C7_presentation - (C7T.3) The general gravitational...

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Unformatted text preview: (C7T.3) The general gravitational potential energy formula, V(r) = - Gm1m2 / r, is always negative for r > 0, but the empirical formula V(z) = mgz is always positive for z > 0. Why are the signs in these expressions different? (A) The empirical formula is wrong. (B) The equations refer to different kinds of interactions. (C) The first equation does not apply to objects that are not point particles. (D) The equations assume different reference separations. (E) Other. Specify: _____________________________ Gravitational Potential Energy: General Formula r m1 m2 m1m2 Vgrav (r ) = -G r Note that G = 6.67 x 10-11 N m 2 /kg 2 m1m2 Vgrav () = -G =0 Suppose we want to use the exact gravitational potential energy formula, V(r) = - Gm1m2/r, to determine the gravitational interaction between a communications satellite and the earth. In this case, the symbol r corresponds to (A) the distance between the satellite and the center of the earth. (B) the radius of the earth. (C) the distance from the satellite to the surface of the earth. (D) the location of the satellite relative to infinity. Two charged particles experience an electromagnetic interaction. Associated with this interaction is an electrostatic potential energy which has the same form as the gravitational potential energy between two masses. r q1 q2 q1q2 Velec (r ) = k r where k is a constant called the Coulomb constant Variation of gravitational potential energy with separation V(r) Vgrav (r ) = -G r m1m2 r As r decreases, the potential energy of the system decreases. This implies that the kinetic energy is increasing: the objects are attracting each other. (C7T.4) The graph at right shows the potential energy function of a certain interaction. This interaction is (A) always attractive (B) always repulsive (C) attractive for small r, repulsive for large r (A) repulsive for small r, attractive for large r (A) there is not enough information to give a meaningful answer r General rule: Positive slope = attraction Negative slope = repulsion m2 = m For an object of mass m interacting with the earth, r with m1m2 Vgrav (r ) = -G r M em = -G r GM e = 4.0 x 1014 N m 2 / kg m1 = M e GM e = 9.8 m/s 2 Note: 2 re m z Me V=0 If we lift an object of mass m through some vertical distance z near the surface of the earth, the increase in the system's gravitational potential energy is V = mgz Is this result compatible with the exact formula m1m2 Vgrav (r ) = -G ? r m z V f = -G M em re + z M em re Me Vi = -G 1 M em M em 1 = -GM e m V = V f - Vi = -G --G r +z -r re + z re e e But 1 1 z r +z -r -r2 e e e for z << re Therefore GM e GM e mz V = m 2 z = mgz 2 r re e (C7T.6) On a planet with twice the mass and twice the diameter of the earth, the value of g at its surface would be what factor times g at the surface of the earth? (A) (B) (C) (D) (E) (F) (T) 2 times larger 4 times larger 8 times larger 2 times smaller 4 times smaller 8 times smaller Other. Specify: _______________________ g planet GM planet = 2 rplanet (C7S.4) A rocket is fired vertically from the surface of the earth. Its engines fire only briefly, and then the rocket continues to coast upward. What must the rocket's initial speed be if its speed is to be not less than 5.0 km/s when it is "very far" from the earth? m vi = ? r i = re vf = 5.0 km/s rf = "very far" ~ System: rocket + Earth Floats in space Energy is conserved E = 0 = K rocket + K earth + Vgrav 1 2 1 2 - GmM e - GmM e 0 = mv f - mvi + - 2 rf ri 2 1 2 1 2 GM e 0 = v f - vi + r 2 e 2 2GM e 2(4.0 x 1014 N m 2 /kg) vi = v 2 + = (5000 m/s) 2 + = 12300 m/s f 6 re 6.38 x 10 m Spring Potential Energy 8 V( x) 0 x -4 x 4 1 Vspring ( x) = k s x 2 2 Assumptions: one end of the spring is attached to a very massive object; the other end of the spring is at x = 0 when the spring is relaxed. The potential energy of a certain spring increases by 50 joules when it is stretched 4.0 cm from its equilibrium length. What is the spring constant ks, in units of J/m2, of the spring? Vspring 1 = ks x 2 2 2 50 J = = 6.25 x 10 4 J/m 2 (0.040 m) 2 ks = 2 Vspring x2 You are asked to design a spring gun that is able to fire a 50-gram ball horizontally at a speed of 20 m/s. The specifications for the gun state that when it is loaded, the spring will be compressed 10.0 cm from its relaxed length. (When the gun is fired, the ball is released when the spring reaches its relaxed length.) What should the spring constant, ks, be? Assume the barrel of the gun has negligible friction. 10 cm initial final 20 m/s System: ball and earth Floats in space Energy is conserved E = 0 = K ball + Vspring E = 0 = K ball + Vspring 1 2 1 0 = ( mv f - 0) + (0 - k s x 2 ) 2 2 ks = mv x2 2 f (0.05 kg)(20 m/s) = = 2000 J/m 2 (0.10 m) 2 2 ...
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This note was uploaded on 04/17/2008 for the course PHYS 120 taught by Professor Decarlo during the Spring '08 term at DePauw.

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