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Chapter 7 Chapter 7 Gravitation Gravitation Basic Requirements 1. Concept of the gravitational force; 2. Newton's law of gravitation; 3. Gravitation and the principle of superposition; 4. Gravitation near Earth's surface; 5. Gravitation inside Earth; 6. Gravitational potential energy; 7. Kepler's Laws; 8. Satellites: orbits and energy. Review and Summary The Law of Gravitation Any particle in the universe attracts any other particle with a gravitational force whose magnitude is F G m1m2 r2 (Newton’s law of gravitation) (7-1) where m1 and m2 are the masses of the particles, r is their separation, and G ( 6.67 1011 N m2 / kg 2 ) is the gravitational constant. Gravitational Behavior of Uniform Spherical Shells Equation 7-1 holds only for particles. The gravitational force between extended bodies must generally be found by adding (integrating) the individual forces on individual particles within the bodies. However, if either of the bodies is a uniform spherical shell or a spherically symmetric solid, the net gravitational force it exerts on an external object may be computed as if all the mass of the shell or body were located at its center. Superposition Gravitational forces obey the principle of superposition; that is, if n particles interact, the net force F1, net on a particle labeled as particle 1 is the sum - 168 - Chapter 7 Gravitation of the forces on it from all the other particles taken one at a time: n F1, net F1i (7-5) i2 in which the sum is a vector sum of the forces F1i on particle 1 from particles 2,3,..., n . The gravitational force F1 on a particle from an extended body is found by dividing the body into units of differential mass dm, each of which produces a differential force dF on the particle, and then integrating to find the sum of those forces: F1 dF Gravitational Acceleration (7-6) The gravitational acceleration a g of a particle (of mass m ) is due solely to the gravitational force acting on it. When the particle is at distance r from the center of a uniform, spherical body of mass M , the magnitude F of the gravitational force on the particle is given by Eq.7-1. Thus, by Newton’s second law, F mag (7-8) GM r2 (7-9) which gives ag Free-Fall Acceleration and Weight The actual free-fall acceleration g of a particle near Earth differs slightly from the gravitational acceleration a g , and the particle’s weight (equal to mg) differs from the magnitude of the gravitational force acting on the particle as computed with Eq.7-1, because Earth is not uniform or spherical and because Earth rotates. Gravitational Within a Spherical Shell A uniform shell of matter exerts no net gravitational force on a particle located inside it. This means that if a particle is located inside a uniform solid sphere at distance r from its center, the gravitational force exerted on the particle is due only to the mass M ins that lies inside a sphere of radius - 169 - Chapter 7 Gravitation r . This mass is given by M ins 4r 3 3 where is the density of the sphere. Gravitational Potential Energy The gravitational potential energy U ( r ) of a system of two particles, with masses M and m and separated by a distance r , is the negative of the work that would be done by the gravitational force of either particle acting on the other if the separation between the particles were changed from infinite (very large) to r . This energy is U GMm r Potential Energy of a System (gravitational potential energy) (7-17) If a system contains more than two particles, its total gravitational potential energy U is the sum of terms representing the potential energies of all the pairs. As an example, for three particles, of masses m1 , m2 , and m3 , Gm1m2 Gm1m3 Gm2 m3 U r13 r23 r12 (7-18) Escape Speed An object will escape the gravitational pull of an astronomical body of mass M and radius R (to reach an infinite distance) if the object’s speed near the body’s is at least equal to the escape speed, given by v 2GM R (7-24) Kepler’s Laws Gravitational attraction holds the solar system together and makes possible orbiting Earth satellites, both natural and artificial. Such motions are governed by Kepler’s three laws of planetary motion, all of which are direct consequences of Newton’s laws of motion and gravitation: - 170 - Chapter 7 Gravitation 1. The law of orbits. All planets move in elliptical orbits with the Sun at one focus. 