course PHY 121 1/7 027. `query 27 Question: Knowing the 9.8 m/s^2 gravitational field strength of the Earth's field at the surface of the Earth, and knowing the radius of the Earth, how do we find the gravitational field strength at a given distance 'above' the surface of the Earth?
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Given Solution: `a** You have an inverse square force. Square the ratio of Earth radius to orbital radius and multiply by 9.8 m/s^2: Field strength=(Re/r)^2*9.8m/s^2 ** STUDENT COMMENT: I used G M_earth / r^2. Wonder if they get the same result. INSTRUCTOR RESPONSE: For r > Re, the expressions G M_earth / r^2 and (Re/r)^2*9.8m/s^2 give the same results, to the number of significant figures dictated by the known quantities. The first formula is inherently more accurate, because the radius of the Earth is not the same at the poles as at the equator, with the result that Re is not known as precisely as G and the mass of the Earth. STUDENT QUESTION: ???? I don't think I understand the answer. Is mine correct? ????? INSTRUCTOR RESPONSE: Your method is completely equivalent to this one, though you didn't actually show the expressions you would get • a_grav = k / r^2, and a_grav = g =9.8 m/s^2 when R = R_e, the radius of the Earth. Thus • 9.8 m/s^2 = k / R_e^2 and • k = 9.8 m/s^2 * R_e^2. The proportionality becomes • a_grav = 9.8 m/s^2 * R_e^2 / r^2 = 9.8 m/s^2 * (R_e / r)^2. Any proportionality of the form • y = k x^p implies that if y1 = k x1^p and y2 = k x2^p, we have • y2 / y1 = k x1^p / (k x2^p) = (x1 / x2)^p. In the current case p = -2, the y quantity would be the acceleration of gravity and the x quantity would be the distance from the center of the Earth. Using y1 = g and x1 = R_e, with y2 = a_grav and x2 = r, we have • a_grav / g = (r / R_e)^(-2) so that • a_grav = g * (R_e / r)^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ---- ********************************************* Question: `qIf we double our distance from the center of the Earth, what happens to the gravitational field strength we experience? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since the field strength is inversely proportional to distance, we would feel 1/(2)^2 or 1/4th the field strength. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** We have an inverse square force so if r2 = 2 * r1 the ratio of the gravitational field will be g2 / g1 = (1 / r2^2) / (1 / r1^2) = r1^2 / r2^2 = (r1 / r2)^2 = (r1 / (2 * r1))^2 = r1^2 / 4 r1^2 = 1/4. In a nutshell double the radius gives us 1 / 2^2 = 1/4 the gravitational field. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ---- ********************************************* Question: `qHow do we approximate the energy required to move a given mass from the surface of the Earth to a given height 'above' the Earth, where the field strength at the given height differ significantly from that at the surface? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Wouldn’t we use the same equation as before? Field strength=(Radius ‘at’earth/radius off earth)^2*9.8m/s^2. Then you could use `dKE (or `dWnet) = Fnet *`ds to find the energy needed. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a STUDENT SOLUTION AND INSTRUCTOR RESPONSE: mass*[(Re + distance)/Re]^2=force Force*distance=KE INSTRUCTOR RESPONSE: The first approximation would be to average the force at the surface and the force at the maximum altitude, then multiply by the distance. The result would give you the work necessary to 'raise' the object against a conservative force, which would be equal to the change in PE. STUDENT COMMENT: OK. `dPE = F_net * `ds INSTRUCTOR RESPONSE: PE = F_cons_BY * `ds, where F_cons_BY is the force exerted BY the system against conservative forces. There could be nonconservative forces present; they would have no effect on the PE change but would have to be included in F_net, so F_net * `ds wouldn't be correct. STUDENT COMMENT: im not sure i understand this answer I know I need to find KE and then the different PE from the min and max altitudes but I dont know if my equation represents the same one shown here INSTRUCTOR RESPONSE: No KE difference is assumed. We are looking only at the work required to 'lift' the object without speeding it up. Energy is required because, in order to 'lift' a mass, a force is required to counter the gravitational pull of the Earth. If we can find the average force required, we need only multiply by the distance. This only gives an approximation (see more about this below). The work done against gravity is done against a conservative force, and is therefore equal to the change in gravitational PE. You can't use the book's m g y formula because in this case the acceleration of gravity changes significantly from the initial point to the final point. The work required is equal to the area beneath the graph of F vs. r. The graph is decreasing at a decreasing rate (i.e., decreasing and concave up), and is asymptotic to the positive r axis. See the Introductory Problem Sets for worked problems of this nature. The approach used there is to approximate the force at the two distances from the center of the Earth, average the two and use this as an approximation to the average force. This approximation is accurate only to the extent that the slope of the F vs. r graph is constant. For a given mass m we have F = m * (r_earth / r)^2 * 9.8 m/s^2; simce m * 9.8 m/s^2 is the weight at the surface of the earth we could write this as F = weight_surface * (r_earth / r)^2, where weight_surface is the weight of the mass at the surface of the Earth.. Either way, this expression gives the force at distance r from the center of the Earth. So you would plug in the initial distance from the center of the Earth, and the final distance, obtaining two values for the force. Averaging these two values you would have an approximate value for the average force, which would then be multiplied by the distance to get the approximate work. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ---- ********************************************* Question: `qQuery class notes #24 Describe the paths of various particles 'shot' parallel to the surface of the Earth from the top of a very high tower, starting with a very small velocity and gradually increasing to a velocity sufficient to completely escape the gravitational field of the Earth. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The particles shot at a very small velocity would succumb to gravity and eventually land on the surface of the earth because their initial velocity wasn’t enough to get them going into an ‘orbit’. The higher velocity particles would orbit around the earth once they reach a certain point in the atmosphere where they are comfortable in their relationship to the gravitational pull of the earth (I can’t think of a better way to phrase that). The really high velocity particles would overcome the gravitational field until they reached a point in “space” where their relationship with earth matched in terms of distance and mass proportions. