#### Transcript Fields - Univerzita Karlova v Praze

Fields Essential idea: Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields. Nature of sciences: Paradigm shift: The move from direct, observable actions being responsible for influence on an object to acceptance of a field’s “action at a distance” required a paradigm shift in the world of science. Understandings: • Gravitational fields • Electrostatic fields • Electric potential and gravitational potential • Field lines • Equipotential surfaces Applications and skills: • Representing sources of mass and charge, lines of electric and gravitational force, and field patterns using an appropriate symbolism • Mapping fields using potential • Describing the connection between equipotential surfaces and field lines Guidence: • Electrostatic fields are restricted to the radial fields around point or spherical charges, the field between two point charges and the uniform fields between charged parallel plates • Gravitational fields are restricted to the radial fields around point or spherical masses and the (assumed) uniform field close to the surface of massive celestial bodies and planetary bodies • Students should recognize that no work is done in moving charge or mass on an equipotential surface Data booklet reference: • W = q ΔVe • W = m ΔVg Theory of knowledge • Although gravitational and electrostatic forces decrease with the square of distance and will only become zero at infinite separation, from a practical standpoint they become negligible at much smaller distances. How do scientists decide when an effect is so small that it can be ignored? Utilization: • Knowledge of vector analysis is useful for this sub-topic (see Physics sub-topic 1.3) Aims: • Aim 9: models developed for electric and gravitational fields using lines of forces allow predictions to be made but have limitations in terms of the finite width of a line. Topic 10: Fields 10.1 Describing fields Gravitational fields The Earth has a radial field of gravity, which means that the gravitational field is circular and acts from the centre point. You can see on the diagram that near the Earth's surface the lines are closer together than higher up. The closeness of the lines represent the relative strength of the field, so from the diagram, you can tell that the strength of the field decreases with altitude. Further apart lines represent points where the field is weaker. Topic 10: Fields 10.1 Describing fields Gravitational fields • A uniform field, however, has the lines perfectly parallel. The Earth's gravitational field can be considered to be uniform on the scale of small things such as cars, balls, and planes. • For small heights at this scale (a few dozen kilometres), the strength of the field doesn't change enough to be noticeable. Again, the arrows point towards the centre of the Earth, since that is the way objects fall.. The Earth's gravitational field is represented by parallel lines on small scales. Topic 10: Fields 10.1 Describing fields Gravitational fields The normal force, the force, tension drag FYI: School A must assume thatfriction the force "signal" travels infinitely and fast, to are all classified as contact forces, occurring where react to relative motion between the masses. two objects contact each other. FYI: gravitational School B assumes that the forceand "signal" already in place in the space The force, theis electric force, surroundingdo thenot source of theobjects force. Thus force does NOT have to however, need tothebe insignal contact (or even keep traveling to the receiving mass. close proximity). These two forces are sometimes called action a distance FYI: Relativity has at determined that the forces. absolute fastest ANY signal can There are schools ofspeed thought 8 m/s, the propogate is ctwo = 3.010 of light.on action at a distance. School A: The masses know where each other are at all times, and the force is instantaneously felt by both masses at all times. School B: The masses deform space itself, and the force is simply a reaction to the local space, rather than the distant mass. Topic 10: Fields 10.1 Describing fields Gravitational fields Thus School B is currently the "correct" view. If, for example, the gravitational forces were truly "action at a distance," the following would occur: SUN c c The planet and the sun could not "communicate" quickly enough to keep the planet in a stable orbit. Topic 10: Fields 10.1 Describing fields