2. The law of areas. A line joining any planet to the Sun sweeps out equal areas in equal times as the planet orbits the Sun. (This statement is equivalent to conservation of angular momentum.) 3. The law of periods. The square of the period T of any planet about the Sun is proportional to the cube of the semimajor axis a of the orbit. For circular orbits with radius r , the semimajor axis a is replaced by r and the law is written as 4 2 3 r T 2 GM (law of periods), (7-30) where M is the mass of the attracting body-the Sun in the case of solar system. This equation is generally valid for elliptical planetary orbits, when the semimajor axis a is inserted in place of the circular radius r . Energy in Planetary Motion When a planet or satellite with mass m moves in a circular orbit with radius r , its potential energy U and kinetic energy K are given by GMm GMm and K r 2r The mechanical energy E K U is then GMm E 2r For an elliptical orbit of semimajor axis a , GMm E 2a U (7-33) (7-35) (7-37) Einstein’s View of Gravitation Einstein pointed out that gravitation and acceleration are equivalent. This principle of equivalence led him to a theory of gravitation (the general theory of relativity) that explains gravitational effects in terms of a curvature of space. - 171 - Chapter 7 Gravitation Examples Example 1 Assume a space flyer circles around the Earth in a circular orbit with speed v0 3.00 104 km / s , the radius of the Earth is R 6.40 103 km , and the height of the flyer above the Earth is h 800km . Now we want to change the orbit to an ellipse. In a very short period of time the flyer ejects gases toward the outer side of the orbit in the direction of the radius vector, then the flyer gains a speed vr 800km / s pointing toward the Earth. Assume that the mass of the flyer remain the same before and after the ejection of the gases, after the flyer orbits around the Earth in the final elliptic orbit, what is the distance between its perigee and the center of the Earth, and what is the distance between its apogee and the center of the Earth? Solution: The key idea is that, before and after the ejection of the gases the angular momentum of the flyer remains the same: mv0 r0 mvr (1) where r0 is the distance between the flyer and the center of the Earth when the flyer orbited the Earth in a circular orbit, i.e., r0 R h ; and r is the distance between the apogee (or perigee) of its elliptic orbit and the center of the Earth, v is the speed at that point, and m is the mass of the flyer. For the flyer on the circular orbit, we have G mE m v02 m r02 r0 (2) After the alternation of the orbit, we take the Earth and the flyer as a system and we have the conservation of mechanical energy, therefore: 1 m m 1 m m m v02 vr2 G E mv2 G E 2 r0 2 r Eliminating mE and v from Eqs.(1), (2) and (3), we have 2 1 1 v02 r02 2v02 r0 v02 vr2 0 r r - 172 - (3) Chapter 7 Gravitation Solving it, we get two roots r1 and r2 , i.e. , r1 v0 r0 v R h vr v R h 0 , r2 0 0 0 v0 vr v0 vr v0 vr v0 vr (Answer) After substituting the known numbers into it, we have r1 7.0110 km and 3 r2 7.40km . These are the distances between the apogee and the perigee and the center of the Earth respectively. Example 2 A geosynchronous satellite is one that stays above the same point on the Earth, which is possible only if it is above a point on the equator. Such satellites are used for such purpose as cable TV transmission, for weather forecasting, and as communication relays. Determine (a) the height above the Earth’s surface such a satellite must orbit and (b) such a satellite’s speed. (c) Compare to the speed of a satellite orbiting 200km above Earth’s surface. Solution: (a) The only force on the satellite is gravity, assuming the satellite moves in a circle, then G mSat mE v2 m Sat r2 r This equation seems to have two unknowns, r and v . But we know that v must be such that the satellite revolves around the Earth with the same period that the Earth rotates on its axis, namely once in 24 hours. Thus the speed of the satellite must be 2r T where T 1day (24h)(3600s / h) 86400 s . We put this into the first equation above and obtain (after canceling mSat on both sides): v mE 2r r2 rT 2 2 G We solve for r: GmET 6.67 10 r 4 2 3 11 N m2 / kg 2 5.98 1024 kg 864000 s 4 2 - 173 - 7.54 1022 m3 Chapter 7 Gravitation and, taking the cube root, r 4.23 10 m , or 42300km from the Earth’s center. 