course phy 201 I am still a little confus. with some of the equations. In order to find the grav. force of 2 objects on each other we will used F=GMm/r^2 which is the same as F = ma where a = v^2/r? But if we want to find the gravit. field on an object within earths field we use g= 9.8m/s/s *rE/r ^2 but this is not a Force...so again we must use F = ma=mg?
......!!!!!!!!...................................
17:55:52 `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?
......!!!!!!!!...................................
RESPONSE --> m1 = 6*10^24kg m2 = 3000kg d = 10000km F =g*m1*m2/r^2 F = 6.67e^-11Nm^2/kg^2 * 6E^24 *3000kg /10000km^2 F = 1.2 E10N confidence assessment: 3
.................................................
......!!!!!!!!...................................
18:00:02 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.
......!!!!!!!!...................................
RESPONSE --> 10000km * 1000 = 1.0e^7 F = 6.67*10^-11N m^2/kg^2 *6*10^24 *3000/(10*10^7)^2 F = 12000N self critique assessment: 2
.................................................
......!!!!!!!!...................................
18:02:06 `q002. What force would the same satellite experience at the surface of the Earth, about 6400 km from the center.
......!!!!!!!!...................................
RESPONSE --> 10000/6400 = 1.56 = 1.6 F = 12000 * (1/1.6^2) F = 4687.5N confidence assessment: 3
.................................................
......!!!!!!!!...................................
18:06:44 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.
......!!!!!!!!...................................
RESPONSE --> I dont understand why I couldnt find the ratio of change and use it as the previous assignment ? F = 6.67E^-11 * 6E^24 * 3000/ 6400000m F = 29000 F=ma F = M * g at earth surface = 3000 * 9.8 = 29400 self critique assessment: 2
.................................................
......!!!!!!!!...................................
18:08:27 `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.
......!!!!!!!!...................................
RESPONSE --> F = ma a = F/m a = 12000/3000 a = 4m/s confidence assessment: 2
.................................................
......!!!!!!!!...................................
18:08:37 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
18:16:44 `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?
......!!!!!!!!...................................
RESPONSE --> a = v^2/r a = 5000m/s ^2 /1.0E^7m a = 2.5 vs a = 4 there is a 1.6 differece in the centripetal accel vs. the acceleration towards earth of an object. This makes sense bec. in cent. acceleration the perpindicular and parallel components are constantly pulling the object toward the center of the earth and around it vs. the acceleration towards the center of the earth due to gravity . Also since the centrip. accel. is for an object in motion it is not being pulled closer and closer like the falling object. confidence assessment: 2
.................................................
......!!!!!!!!...................................
18:17:02 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
18:18:40 `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?
......!!!!!!!!...................................
RESPONSE --> a = v^2/r a = 10000m/s ^2 / 1.0E67 a = 10m/s This is more which is understandable because the velocity is greater. confidence assessment: 3
.................................................
......!!!!!!!!...................................
18:18:50 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
18:29:18 `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?
......!!!!!!!!...................................
RESPONSE --> d = 1E^7m a = 4m/s/s m = 3000 v^2/r = 9.8m/s^2 * (rE/r)^2 *r v^2 = 9.8m/s^2 * rE^2/r v = sqrt 9.8m/s/s * (6.4*10^6 ^2 /1E^7) v = 6335m/s confidence assessment: 2
.................................................
......!!!!!!!!...................................
18:32:01 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.
......!!!!!!!!...................................
RESPONSE --> I had assumed I could not use the centrip. accel equation because the object was not in orbit however this way was much easier. self critique assessment: 3
.................................................
......!!!!!!!!...................................
18:40:55 `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?
......!!!!!!!!...................................
RESPONSE --> 5kg d = 10000km F = gm1*m2 /r^2 F = 21.3 a = 21.3/5 a = 4.3 m/s/s confidence assessment: 3
.................................................
......!!!!!!!!...................................
18:44:40 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.
......!!!!!!!!...................................
RESPONSE --> gravity is constant!!!! the force will change due to the mass of the object however the accel will not. I notice these answers were almost the exact same and it is interesting to realize the reason behind this concept self critique assessment: 2
.................................................
......!!!!!!!!...................................
18:47:07 `q008. What therefore will be the orbital velocity of the 5-kg piece of junk?
......!!!!!!!!...................................
RESPONSE --> a = v^2/r v = sqrt a*r v = sqrt 4m/s * 10000000 v = 6324 same as the other object. confidence assessment: 2
.................................................
......!!!!!!!!...................................
18:47:24 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
18:58:36 `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).
......!!!!!!!!...................................
