000 000 J. However when changing orbits the satellite also slows, so it loses some of its KE. The lost KE is converted into gravitational PE. So the force required to change the orbit does less work than it would if we simply changed the distance from the center of the planet. Conservation of energy tells us that `dW_nc_ON = `dPE + `dKE, where `dW_nc_ON is the nonconservative force acting on the system. Since `dKE is negative, the work done by the nonconservative force is less that the 24 000 000 J change in PE. STUDENT COMMENT That is very confusing. I do not really understand this INSTRUCTOR RESPONSE (Remember to be specific about what you do and do not understand.) `dPE is equal and opposite to the work done by gravity on the satellite, so is equal to the work required to counter gravity as the satellite moves 2 km further from the Earth. Over this 2 km distance the force stays close to 12,000 N. `dW = F `ds = 12,000 N * 2 km = 12,000 N * 2000 m = 24,000,000 J. This force would be in the direction of motion, so the work is positive. Therefore `dPE = +24,000,000 J. KE = 1/2 m v^2, which decreases because the speed of the satellite decreases. Thus `dKE is negative. We could calculate the velocity of a circular orbit at each distance. If we did this would find that `dKE = -12,000,000 J, half the magnitude of the PE change and of opposite sign. Recall that conservation of energy requires that `dW_noncons_ON = `dPE + `dKE. In the present example, `dPE + `dKE = 24,000,000 J - 12,000,000 J = 12,000,000 J. It follows that `dW_noncons_ON = 12,000,000 J. That is, to move from the first circular orbit to the second, we would require 12,000,000 J of work from a nonconservative source (e.g., from the satellite's thrusters; it takes fuel to change the orbit of the satellite). This situation generalizes. The KE change from one circular orbit to another orbit of greater radius always has half the magnitude of the PE change, and the opposite sign. The work required from nonconservative forces is therefore half the magnitude of the PE change. STUDENT COMMENT since we are moving further away from the earths pull there is less KE needed? So there would be less work required at 10002 vs 10000 and the same would be for less work at 10004 vs 10002 INSTRUCTOR RESPONSE The force exerted by gravity doesn't change much between r = 10000 km and r = 10004 km, so it's easy to calculate a very close approximation to the PE change--just multiply the (nearly constant) force by the additional distance you need to move away from the planet. The velocity required for a circular orbit is v = sqrt( G M / r), obtained by setting centripetal force equal to gravitational force and solving for v. It's clear from the fact that r is in the denominator that v is smaller when r is larger, so the orbit with greater radius has the smaller velocity and hence the lesser KE. If you can accurately calculate 1/2 m v^2 for both orbits (possible if you use enough significant figures to make the difference in KE accurate to at least a few significant figures), you will find that the decrease in KE is half as great as the increase in PE. Note also that gravitational PE can be calculated using the equation PE = - G M m / r. If you evaluate this for both values of r, and find the difference (again being careful to use plenty of significant figures), you will find that the PE gain agrees with your previous result, and is double the KE loss. Finally, it is possible to do the algebra with the symbols and show that the KE loss is exactly half the PE gain: v = sqrt( G M / r^2) so KE = 1/2 m v^2 = 1/2 G M m / r^2 PE = - G M m / r Except for the - sign, the only difference in the KE and PE expressions is the 1/2. So the magnitude of the PE is always double the magnitude of the KE, and the change in PE always has double the magnitude of the change in KE. STUDENT QUESTION Let me know if this makes sense. I was a bit confused and this is what I took from it. INSTRUCTOR RESPONSE: Overview of the solution: The KE is determined by the orbital radius and the mass of the satellite, so `dKE is determined by the mass and the two orbital radii. The work done against gravity to move from one orbital radius to another is `dPE, and is found by multiplying F_ave * dist, which is F_ave * `dr. When these two sets of calculations are done, we find the | `dKE | = 1/2 | `dPE |. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ?????????????So is this just saying that the KE that we calculated to be 24,000,000 is not totally the amount joules nned to move the force and that some of this is lost to the gravitational PE, making the 24,000,000 much less. ?????????????????Is this because the gravitational force is lessened a bit at the object travles further from the earth, which is why the gravitational PE of the earth is increased at the KE decreases.??????????????????????????????????????????????