Kepler's and Newton's Laws of Planetary Motion: Periods, Speeds, and Accelerations - Prof., Assignments of Physics

Download Kepler's and Newton's Laws of Planetary Motion: Periods, Speeds, and Accelerations - Prof. and more Assignments Physics in PDF only on Docsity! Assignment #6 Chapter 6 Questions: 6.26 Let the planet at distance r be labeled #1 and that at 2r be labeled #2. Periods (using Kepler’s third law) (T1/T2) 2 = (r1/r2) 3 T1/T2 = (r1/r2) 3/2 = (1/8)1/2 Speeds (use Newton’s 2nd law) GMm/r2 = v2/r v = (GMm/r)1/2 v1/v2 = √2 Accelerations (use Newton’s 2nd law with centripetal acceleration) GMm/r2 = ma a= GM/r2 a1/a2 = 4 6.31 Each orbital plane must contain the center of the earth at on focus of the orbital ellipse. The centers of the Artic and Antarctic circles and the circles of the Tropics of Cancer and Capricorn are not the center of the earth. Chapter 7 Questions: 7.14 a) The frequency is independent of the amplitude, A, so the frequency is unaffected b) The maximum speed is Aω, so increasing A by a factor of 3 triples the maximum speed. c) The magnitude of the maximum acceleration is ω2A, so increasing A by a factor of 3 triples the magnitude of the maximum acceleration 7.21 a) Nothing happens b) Nothing happens c) The effective value of the acceleration magnitude g (in Equation 7.22) increases, so the period decreases. d) The effective value of the acceleration magnitude g (in Equation 7.22) decreases, so the period increases. Chapter 6 Problems: 6.22 a) Use Kepler’s third law of planetary motion dn? "GM f the Earth and r is the radius of the orbit of the satellite. Solve for the radius. ve where M is the mass 0: [Care TS VG ¥y = 8.6400 x 10? s. ‘Therefore ‘The period T of the satellite is one ct {GMT _ rey 23 x 107 m} “ae < = 3.08 x 10" m/s. fF ~ "8.6400 x 10% m x Roms (3.08 x 107 m/ — eae =: ().224 m/s” 4.23.x 107 m Qcentripetal = 6.58 Here's the picture. Barth ns a) From the diagram, the major axis of the spacecraft orbit is the sum of the semimajor axes of the orbits of Venus and the Earth. Using the data on the inside cover, we have 2a = 1.00 AU 40.723 AU = a= 0.862 AU. Now use Kepler’s third law with customized units to find the spacecraft orbital period. 24 _¥? 3 y? 3 ‘ 7? = 1 et = 1apg (0.862 AUYS > T= 0.800 ‘Phe spacecraft only traverses half of its orbit, so the flight time is half the period. Thus T ffigu = = 0.400 y b) The aphelion distance of the spacecraft is 1.00 AU. The aphelion distance is Taphelion = (1 +e) => 1,00 AU = (0.862 AU)(1 +e) = > ¢= 0.160. The maximum magnitude of the acceleration is therefore dmax = Aw?. Hence, if m is not slipping but is on the verge of slipping femax = ™/4mar => feN =mAw? => pemg = mAw? => pg = Aw’. The total mass on the end of the spring is M-+m. The angular frequeney of the oscillation is Hence 42 k 1, Hel M+ m)g Meg = Aw =A = be If m is inereased and / is held constant, the angular frequency of the oscillation will decrease, so the maximum magnitude of acceleration will also decrease, and the force of static friction needed to prevent slippage will be less than f max» Hence, increasing m will mean that m is no longer on the verge of slipping. TAT a) The angular frequency is found from m ——— =11 5 rad/s m \ 150kg rays When t = 0s, the mass is released at 2 = 0.100 m, so the general equation for x becomes (1) 0.100 m = A cos[w(O0s) + 6] = A cosa, Likewise, when t = 0s, the velocity component is v, = 2.00 m/s. The velocity at any instant is d . ; v(t) = qt = —Aw sin(wt +4). (2) 2.00 m/s = —Aw sin[u(0 8) + 4] = —A(11.5 rad/s ) sin d. Divide equation (2) by equation (1). 2.00m/s — —A(11.5 rad /. 's) sing 0.100m Acos == 20.087! =-11.5 rad/s tang = tang = —-1.74 = = -1.05 rad. Now use this value for ¢ in equation (1) to find A: 0.100 m = Acos@ = Acos(—-1.05rad) == A=0.201 m. Hence x(t) = (0.201 mj cos[(11.5 rad/s jt — 1.05 rad]. b) The period T of the oscillation is the inverse of the frequency v: peta 2 asus, ce) The velocity component at any time is da(t) . vy(t) = —— = —Aw sin(wt + o). dt . Since the maximum magnitude of the sine is 1, the maximum speed is Unex =WA = (11.5 rad/s )(0.201 m) =2.31 m/s The acceleration component at any time is Since the maximum magnitude of the cosine is 1, the maximum magnitude of the acceleration is Gnas = WA = (11.5 rad/s }?(0.201 m ) = 26.6 m/s’ d) A plot of a(t) versus ¢ for two periods is shown below x(m) 0.2 O41 f(s) 7 <t a 46 B19 92 -0.2 7.30 a) Before the turkey is pulled down below its equilibrium position, the forees on it are: 1. its weight W, of magnitude mg, directed downward; and 2. the Hooke’s law force of the spring, of magnitude kx, where x is the amount by which the spring has stretched, directed upward Choose a coordinate system with i pointing down and origin at the point where the turkey would be if the spring were not extended In its equilibrium position, the turkey is not accelerating, so the total force on it is zero. mg — (8.00 kg )(9.81 my' 0.10 cm. 7.8 10? N/m Fe total =ON => mg —ke b) The amplitude of the oscillation is the additional distance the turkey is pulled down below its equilibrium position, A = (0.040 m. The angular freqnency of the oscillation is a7 300k When ¢ = 0s, the turkey is at r = 0,040 m, hence 0.040 m ) cos[w(Os)+o] == L=cosd == @ Therefore x(t) = (0.040 m ) cos[(9.9 rad fs )é] ocity component is = —Awsin(wt + 2) Since the maximum magnitude of the sine is 1, the maximum speed is Aw (0.040 m (9.9 rad/s) = 0.40 m/s Umax d) The acceleration component is du(t) dt a(t) = coslwt +). Since the maximum magnitude of the cosine is 1, the maximum magnitude of the acceleration is 2 Omax = Aw? = (0.040 m )(9.9 rad/s )? = 3.9m/s? TAG a) Spring 1 exerts a foree on the hero equal to (—kyr)i. (Directed toward the equilibrium position) b) Sprin; 2 also exerts a foree on the hero toward the equilibrium position. This force is (—h2 c) The total force on the hero is the vector sum of the two forces. B votat = (—kaw i+ (- —(hy + ka)ai. d) Apply Newton's sceand law to the hero in the horizontal direetion aS 4/2 m e) The coefficient of z(t) in the differential equation for a simple harmonic oscillator is the square of the angular frequency. Hence f 2 a(t) =0 m/s Pe total = md, => —(hy + ko) =m