7 We substrate the Earth’s radius of 6380km to find that the satellite must orbit about 36000km (about 6rE ) above the Earth’s surface. (b) v GmE r 6.67 10 11 N m2 / kg2 5.98 1024 kg 3070m / s 4.23 107 m We get the same result if we use v 2r / T . (Answer) (c) Since v 1 / r , for r rE h 6380km 200km 6580km , we get v' v r / r ' 3070m / s 42300km /6580km 7780m / s (Answer) Example 3 There is a thin rod with mass m , length l and linear density of mass , if a mass point with mass m is placed on the extended line of the rod and distance r away from the end of the Fig.7-1 Example 3. rod as shown in Fig.1, what is the gravitational potential energy and the gravitational force on the mass point by the rod? Solution: As shown in the figure, take an element of the rod d x and its mass is dm dx . The distance between the rod element and the mass point m is x , and the gravitational potential energy between the mass element dm’ and the mass point m is: dE p Gmdm' Gmdx x x Adding up all the gravitational potential energies of mass elements dm dm’ and mass point m , i.e., integrating from x r to x r l , we have the gravitational potential energy of mass point m at point P as follow: Ep r l r dE Gm p After the integration, we have - 174 - r l r dx x Chapter 7 Gravitation E p Gm ln r rl (Answer) The gravitational force on mass point m can be directly calculated by the universal gravitational law. In Fig.1, mass element dm dx has the following gravitational force on mass point m : mdm' dF G 2 i x The direction of dF is pointing to the positive ox axis, then the gravitational force of the rod on mass point m is: r l dx 1 1 F dF Gm i G m i r x2 r lr i.e., mm' F G i r l r (Answer) Problem Solving 1 (2) In Fig. 2a, particle A is fixed in place at x 0.20m on the x axis and particle B, with a mass of 1.0kg , is fixed in place at the origin. Particle C (not shown) can be moved along the x axis, between particle B and x . Figure 2b shows the x component Fig.7-2 Problem 1. Fnet , x of the net gravitational force on particle B due to particle A and C, as a function of position x of particle C. The plot actually extends to the right, approaching an asymptote of 4.7 1010 N as x . What are the masses of (a) particle A and (b) particle C? Solution: (a) The component Fnet , x of the net gravitational force on particle B is: - 175 - Chapter 7 Fnet , x G Gravitation mB mC mAmB G 2 x 0.20m2 The key idea is that as x , the net gravitational force on particle B is only due to particle A: 4.17 1010 N G Substituting mB 1.0kg and G 6.67 10 11 mAmB 0.20m 2 N m2 kg 2 into the above equation, yields mA 0.25kg (Answer) (b) The key idea is that from the Fnet , x ( x ) curve, as x 0.4m , Fnet , x 0 , so 0G mB mC 0.4m 2 G mA mB 0.2m 2 Substituting mA 0.25kg , mB 1.0kg into the above equation, and mC 1.0kg (Answer) 2 (5) Two dimensions. In Fig.3, three point particles are fixed in place in xy plane. Particle A has mass mA , particle B has mass 2.00m A , and particle C has mass 3.00m A . A fourth particle D, with mass 4.00m A , is to be placed near the other three particles. In terms of distance d , at what (a) x coordinate and (b) y coordinate should particle D be placed so that the net gravitational force on particle A Fig.7-3 Problem 2. from particle B, C, and D is zero? Solution: The key idea is that, as the net gravitational force on particle B from the other three particles is zero, the gravitational force on particle A from D is the same in - 176 - Chapter 7 Gravitation magnitude and opposite in direction to the composite gravitational force on article A from B and C. As shown in the diagram, there is: 2 2 FAD FAB FAC y That is, mm mm mm G 2 A D2 G A 2 B G A C2 x y d 1.5d 2 α 2 (1) C A B α And y d (2) x 1.5d Substituting mB 2.00mA , mC 3.00mA , and mD 4.00mA into Eqs.