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aGOOD STUDENT ANSWER: Each particle sets out to follow an orbit around the center of mass of the earth. But for particles shot at slower speeds, this path is interupted by the surface of the eath and simply stops there. The faster it is shot, the further x distance becomes before the particle lands. However, if it given a great enough velocity, it will fall around the curviture of the earth. If is shot even faster than that, it will follow an eliptical oribit with varying speeds and distances from center of earth. GOOD STUDENT ANSWER: With a very low velocity the projectile will not traveled as far. It will fall to earth in a nearly parabolic fashion since it gains vertical velocity as it travels horizontally at a steady pace. If the projectile is fired at a very strong velocity it will leave the earths vacinity but will still be pulled by the forces acting on it from the earths center. This will cause it to go only so far at which point it has slowed down considerabley, since it has lost most of its kinetic energy. It turns and begins to gain energy as it approaches the earths area, using the potential energy it gained on the trip out. (Causing it to speed up). The path that this projectile will take will be eliptical, and it will continue to loop around the earth. If the projectile is fired at the correct velocity to form a circular orbit, it will also fall at a parabolic fashion, although the earth's surface will also be descending at the same rate so that the object will appear to be 'not falling'. It is falling but at the same rate the earth is 'falling' under it. It will circle the earth until something causes it to stop. INSTRUCTOR RESPONSE: The path of the projectile will always be an ellipse with the center of the Earth at one focus. For low velocities and low altitude this path is very nearly parabolic before being interrupted by the surface of the Earth. One of these ellipses is a perfect circle and gives us the circular orbit we use frequently in this section. ** STUDENT COMMENT: I thought this object left the earths force...if not i understand it will fall back to earth as a normal projectile because of a small velocity and have a parabolic path because gravity will take over
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Given Solution: `a** For a given distance from the center of the Earth, there is only one velocity for which centripetal acceleration is equal to gravitational acceleration, so there is only one possible velocity for a circular orbit of given orbital radius. The orbital radius is determined by the height of the 'tower', so for a given tower there is only one velocity which will achieve a circular orbit. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ---- ********************************************* Question: `qIs it necessary in order to achieve a circular orbit to start the object out in a direction parallel to the surface of the Earth? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes; If you shoot the object at an angle, it’s relationship to the radius of the earth constantly changes and will quickly be knocked out contention of a circular orbit. Like the earth would try to correct it at the same time the object is trying to correct itself and the oppositional forces would be overkill, so it would miss its chance to orbit. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** If you have just one 'shot' then you must start out parallel to the surface of the Earth. The reason is that any circle about the center must be perpendicular at every point to a radial line--a line drawn from the center to the circle. Any radial line will intercept the surface of the Earth and must be perpendicular to it, and the circular orbit must also be perpendicular to this line. Therefore the orbit and the surface are perpendicular to the same line and are therefore parallel. ** STUDENT COMMENT: im now seeing the idea of a launch paralell to earth. the tower needs to be pretty tall then. the velocity needs to be headed in the direction of the orbit it will take. INSTRUCTOR REPSONSE: The main reason it has to be very tall is because of the atmosphere. If you shoot the projectile with sufficient speed while it's in the atmosphere, it will quickly lose most of its kinetic energy to air resistance; the energy goes mostly into heating the object, which as a result proceeds to melt as its orbit decays. Whether it melts before hitting the ground or not depends on how quickly the orbit decays and how high it was in the first place. If there was no atmosphere, you still would need to be careful about the flattening of the Earth at the equator (the radius at the equator is about 20 km greater than the radius at the poles, which means that if you wanted an orbit that took you over the equator, then even in the absence of atmosphere a tower at the pole would have to be at least 20 km high. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ---- ********************************************* Question: `qPrinciples of Physics and General College Physics Problem 5.2: A jet traveling at 525 m/s moves in an arc of radius 6.00 km. What is the acceleration of the jet? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 6km * 1000m/km = 6000m. A = v^2 /(r) = (525m/s^2)^2 / (6000m) = 3,150,000m/s^2. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe jet will have centripetal acceleration a_cent = v^2 / r, where v is its speed and r the radius of the circle on which it is traveling. In this case we have v = 525 m/s and r = 6.00 km = 6000 meters. The centripetal acceleration is therefore a_cent = v^2 / r = (525 m/s)^2 / (6000 m) = 45 m/s^2, approx.. One 'g' is 9.8 m/s^2, so this is about (45 m/s^2) / (9.8 m/s^2) = 4.6 'g's'. STUDENT QUESTION: there are roughly 1600m in a mile I believe, so 45m/s would be 2700m a minute and 162,000m/hr. so that would be roughly 100mi/hour, am I making a math error here or is that speed really 4.6g’s??? INSTRUCTOR RESPONSE: That speed would be about 100 mph. However 100 mph is not an acceleration, but as you say, a speed. g's measure acceleration, not speed. Nothing here is moving at 45 m/s. There is a centripetal acceleration of 45 m/s^2. One 'g' is 9.8 m/s^2. So 45 m/s^2 is somewhat more than 4 'g's'. It is in fact 4.6 g's &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: "