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Ch 11 HW Day 4, p 362 – 369 #’s 37, 44, 53, 55, 56, 61, 69, 74, 75, 82, 92 37 • Picture the Problem Newton’s 2nd law of motion relates the masses and accelerations of these objects to their common accelerating force. (a) Apply Newton’s 2nd law to the standard object: F m1a1 Apply Newton’s 2nd law to the object of unknown mass: F m2 a2 Eliminate F between these two equations and solve for m2: m2 Substitute numerical values and evaluate m2: 2.6587 m/s 2 1 kg 2.27 kg m2 1.1705 m/s 2 (b) a1 m1 a2 It is the inertial mass of m2 . 44 •• Picture the Problem Let the zero of gravitational potential energy be at infinity, m represent the mass of the object, and h the maximum height reached by the object. We’ll use conservation of energy to relate the initial potential and kinetic energies of the object-earth system to the final potential energy. Use conservation of energy to relate the initial potential energy of the system to its energy as the object is about to strike the earth: K f Ki U f U i 0 or, because Kf = 0, (move all else to the other side), K RE U RE U RE h 0 (1) where h is the initial height above the earth’s surface. Express the potential energy of the object-earth system when the object is at a distance r from the surface of the earth: Substitute in equation (1) to obtain: U r Solve for h: Little m cancels out of all 3 terms. Move the first two terms to the other side of the eq. Get a common denominator of 2RE. Then, you’ll have a single fraction on both the left and right. This way, you can cross-multiply, and then subtract RE from both sides to isolate h. Don’t forget that ALL units need to be SI. 1 2 GM E m r GM E m GM E m mv 0 RE RE h h 2 2GM E RE RE 2GM E RE v 2 Or, you can use their derivation below. h RE 2 gRE 1 v2 Substitute numerical values and h evaluate h: 6.37 106 m 2 9.81m/s 2 6.37 106 m 4 10 m 3 2 1 935 km 53 •• Picture the Problem We can use its definition to express the period of the spacecraft’s motion and apply Newton’s 2nd law to the spacecraft to determine its orbital velocity. We can then use this orbital velocity to calculate the kinetic energy of the spacecraft. We can relate the spacecraft’s angular momentum to its kinetic energy and moment of inertia. (a) Express the period of the spacecraft’s orbit about the earth: Use Newton’s 2nd law to relate the gravitational force acting on the spacecraft to its orbital speed: Solve for v to obtain: T 2 R 2 3RE 6 RE v v v where v is the orbital speed of the spacecraft. NOTE: 3RE because you have to use 1 RE + a height of 2 RE for a total of 3 RE. F = ma, where a = v2/r, so, Fradial v GM E m 3RE 2 v2 m 3 RE gRE 3 NOTE: normally, v = gRE , but since this orbits at a height of 3RE (total), after you divide both sides by 3 and take the square root, you end up with the above expression. Substitute for v in our expression for T to obtain: RE T 6 3 g Substitute numerical values and evaluate T: 6.37 106 m T 6 3π 9.81 m/s 2 1h 7.31 h 3600 s NOTE: I would have left it as 2.6 x 104 s, since that’s SI. 2.631 10 4 s (b) Using its definition, express the spacecraft’s kinetic energy: K 12 mv 2 12 m 13 gRE 100 kg 9.81m/s 2 6.37 106 m Substitute numerical values and evaluate K: K (c) Express the kinetic energy of the spacecraft in terms of its angular momentum: L2 K 2I Solve for L: L 2IK 1 6 1.04 GJ Express the moment of inertia of the spacecraft with respect to an axis through the center of the earth: I m3RE Substitute and solve for L: L 18mRE2 K 3RE 2mK 2 9mRE2 Substitute numerical values and evaluate L: L 3 6.37 106 m 2100 kg 1.04 109 J 8.72 1012 J s 55 •• Picture the Problem We can express the energy difference between these two orbits in terms of the total energy of a satellite at each elevation. The application of Newton’s 2nd law to the force acting on a satellite will allow us to express the total energy of each satellite as function of its mass, the radius of the earth, and