Gravitational fields Instead, look at the School B view of the space surrounding the sun: FYI: You have probably seen such a model at the museum. A coin is allowed to roll, and it appears to be in orbit. FYI: Of course, if the slope isn't just right to match the tangential speed, the coin will spiral into the central maw. The planet "knows" which way the the the to "roll" because of local curvature of space surrounding sun. Long distance communication is not needed. FYI: The view of School B is called the FIELD VIEW. Topic 10: Fields 10.1 Describing fields Gravitational fields Of course, if the mass isn't moving, it will roll downhill if placed on the grid: m0 m0 Depending on where we place the "test" mass m0, it will roll differently. Question: Why is the arrow at the location of the first test mass smaller than that at the second one? Topic 10: Fields 10.1 Describing fields Gravitational fields We can assign an arrow to each position surrounding the sun, representing the direction the test mass will go, and how big a force the test mass feels. We represent fields with vectors of scale length. In the case of the gravitational field, the field vectors all point toward the center: Topic 10: Fields 10.1 Describing fields Electrostatic fields If we view our gravitational field arrows from above, we get a picture that looks like this: What the field arrows tell us is the magnitude and the direction of the force on a particle placed anywhere in space: The blue particle will feel a "downward" force. The red particle will feel SUN FYI: We don't even have to draw the object that is creating the field. a "leftward" force whose magnitude is LESS than that of the blue particle. Topic 10: Fields 10.1 Describing fields Electrostatic fields FYI: Inward-pointing fields are called SINKS. Think of the field arrows as Consider, now, a water flowing into a hole. negative charge: have is an attractive force. FYI: Test charges are by convention POSITIVE. Therefore, field vectors around a charge show the direction a POSITIVE charge would want to go if placed in the field. For masses, all we SUN - All of the field If we place a very small arrows point inward. POSITIVE test charge in the vicinity of our negative charge, it will be attracted to the center: We can map out the field vectors just as we did for the sun. Topic 10: Fields 10.1 Describing fields Outward-pointing fields called SOURCES. thetwo field arrows as FYI: We call such a system anare ELECTRIC DIPOLE. A Think dipoleofhas poles water flowing a fountain. (charges) whichout areofopposite in sign. Electrostatic fields Question: T or F: Gravitational fields alwaysand sinks. FYI: The ELECTRIC DIPOLE consists ofare a source a sink. We can place a negative Question: is T orpositive. F: Electric fields are always sinks. here. charge charge Now suppose our + + - If we place a very small The fields of the two POSITIVE test charge in the vicinity of our positive charge, it will be repelled from the center: We can map out the field vectors just as we did for the negative charge. charges will interact: Keep in mind that the field lines show the direction a POSITIVE charge would like to travel if placed in the field. A negative charge will do the opposite: Topic 10: Fields 10.1 Describing fields Electrostatic fields You may have noticed that in the last field diagram some of our arrows were unbroken. We can draw solid field arrows as long as we note that THE CLOSER THE FIELD LINES ARE TO ONE ANOTHER, THE STRONGER THE FIELD. + - FYI: There are some simple rules for drawing these solid "electric lines of force." Rule 1: The closer together the electric lines of force are, the stronger the electric field. strong Rule 2: Electric lines of force originate on positive charges and end on negative charges. weak Rule 3: The number of lines of force entering or leaving a charge is proportional to the magnitude of that charge. Topic 10: Fields 10.1 Describing fields Field lines A convenient way of visualizing electric field patterns is to draw lines that follow the same direction as the electric field vector at any point. These lines, called electric field lines, are related to the electric field in any region of space in the following manner: – The electric field vector E is tangent to the electric field line at each point. – The number of lines per unit area through a surface perpendicular to the lines is proportional to the magnitude of the electric field in that region. Thus, E