(1) and (2) and solving for x and y : tan x 0.99d , y 0.66d (Answer) 3 (10) A projectile is shot directly away from Earth’s surface. Neglect the rotation of Earth. What multiple of Earth’s radius RE gives the radial distance a projectile reaches if (a) its initial speed is 0.500 of the escape speed from Earth and (b) its initial kinetic energy is 0.500 of the kinetic energy required to escape Earth? (c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth? Solution: The escape speed is: v 2GM RE (a) As the projectile’s initial speed is 0.500v , the key idea is that when it reaches the maximum radial distance R1 , its kinetic energy is zero and potential energy U1 GMm , then R1 - 177 - Chapter 7 Gravitation GMm 1 GMm 2 m0.500v 2 RE R1 yields R1 4 RE 3 (Answer) (b) As the projectile’s initial kinetic energy is 0.500 of 1 2 mv , the maximum 2 radial distance it reaches is R2 , so GMm 1 GMm 0.500 mv2 RE R2 2 yields R2 2RE (Answer) (c) When the projectile escapes Earth, its least required kinetic energy is zero and the potential energy is zero, too. As the mechanical energy is conserved, the least initial mechanical energy required at launch is zero. 4 (12) Zero, a hypothetical planet, has a mass of 5.0 1023 kg , a radius of 3.0 106 m , and no atmosphere. A 10kg space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial energy of 5.0 107 J , what will be its kinetic energy when it is 4.0 106 m from the center of Zero? (b) If the probe is to achieve a maximum distance of 8.0 106 m from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero? Solution: (a) The key idea is that, the mechanical energy of the space probe is conserved, then: K f U f Ki U i GMm GMm Ki K f R R 1 0 R0 3.0 106 m is the radius of Zero and R1 4.0 106 m , K i 5.0 107 J . Substituting G 6.67 1011 N m2 kg 2 , M 5.0 1023 kg where - 178 - Chapter 7 Gravitation and m 10kg into the above equation, yields: K f 2.33 107 J (Answer) (b) The key idea is that, as the probe achieves the maximum distance R2 8.0 106 m , its minimum final kinetic is zero, we have: GMm GMm Ki ' R R 2 0 Solving for Ki ' , yields K i ' 6.94 10 J 7 (Answer) So it must be launched with at least 6.94 10 J kinetic energy. 7 5 (16) A 20kg satellite has a circular orbit with a period a 2.4h and a radius of 8.0 106 m around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface on the planet is 8.0m / s 2 , what is the radius of the planet? Solution: The gravitational acceleration of the satellite: v 2 2r / T 4 2 r 2 r r T 2 ag Substituting T 2.4h and r 8.0 106 m into the above equation and yielding: ag 4.2m / s 2 The gravitational acceleration at the surface: ag 0 GM R2 (1) where R is the radius of the planet, and ag GM r2 (2) From Eqs. (1) and (2) we have: r ag R ag 0 2 (3) Substituting ag 0 8.0m / s , ag 4.2m / s , r 8.0 10 m into Eq.(3) ,yields 2 2 - 179 - 6 Chapter 7 Gravitation R 5.8 106 m (Answer) 6 (18) Three identical stars of mass M form an equilateral triangle that rotates around the triangle’s center as the stars move in a common circle about the center. The triangle has edge length L . What is the speed of the stars? Solution: Analyze the free-body diagram of one of the stars as shown. The net force is: Fnet 2 F cos 30 3F 3 F F GMM L2 (1) As the star is rotating around the triangle’s center, then: 30° Fnet Mv 2 r where v is the speed of the star and r (2) 3 L is the distance 3 between the star and the triangle’s center. Solving Eqs. (1), (2) for v , and v The speed of the three identical stars is GM L (Answer) GM . L 7 (19) One way to attack a satellite in Earth orbit is to launch a swam of pellets in the same orbit as the satellite but in the opposite direction. Suppose a satellite in a circular orbit 500km above Earth’s surface collides with a pellet having mass 4.0g . (a) What is the kinetic energy of the pellet in the reference frame of the satellite just before the collision? (b) What is the ratio of this kinetic energy to the kinetic energy of a 4.0g bullet from a modern army rifle with a muzzle speed of 950m / s ? Solution: (a) In the reference frame of Earth, since