its orbital radius. Express the energy difference: E Egeo E1000 (1) Express the total energy of an orbiting satellite: E tot K U 12 mv 2 GM E m R where R is the orbital radius. (2) Apply Newton’s 2nd law to a satellite to relate the gravitational force to the orbital speed: Fradial GM E m v2 m 2 R R Which is N’s Law of Gravity, a centripetal force, and N2, F = ma, where a = v2/R. Also, since GME = gRE2, and little m cancels out, this can be simplified to: gRE2 v 2 2 R R Simplify and solve for v2: Substitute in equation (2) to obtain: Substitute in equation (1) and simplify to obtain: 2 gR E v2 R Etot 2 2 2 gR gR m mgR E 12 m E E R R 2R mgRE2 mgRE2 E 2 Rgeo 2 R1000 mgRE2 2 1 1 R 1000 Rgeo Substitute numerical values and evaluate E: 2 1 1 11.1GJ E 12 500 kg 9.81 N / kg 6.37 106 m 6 7 7 . 37 10 m 4.22 10 m 56 •• Picture the Problem. We can use Kepler’s 3rd law to relate the periods of the moon and Earth, in their orbits about the earth and the sun, to their mean distances from the objects about which they are in orbit. We can solve these equations for the masses of the sun and the earth and then divide one by the other to establish a value for the ratio of the mass of the sun to the mass of the earth. Using Kepler’s 3rd law, relate the period of the moon to its mean distance from the earth: Using Kepler’s 3rd law, relate the period of the earth to its mean distance from the sun: Solve equation (1) for ME: Solve equation (2) for Ms: 4 2 3 T rm GM E 2 m (1) where rm is the distance between the centers of the earth and the moon. 4 2 3 T rE GM s 2 E (2) where rE is the distance between the centers of the earth and the sun. 4 2 3 ME r 2 m GTm (3) 4 2 3 Ms r 2 E GTE (4) Divide equation (4) by equation (3) and simplify to obtain: M s rE M E rm 2 Substitute numerical values and evaluate Ms/ME: M s 1.5 1011 m M E 3.82 108 m 3 3 Tm TE 3.38 105 27.3 d 365 . 24 d 2 Express the difference between this value and the measured value of 3.33105: 3.38 105 3.33 105 % diff 3.33 105 1.50 % 61 •• Picture the Problem The configuration of point masses is shown to the right. The gravitational field at any point can be found by superimposing the fields due to each of the point masses. (a) Express the gravitational field at x = 2 m as the sum of the fields due to the point masses m1 and m2: g g1 g2 (1) Express g1 and g 2 : N2 = N’s LoG, cxl m’s Gm Gm g1 2 1 iˆ and g 2 2 2 iˆ x2 x1 Substitute in equation (1) to obtain: Gm Gm g 2 1 iˆ 2 2 iˆ x1 x2 Gm1 ˆ Gm2 ˆ i i 2 x12 2 x1 G m1 14 m2 iˆ 2 x1 Substitute numerical values and evaluate g : 6.6726 10 11 N m 2 /kg 2 g 2 m 2 2 kg 1 4 kg iˆ 4 (b) Express g1 and g 2 : Substitute in equation (1) to obtain: Substitute numerical values and evaluate g : 1.67 10 11 N/kg iˆ Gm Gm g1 2 1 iˆ and g 2 2 2 iˆ x1 x2 Gm Gm g 2 1 iˆ 2 2 iˆ x1 x2 Gm1 ˆ Gm2 ˆ i 2 i 2 x2 2 x2 G 1 m m2 iˆ 2 4 1 x2 6.6726 10 11 N m 2 /kg 2 g 6 m 2 1 2 kg 4 kg iˆ 4 (c) Express the condition that g = 0: 8.34 10 12 N/kg iˆ Gm1 Gm2 0 2 2 x 6 x or 2 4 0 2 2 x 6 x Express this quadratic equation in standard form: x 2 12 x 36 0 , where x is in meters. Solve the equation to obtain: x 2.48 m and x 14.5 m From the diagram it is clear that the physically meaningful root is the positive one at: x 2.48 m 69 •• Picture the Problem: The magnitude of the gravitational force is F = mg where g inside a spherical shell is zero and outside is given 2 g GM r . This is N2 set equal to N’s Law of Gravity (Eq 11by 29). (a) At r = 3a, the masses of both spheres contribute to g: F mg m G M 1 M 2 3a 2 GmM 1 M 2 9a 2 (b) At r = 1.9a, g due to M2 = 0: GM1 GmM1 F mg m 2 3.61a 2 1.9a (c) At r = 0.9a, g = 0: F 0 74 ••• Picture the Problem The diagram shows the portion of the solid sphere in which the hollow sphere is embedded. g1 is the field due to the solid sphere of radius R and density 0 and g 2 is the field due to the sphere of radius ½R and negative density 0 centered at ½R. We can find the resultant field by adding the x and y components