is great when the field lines are close together and small when they are far apart. FYI: The same thing pays for field lines of gravitational field. Question: Which charge is the strongest negative one? Topic 10: Fields Question: Which charge is the weakest positive one? 10.1 Describing fields Question: Which charge is the strongest positive one? Electrostatic fields Question: Which charge is the weakest negative one? Example 1. D E A Determine the sign and the magnitude of the charges by looking at the electric lines of force. C A B C D E F Topic 10: Fields 10.2 Fields at work Potential and potential energy You can think of potential energy as the capacity to do work. And work is a force times a distance. Recall the gravitational force (from Newton): F = Gm1m2/r2 universal law of gravitation where G = 6.67×10−11 N m2 kg−2 If we multiply the above force by a distance r we get EP = -Gm1m2/r gravitational where G = 6.67×10−11 N m2 kg−2 potential energy FYI The actual proof is beyond the scope of this course. Note, in particular, the minus sign. Topic 10: Fields 10.2 Fields at work Gravitational fields EP = -Gm1m2/r gravitational where G = 6.67×10−11 N m2 kg−2 potential energy EXAMPLE 1. Find the gravitational potential energy stored in the Earth-Moon system. M = 5.981024 kg m = 7.361022 kg d = 3.82108 m SOLUTION: Use EP = -Gm1m2/r. EP = -Gm1m2/r = -(6.67×10−11)(7.36×1022)(5.98×1024)/3.82×108 = -7.68×1028 J. Topic 10: Fields 10.2 Fields at work Gravitational fields The previous formula is for large-scale gravitational fields (say, some distance from a planet). Recall the “local” formula for gravitational potential energy: local ∆EP ∆EP = mg∆y where g = 9.8 m s-2. The local formula treats y0 as the arbitrary “zero value” of potential energy. The general formula treats r = as the “zero value”. FYI The local formula works only for g = CONST, which is true as long as ∆y is relatively small (say, sea level to the top of Mt. Everest). For larger distances use ∆EP = -Gm1m2(1/rf – 1/r0). Topic 10: Fields 10.2 Fields at work Gravitational fields Recall the definition of work: work W = Fd cos where is the angle between the force F and the displacement d. Consider the movement of a ball from the ground to the table via displacements d1 and d2: Along the displacement d1, gravity does work given by W1 = mgd1 cos 180° = -mgd1. Along the displacement d2, gravity does work given by W2 = mgd2 cos 90° = 0. d2 d1 The total work done by gravity is thus Wg = -mgd1. Topic 10: Fields 10.2 Fields at work Consider the movement of a ball from the ground to the table via the 6 new displacements s: Along s1, s3 and s5 gravity does work given by Wg = -mgs1 + -mgs3 + -mgs5 = -mg(s1 + s2 + s3). Along s2, s4 and s6 gravity does no work. (Why?) Superimposing d1 and d2 from the previous path we see that Wg = -mg(s1 + s2 + s3) = -mgd1. This is exactly the same as we got before! s6 d2 s4 d1 s1 s2 s3 s5 FYI The work done by gravity is independent of the path followed. Topic 10: Fields 10.2 Fields at work We call any force which does work independent of path a conservative force. Thus gravity is a conservative force. FYI Only conservative forces have associated potential energy functions. Path 2 PRACTICE: Show that friction is not a conservative force. SOLUTION: Suppose you slide a crate across the floor along paths 1 and 2: Clearly the distance along Path 2 is greater than on Path 1. The work is different. Thus its work is not path-independent. Thus it is not conservative. A Path 1 B Topic 10: Fields 10.2 Fields at work If work is done in a conservative force field then there is an associated potential energy function. EXAMPLE: Show that for a conservative force ∆EP = -W = -Fd cos potential energy function where F is a conservative force SOLUTION: From conservation of mechanical energy we have ∆EP + ∆EK = 0 so that ∆EP = -∆EK. From the work-kinetic energy theorem we have W = ∆EK. Thus ∆EP = -W = -Fd cos . Topic 10: Fields 10.2 Fields at work If work is done in a conservative force field then there is an associated potential energy function. ∆EP = -W = -Fd cos where F is a conservative force potential energy function EXAMPLE: yf Show that for the gravitational force F = -mg that the potential energy function is ∆EP = mg∆y. SOLUTION: d Consider lifting a weight. The work done by gravity on the weight is independent of the path, so let us lift it straight up for simplicity. Observe: = 180º, d = ∆y, F = -mg. Thus yo mg ∆EP = -W = -Fd cos = -mg∆y cos 180º = mg∆y. Topic 10: Fields 10.2 Fields at work Define gravitational potential and gravitational potential energy. EP = -Gm1m2/r Gravitational where G = 6.67×10−11 N m2 kg−2 potential energy We now define gravitational potential as gravitational potential energy per unit mass: ∆V = ∆EP/m V = -Gm/r where G = 6.67×10−11 N m2 kg−2 Gravitational potential FYI The units of V are J kg-1. Gravitational potential is the work done per unit mass done in moving a small mass from infinity to r. (Note that V = 0 at r = .) Topic 10: Fields 10.2 Fields at work State and apply the expression for gravitational potential due to a point mass. ∆V = ∆EP/m Gravitational potential V = -Gm/r where G = 6.67×10−11 N m2 kg−2 EXAMPLE: Find the change in gravitational potential in moving from Earth’s surface to 5 Earth radii (from Earth’s center). SOLUTION: Use ∆V = -Gm/r2-(-Gm/r1), m = 5.98×1024 kg. At Earth’s surface r1 = 6.37×106 m. But then r2 = 5(6.37×106) = 3.19×107 m. Thus ∆V = -Gm(1/r2 - 1/r1) = -Gm(1/3.19×107 - 1/6.37×106) = -Gm(-1.26×10-7) = -(6.67×10−11)(5.98×1024)(-1.26×10-7) = +5.01×107 J kg-1. Addition of potential • Potential is a scalar quantity, so adding the magnitudes. If we tak the example: Potential at P: GM A GM B V rA rB Topic 10: Fields 10.2 Fields at work Define gravitational potential and gravitational potential energy. FYI A few words clarifying the gravitational potential energy and gravitational potential formulas are in order. EP = -Gm1m2/r gravitational potential energy V = -Gm/r gravitational potential Be aware of the difference in name. Both have “gravitational potential” in them and can be confused during problem solving. Be aware of the minus sign on both formulas. The minus sign is there so that as you separate two masses, or move farther out in space, their values increase (as in the last example). Both formulas become zero when r becomes infinitely large. Topic 10: Fields 10.2 Fields at work State and apply the formula relating gravitational field strength to gravitational potential gradient. The gravitational potential gradient is the change in gravitational potential per unit distance. Thus the GPG = ∆V/∆r. EXAMPLE: Find the GPG in moving from Earth’s surface to 5 radii from Earth’s center. SOLUTION: In a previous problem (slide 11) we found the value for the change in gravitational potential to be ∆V = + 5.01×107 J kg-1. 5RE: r2 = 3.19×107 m. Earth: r1 = 6.37×106 m. ∆r = r2 – r1 = 3.19×107 - 6.37×106 = 2.55×107 m. Thus the GPG = ∆V/∆r = 5.01×107 / 2.55×107 = 1.96 J kg-1 m-1. Topic 10: Fields 10.2 Fields at work State and apply the formula relating gravitational field strength to gravitational potential gradient. The gravitational potential gradient is the change in gravitational potential per unit distance. Thus the GPG = ∆V/∆r. PRACTICE: Show that the units for the gravitational potential gradient are the units for acceleration. SOLUTION: The units for ∆V are J kg-1. The units for work are J, but since work is force times distance we have 1 J = 1 N m = 1 kg m s-2 m. Therefore the units of ∆V are (kg m s-2 m)kg-1 or [∆V] = m2 s-2. Then the units of the GPG are [GPG] = [∆V/∆r] = m2 s-2/m = m/s2. Topic 10: Fields 10.2 Fields at work State and apply the formula relating gravitational field strength to gravitational potential gradient. EXAMPLE: Show that for the gravitational field near the surface of Earth g = ∆V/∆y local potential gradient SOLUTION: Recall that ∆V = ∆EP/m and ∆EP = mg∆y. Thus ∆V = ∆EP/m ∆V = mg∆y/m ∆V = g∆y g = ∆V/∆y. FYI From the last example we already know that the units work out. The above formula only works where g is constant (for small ∆y’s). Topic 10: Fields 10.2 Fields at work State and apply the formula relating gravitational field strength to gravitational potential gradient. The following general potential gradient (which we will not prove) works over greater range: g = -∆V/∆r general potential gradient EXAMPLE: The gravitational potential in the vicinity of a planet changes from -6.16×107 J kg-1 to -6.12×107 J kg-1 in moving from r = 1.80×108 m to r = 2.85×108 m. What is the gravitation field strength in that region? SOLUTION: g = -∆V/∆r g = -(-6.12×107 - -6.16×107)/(2.85×108 - 1.80×108) g = -4000000/105000000 = -0.0381 m s-2. Topic 10: Fields 10.2 Fields at work Determine the potential due to one or more point masses. Gravitational potential is derived from gravitational potential energy and is thus a scalar. There is