the pellet is in the circular orbit with the speed of v , there is: - 180 - Chapter 7 m Gravitation v2 Mm G 2 r r (1) Substituting the mass of Earth M 5.98 10 kg , G 6.67 10 24 11 N m2 kg 2 and r R 500km (the radius of Earth R 6.37 10 m ) into Eq. (1), yields 6 v 7.6 103 m / s In the reference frame of the satellite, since the pellet moves in the opposite direction to the satellite, the speed of the pellet is then 2v before the collision. Then the kinetic energy of the pellet: K 2 1 2 m2v 2mv2 2 4.0 10 3 kg 7.6 103 m / s 4.6 105 J 2 (Answer) (b) The kinetic energy of the bullet: Kb 1 1 2 2 mb vb 4.0 10 3 kg950m / s 1.8 103 J 2 2 So the ratio is: K 4.6 105 256 K b 1.8 103 (Answer) 8 (24) A typical neutron star may have a mass equal to that of the Sun but a radius of only 10km . (a) What is the gravitational acceleration at the surface of such a star? (b) How fast would an object be moving to the surface if it fell from rest through a distance of 1.0m on such a star? (Assume the star does not rotate.) Solution: (a) The gravitational acceleration is ag 11 GM R2 Substituting G 6.67 10 N m kg , the mass of the sum M 1.99 10 kg , and R 10km into the above equation and yielding: 2 2 ag 1.33 1012 m / s 2 - 181 - 30 (Answer) Chapter 7 (b) v Gravitation 2ag h 2 1.33 1012 m / s 2 1.0m 1.63 106 m / s (Answer) 9 (25) A very early, simple satellite consisted of an inflated spherical aluminum balloon 30m in diameter and of mass 20kg . Suppose a meteor having a mass of 7.0kg passes within 3.0m of the surface of the satellite. What is the magnitude of the gravitational force on the meteor from the satellite at the closest approach? Solution: The gravitational force: F G Mm r2 (1) Since the radius of the satellite is 3.0m , then r 30m 3.0m 18m 2 Substituting Eq.(2) and G 6.67 10 11 (2) N m2 kg 2 , M 20kg , m 7.0kg into Eq.(1), then F 2.88 1011 N (Answer) 10 (28) The orbit of Earth around the sun is almost circular: The closest and farthest distances are 1.47 108 km and 1.52 108 km respectively. Determine the corresponding variations in (a) total energy, (b) gravitational potential energy, (c) kinetic energy, and (d) orbital speed. (Hint: Use conservation of energy and conservation of angular momentum.) r2 r1 Solution: (a) Since the only force acting on Earth is the gravitational force, the total energy of Earth is conserved, so E 0 (Answer) (b) Fig.7-4 Problem 10. GMm GMm 1 1 GMm U U 2 U1 r2 r1 r1 r2 where the subscripts 1 and 2 refer to the parameters of the closest and farthest points. - 182 - Chapter 7 Substituting G 6.67 10 11 Gravitation N m kg , the mass of the Sun M 1.99 1030 kg , 8 8 the mass of Earth m 5.98 1024 kg , and r1 1.47 10 km , r2 1.52 10 km 2 2 into the above equation , yields: U 1.78 1032 J (Answer) (c) Since E K U 0 K U 1.78 1032 J (Answer) (d) Because the gravitational force acting on the Earth passes through the center of the Sun, there is no external torque acting on the Earth. The angular momentum is conserved, so we can get I11 I 22 mr12 That is v1 v mr22 2 r1 r2 So v2 r1 1.47 v1 r2 1.52 From m So And v1 v12 Mm G 2 r1 r1 GM 6.67 1011 1.99 1030 3.0 104 m / s r1 1.47 1011 v v2 v1 1.47 v1 v1 990m / s 1.52 (Answer) 11 (32) Several planets (Jupiter, Saturn, Uranus) are encircled by rings, perhaps composed of material that failed to form a satellite. In addition, many galaxies contain ring-like structures. Consider a homogeneous thin ring of mass M and outer radius R (Fig. 5). (a) Fig.7-5 Problem 11. - 183 - Chapter 7 Gravitation What gravitational attraction does it exert on a particle of mass n located on the ring’s central axis a distance x from the ring center? (b) Suppose the particle falls from rest as a result of the attraction of the ring of matter. What is the speed with which it passes through the center of the ring? Solution: (a) The key idea is to find the net gravitational force of the ring on the particle. Take an element of the ring dr as shown, and its mass is dM M dr , the gravitational 2R force on the particle is: dF GmdM Gm M 2 dr 2 2 x R x R 2 2R The component of the force along x and y axis is respectively: GmM x dr 2 2 x R 2R x R2 GmM R dFy 2 dr 2 2 x R 2R x R2 dFx 2 For arbitrary