RESPONSE --> mObj G r mEarth a= f/m a = G*mEarth*mObj/r^2 /m confidence assessment: 1
.................................................
......!!!!!!!!...................................
19:00:48 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.
......!!!!!!!!...................................
RESPONSE --> the constant G will alway have an effect on the mEarth there fore giving us a constant GmEarth self critique assessment: 2
.................................................
......!!!!!!!!...................................
19:04:19 `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?
......!!!!!!!!...................................
RESPONSE --> W = F * ds W = 6300N * 2000m W = 1.26E^7 confidence assessment: 2
.................................................
......!!!!!!!!...................................
19:13:26 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. **
......!!!!!!!!...................................
RESPONSE --> In order to move 2000m the object has to do work agains the pull of earths gravity. I understand this but used the F = 6300N in the last problem instead of 12000N that was my only mistake self critique assessment: 2
.................................................
......!!!!!!!!...................................
19:15:12 `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?
......!!!!!!!!...................................
RESPONSE --> W = F *ds No; the force is greater therefore there will be greater work required to move this object the same distance. confidence assessment: 2
.................................................
......!!!!!!!!...................................
19:25:01 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)
......!!!!!!!!...................................
RESPONSE --> 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
.................................................
course phy 201 Again I am still having trouble with what equations to use...for PE change we would use the equation PE = Gmm/r1 - Gmm/r2 same as the Force equation or do we just find dPE by grav. Force * ds both are listed in my notes. As far as KE I assume m*9.8m/s/s *rE/r^2 ¦¨ÖHÌ¡½Š§ÚÑ›}{^²ÔçÄÛ
......!!!!!!!!...................................
20:31:10 Query class notes #26 Explain how we use proportionality along with the radius rE of the Earth to determine the gravitational acceleration at distance r from the center of the Earth to obtain an expression for the gravitational acceleration at this distance. Explain how we use this expression and the fact that centripetal forces is equal to v^2 / r to obtain the velocity of a satellite in circular orbit.
......!!!!!!!!...................................
RESPONSE --> F = 9.8m/s/s * (rE/r)^2 = v^2/r = a we find the ratio of the distance between earth and given r since gravit. field is equal to centrip. accel we set the equations equal to find v v^2/r = 9.8m/s/s *(rE/r)^2 v = sqrt 9.8m/s/s *rE^2/r
.................................................
......!!!!!!!!...................................
20:33:55 ** The proportionality is accel = k r^2. When r = rE, accel = 9.8 m/s^2 so 9.8 m/s^2 = k * rE^2. Thus k = 9.8 m/s^2 / rE^2, and the proportionality can now be written accel = [ 9.8 m/s^2 / (rE)^2 ] * r^2. Rearranging this gives us accel = 9.8 m/s^2 ( r / rE ) ^2. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
20:39:24 Principles of Physics and Gen Phy problem 5.30 accel of gravity on Moon where radius is 1.74 * 10^6 m and mass 7.35 * 10^22 kg.
......!!!!!!!!...................................
RESPONSE --> r = 1.74E6m m = 7.35 E22kg F = 9.8m/s/s * (6.4E6 /1.74E6m)^2 F = 133 F = ma a = 133N / 1.74E22 a = 7.64E-21
.................................................
......!!!!!!!!...................................
20:48:42 ** The acceleration due to gravity on the Moon is found using the equation g' = G (Mass of Moon)/ radius of moon ^2 g' = (6.67 x 10^-11 N*m^2/kg^2)(7.35 X 10^22 kg) / (1.74 X 10^6 m) = 1.619 m/s^2 **
......!!!!!!!!...................................
RESPONSE --> I used equation for grav. field of the earth which was inapp. for this problem leading to the wrong equation. Since the centripetal force = gravitational force I should have assumed v^2/r = Gmm/r^2 since a = v^2/r then a = Gmm/r^2 6.67E-11 * 7.35E22 /1.74E6^2 a = 1.61ms/s
.................................................
......!!!!!!!!...................................
20:57:19 Query gen phy problem 5.40 force due to planets (Mv, Mj, Ms, are .815, 318, 95.1 Me; orb radii 108, 150, 778, 1430 million km). What is the total force on Earth due to the planets, assuming perfect alignment?
......!!!!!!!!...................................
RESPONSE --> .815 318 95.1 F1 = 6.67E-11 * 6.64E24 .815E24 /108mill km ^2 F2 = 6.67e-11 *6.64E24* 318E24 /150mill km^2 F3 = 6.67E-11 *6.64E24* 95.1E24 / 778millkm^2 Total F = F1+F2+F3 I think I have the right set up I am just not sure about the mechanics
.................................................