no need to worry about vectors. EXAMPLE: Find the gravitational potential r at the midpoint of the circle of masses shown. Each mass is 125 kg and the radius of the circle is 2750 m. SOLUTION: Because potential is a scalar, it doesn’t matter how the masses are arranged on the circle. Only the distance matters. For each mass r = 2750 m. Each mass contributes V = -Gm/r so that V = -(6.6710-11)(125)/2750 = -3.0310-12 J kg-1. Thus Vtot = 4(-3.0310-12) = -1.2110-11 J kg-1. Topic 10: Fields 10.2 Fields at work Determine the potential due to one or more point masses. Gravitational potential is derived from gravitational potential energy and is thus a scalar. There is no need to worry about vectors. EXAMPLE: If a 365-kg mass is brought in r from to the center of the circle of masses, how much potential energy will it have lost? SOLUTION: Use ∆V = ∆EP/m and the fact that gravitational potential is zero at infinity. ∆EP = m∆V FYI 0 Since ∆EP = -W we see = m(V – V0) that the work done in = mV bringing the mass in -11 = 365(-1.2110 ) from infinity is -9 = -4.4210 J. +4.4210-9 J. Topic 10: Fields 10.1 Describing fields Describe and sketch the pattern of equipotential surfaces due to one and two point masses. Equipotential surfaces are imaginary surfaces at which the potential is the same. Since the gravitational potential for a point mass is given by m V = -Gm/r it is clear that the equipotential surfaces are at fixed radii and hence are equipotential surfaces concentric spheres: FYI Generally equipotential surfaces are drawn so that the ∆Vs for consecutive surfaces are equal. Because V is inversely proportional to r the consecutive rings get farther apart as we get farther from the mass. Worked example Reffering to the figure. 1. What is the potential at A? 2. If a body is moved from A do B what is the change in potential? 3. How much work is done moving a 2 kg mass from A do B? 1. VA = gh so potential at A = 10 x 3 = 30 J kg-1 2. VA = 30 J kg-1 VB = 80 J kg-1 Change in potential = 80 – 30 = 50 J kg-1 3. The work done moving from A to B is equal to the change in potential x mass = 50 x 2 = 100 J. Topic 10: Fields 10.1 Describing fields Describe and sketch the pattern of equipotential surfaces due to one and two point masses. EXAMPLE: Sketch the equipotential surfaces due to two point masses. SOLUTION: Here is the sketch: m m Topic 10: Fields 10.1 Describing fields State the relation between equipotential surfaces and gravitational field lines. We know that for a point mass the gravitational field lines point inward. Thus the gravitational field lines are perpendicular to the equipotential surfaces. m FYI A 3D image of the same picture looks like this: Topic 10: Fields 10.1 Describing fields State the relation between equipotential surfaces and gravitational field lines. EXAMPLE: Use the 3D view of the equipotential surface to interpret the gravitational potential gradient g = -∆V/∆r. SOLUTION: We can choose any direction for our r value, say the ∆y y-direction: Then g = -∆V/∆y. ∆V This is just the gradient (slope) of the surface. Thus g is the (-) gradient of the equipotential surface. Topic 10: Fields 10.1 Describing fields State the relation between equipotential surfaces and gravitational field lines. EXAMPLE: Sketch the gravitational field lines around two point masses. SOLUTION: Remember that the gravitational field lines point inward, and that they are perpendicular to m m the equipotential surfaces. Topic 10: Fields 10.1 Describing fields State the relation between equipotential surfaces and gravitational field lines. EXAMPLE: Use a 3D view of the equipotential surface of two point masses to illustrate that the gravitational potential gradient is zero somewhere in between the two masses. SOLUTION: Remember that the gravitational potential gradient g = -∆V/∆r is just the slope of the surface. The slope is zero on the saddle point. Thus g is also zero there. Practice The Moon has a mass of 7.4 x 1022 kg and the Earth has mass of 6.0 x 1024 kg. The average distance between the Earth and the Moon is 3.8 x 105 km. If you travel directly between the Earth and the Moon in a rocket of mass 2000 kg (a) Calculate the gravitational potential when you are 1.0 x 104 km from the Moon. (b) Calculate the rocket‘s PE at the point in part (a). (c) Draw a sketch graph showing the change in potential. (d) Mark the point where the gravitational field strength is zero. Topic 10: Fields 10.2 Fields at work Explain