dr , there is always an element dr on the symmetric place to dr , as shown, and the component of the gravitational force of dr on the particle is same in magnitude and opposite in direction to that of dr , so the net force of the thin ring on the particle along the x axis is: F dFx 2R 0 GmM x GmMx dr 2 2 3/ 2 2 2 2 x R2 x R 2R x R The gravitational attraction on the particle from the ring is x (Answer) GmMx 2 R2 3/ 2 in magnitude. (b) The change in the kinetic energy of the particle comes from the work done by the gravitational force 0 1 2 mv 0 Fdx x 2 Substituting F into the integral for v , yields - 184 - Chapter 7 Gravitation 1 v 2GM R x R 1 2 2 (Answer) That is the speed of the particle when it passes through the center of the ring. 12 (33) The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just barely provides the centripetal force needed for the rotation. (Why?) (a)Show that the corresponding shortest period of rotation is 3 , G T where is the uniform density (mss per unit volume) of the spherical planet. (b) Calculate the rotation period assuming a density of 3.0g / cm3 , typical of many planets, satellites, and asteroids. No astronomical object has ever been found to be spinning with a period shorter than that determined by this analysis. Solution: (a) Assume a body of mass m is at the equator, the mass and the radius of the planet is M and R respectively, the gravitational force is: Fg G Mm R2 (1) For the rotating body, its centripetal force is: Fc m v2 R (2) where v is the linear speed of the body. Since Fg Fc , we have: G Mm v2 m R2 R (3) And the shortest period is : T 2R v The uniform density of the spherical planet is: - 185 - (4) Chapter 7 Gravitation M (5) 4 3 R 3 Combining Eqs.(3), (4), (5) we have T 3 G (6) (b) Substituting G 6.67 1011 N m2 kg 2 , 3.0 g / cm3 into Eq.(6), yields T 6.9 103 s (Answer) 13 (34) The gravitational force between two particles with masses m and M , initially at rest at great separation, pulls them together. Show that at any instant the speed of 2G ( M m) / d , where d is their separation at that instant. (Hint: Use the laws of conservation of energy and conservation of linear momentum.) Solution: Assume the velocities of the two particles are v1 and v2 , respectively, and the linear momentum of the two-particle system is conserved: either particle relative to the other is mv1 mv2 0 (1) The particles are pulled together by the gravitational force and the mechanical energy of the two-particle system is also conserved: K f U f Ki U i i.e. 1 2 1 2 Mm mv1 mv2 G 0 2 2 d (2) From Eqs.(1) and (2) we have: v2 2G M M m d and v1 And the speed of either particle relative to the other is then: - 186 - mv2 M Chapter 7 v2 v1 v2 mv2 M m M M Gravitation 2G M M md 2G M m d (Answer) 14 (37) Some people believe that the positions of the planets at the time of birth influence the newborn. Others deride this belief and claim that the gravitational force exerted on a baby by the obstetrician is greater than the force exerted by the planets. To check this claim, calculate the magnitude of the gravitational force exerted on a 3kg baby (a) by a 70kg obstetrician who is 1m away and roughly approximated as a point mass, (b) by the massive planet Jupiter ( m 2 1027 kg ) at its closest approach to Earth ( 6 1011 m ), and (c) by Jupiter at its greatest distance from Earth ( 9 1011 m ). (d) Is the claim correct? Solution: (a) The gravitational force exerted on the 3kg baby by the obstetrician is F1 G Substituting G 6.67 10 11 Mm0 r12 (1) N m2 kg 2 , M 70kg , m0 3kg , r1 1m into Eq.(1), and F1 1.4 108 N (Answer) (b) The gravitational force exerted on the 3kg baby by the Jupiter planet at its closest approach to Earth is F2 G Substituting G 6.67 10 11 Mm0 r22 (2) N m2 kg 2 , m 2 1027 kg , m0 3kg , r2 6 1011 m into Eq.(2), and F2 1.1 106 N (Answer) (c) The gravitational force exerted on the 3kg baby by the Jupiter planet at its greatest - 187 - Chapter 7 Gravitation distance from Earth is F3 G Substituting G 6.67 10 11 mm0 r32 (3) N m2 kg 2 , m 2 1027 kg , m0 3kg , r3 9 1011 m into Eq.(3), and F3 4.9 107 N (d) Since F2 F1 and even F3 F1 , the claim is not correct. - 188 - (Answer) (Answer)