of g1 and g 2 . Set N2 = N’s L.of G., cancel m; use givens and sub. Use its definition to express g1 : GM G g1 0V r2 r2 40 rG 3 40 r 3G 3r 2 Find the x and y components of g1 : 40Gx x g1x g1 cos g1 3 r and 40Gy y g1 y g1 sin g1 3 r where the negative signs indicate that the field points inward. Use its definition to express g2 : GM 2 G 0V2 40 r23G g2 2 r r22 3r22 where 40 r2G 3 r2 x 12 R 2 y 2 Express the x and y components of g 2 : x 12 R 40 Gx 12 R g 2 r 3 2 g2x g2 y Add the x components to obtain the x component of the resultant field: y 40Gy g 2 3 r2 g x g1x g 2 x 40Gx 40G x 12 R 3 3 20GR 3 where the negative sign indicates that the field points inward. Add the y components to obtain the y component of the resultant field: g y g1 y g 2 y Express g in vector form and evaluate g : 20GR ˆ g g x iˆ G y ˆj i 3 40Gy 40Gy 0 3 3 and 20GR g 3 75 ••• Picture the Problem The gravitational field will exert an inward radial force on the objects in the tunnel. We can relate this force to the angular velocity of the planet by using Newton’s 2nd law of motion. Letting r be the distance from the objects to the center of the planet, use Newton’s 2nd law to relate the gravitational force acting on the objects to their angular velocity: Solve for to obtain: Use its definition to express g: Fnet Fg mr 2 or mg mr 2 g r (1) GM G 0V 40 r 3G g 2 2 r r 3r 2 40 rG 3 Substitute in equation (1) and simplify: 40 rG 3 r 40G 3 82 • Picture the Problem Consider an object of mass m at the surface of the earth. We can relate the weight of this object to the gravitational field of the earth and to the mass of the earth. This is N2 set equal to N’s Law of Gravity. Using Newton’s 2nd law, relate the weight of an object at the surface of the earth to the gravitational force acting on it: Solve for ME: w mg gRE2 ME G GM E m RE2 Substitute numerical values and evaluate ME: ME 2 9.81 N / kg 6.37 106 m 6.6726 1011 N m 2 /kg 2 5.97 1024 kg 92 •• Picture the Problem. Let r represent the separation of the particle from the center of the earth and assume a uniform density for the earth. The work required to lift the particle from the center of the earth to its surface is the integral of the gravitational force function. This function can be found from the law of gravity and by relating the mass of the earth between the particle and the center of the earth to the earth’s mass. We can use the work-kinetic energy theorem to find the speed with which the particle, when released from the surface of the earth, will strike the center of the earth. Finally, the energy required for the particle to escape the earth from the center of the earth is the sum of the energy required to get it to the surface of the earth and the kinetic energy it must have to escape from the surface of the earth. (a) Express the work required to lift the particle from the center of the earth to the earth’s surface: Using the law of gravity, express the force acting on the particle as a function of its distance from the center of the earth: RE W Fdr (1) 0 where F is the gravitational force acting on the particle. GmM F r2 (2) where M is the mass of a sphere whose radius is r. Express the ratio of M to ME: 43 r 3 M 4 M E 3 RE3 r3 M ME 3 RE Substitute for M in equation (2) to obtain: GmM E mgRE2 mg F r r r 3 3 RE RE RE Substitute for F in equation (1) and evaluate the integral: mg E gmRE W rdr RE 0 2 (b) Use the workkinetic energy theorem to relate the kinetic energy of the particle as it reaches the center of the earth to the work done on it in moving it to the surface of the earth: W K mv Substitute for W and solve for v: v (c) Express the total energy required for the particle to escape when projected from the center of the earth: E esc W 12 mve2 Substitute for W and solve for vesc: R 1 2 2 gRE 2 12 mvesc where ve is the escape speed from the surface of the earth. vesc 3gRE Substitute numerical values and evaluate vesc: vesc 39.81 N/kg 6.37 106 m 13.7 km/s