the concept of escape speed from a planet. We define the escape speed to be the minimum speed an object needs to escape a planet’s gravitational pull. We can further define escape speed vesc to be that minimum speed which will carry an object to infinity and bring it to rest there. Thus we see that as r then v0. M R m r=R r= Topic 10: Fields 10.2 Fields at work Derive an expression for the escape speed of an object from the surface of a planet. From the conservation of mechanical energy we have ∆EK + ∆EP = 0. Then ∆EK + ∆EP = 0 EK – EK0 + EP – EP0 = 0 0 0 (1/2)mv2 – (1/2)mu2 + -GMm/r - -GMm/r0 = 0 (1/2)mvesc2 = GMm/R vesc = 2GM/R PRACTICE: Find the escape speed from Earth. SOLUTION: M = 5.981024 kg. R = 6.37106 m. Thus vesc2 = 2GM/R = 2(6.6710-11)(5.981024)/6.37106 vesc = 11200 ms-1 (= 24900 mph!) escape speed Topic 10: Fields 10.2 Fields at work Solve problems involving gravitational potential energy and gravitational potential. The ship MUST slow down and reverse (v becomes -). The force varies as 1/r2 so that a is NOT linear. Recall that a is the slope of the v vs. t graph. Topic 10: Fields 10.2 Fields at work Solve problems involving gravitational potential energy and gravitational potential. Escape speed is the minimum speed needed to travel from the surface of a planet to infinity. It has the formula vesc2 = 2GM/R. Topic 10: Fields 10.2 Fields at work Solve problems involving gravitational potential energy and gravitational potential. To escape we need vesc2 = 2GM/Re. The kinetic energy alone must then be (1/2)m(2GM/Re) = GMm/Re. E = (1/2)mvesc2 = This is to say, to escape E = 4GMm/(4Re). Since we only have E = 3GMm/(4Re) the probe will not make it into deep space. Topic 10: Fields 10.2 Fields at work Solve problems involving gravitational potential energy and gravitational potential. Recall that Ep = -GMm/r. Thus ∆EP = -GMm(1/R – 1/Re). Topic 10: Fields 10.2 Fields at work Solve problems involving gravitational potential energy and gravitational potential. The probe is in circular motion so Fc = mv2/R. But FG = GMm/R2 = Fc. Thus mv2/R = GMm/R2 or mv2 = GMm/R. Finally EK = (1/2)mv2 = GMm/(2R). Topic 10: Fields 10.2 Fields at work Solve problems involving gravitational potential energy and gravitational potential. The energy given to the probe is stored in potential and kinetic energy. Thus ∆EK + ∆EP = E GMm/(2R) - GMm(1/R–1/Re) = 3GMm/(4Re) 1/(2R) - 1/R + 1/Re = 3/(4Re) 1/(4Re) = 1/(2R) R = 2Re Topic 10: Fields 10.2 Fields at work Solve problems involving gravitational potential energy and gravitational potential. Just know it! Topic 10: Fields 10.2 Fields at work Solve problems involving gravitational potential energy and gravitational potential. From ∆V = ∆EP/m we have ∆EP = m∆V. Thus ∆EP = (4)( -3k - -7k) = 16 kJ Topic 10: Fields 10.2 Fields at work Solve problems involving gravitational potential energy and gravitational potential. It is the work done per unit mass in bringing a small mass from infinity to that point. Topic 10: Fields 10.2 Fields at work Solve problems involving gravitational potential energy and gravitational potential. V = -GM/r so that V0 = -GM/R0. But –g0R0 = -(GM/R02)R0 = -GM/R0 Thus V0 = -g0R0. = V0. Topic 10: Fields 10.2 Fields at work Solve problems involving gravitational potential energy and gravitational potential. At R0 = 0.5107 3.8107. V0 = - From previous problem g = -V0/R0 g = - -3.8107 / 0.5107 = 7.6 m s-2. 0.5107 = 5.0106 = R0. Topic 10: Fields 10.2 Fields at work Solve problems involving gravitational potential energy and gravitational potential. 0 ∆EK = -∆EP EK – EK0 = -∆EP (1/2)mv2 = ∆EP v2 = 2∆EP/m v2 = 2∆V This solution assumes probe not in orbit but merely reaches altitude (and returns). v2 = 2(3.0107) v = 7700 ms-1. ∆V = (-0.8 - -3.8)107 ∆V = 3.0107 Topic 10: Fields 10.2 Fields at work Solve problems involving gravitational potential energy and gravitational potential. ∆EP = 3.0107m. The probe is in circular motion so Fc = mv2/R. But FG = GMm/R2 = Fc. Thus mv2/R = GMm/R2 or mv2 = GMm/R. Then EK = (1/2)mv2 = GMm/(2R). From energy ∆EK = -∆EP or EK – EK0 = -∆EP. Then EK0 = ∆EP + EK = ∆EP + GMm/(2R). From g0 = GM/R02 we have GM = g0R02. Then EK0 = ∆EP + g0R02m/(2R) so that EK0 = m[3107 + 7.6(0.5107)2/(22107)] = 3.5107m. Then (1/2)mv2 = 3.3107m and v = 8100 ms-1. This solution assumes the satellite is in orbit. Topic 10: Fields 10.2 Fields at work Solve problems involving gravitational potential energy and gravitational potential. Just know it!