solve parts 3.5, 3.6 and section 4

Anonymous
timer Asked: Apr 25th, 2017

Question description

attached the whole project with the solutions of section 2 and 3 except 3.6

please find section 4 and part 3.6

solve parts 3.5, 3.6 and section 4
12.png
solve parts 3.5, 3.6 and section 4
media_2fa59_2fa59da76a_90b7_4dfc_8aae_98e6f89373d9_2fphphchivi.png
solve parts 3.5, 3.6 and section 4
7.png
solve parts 3.5, 3.6 and section 4
8.png
solve parts 3.5, 3.6 and section 4
9.png
solve parts 3.5, 3.6 and section 4
10.png
solve parts 3.5, 3.6 and section 4
7.png
solve parts 3.5, 3.6 and section 4
8.png
solve parts 3.5, 3.6 and section 4
9.png
solve parts 3.5, 3.6 and section 4
10.png
solve parts 3.5, 3.6 and section 4
9.png
solve parts 3.5, 3.6 and section 4
10.png
solve parts 3.5, 3.6 and section 4
11.png
solve parts 3.5, 3.6 and section 4
12.png
solve parts 3.5, 3.6 and section 4
media_2fa59_2fa59da76a_90b7_4dfc_8aae_98e6f89373d9_2fphphchivi.png
Project Overview The goal of this project is to find an approximate — but highly accurate — solution to the nonlinear differential equation that governs the orbits of planets around the sun according to general relativity. • In Section 1, you will derive the governing differential equation for the orbit of a planet around the sun assuming Newtonian mechanics. • In Section 2, you will solve this governing differential equation. You will show that, according to Newtonian mechanics, planets indeed orbit in ellipses with the sun at one focus. • In Section 3, you will consider the solution of the governing differential equation for orbits according to Einstein’s theory of general relativity — which is a slight modification of Newton’s Law of Gravitation. You will see general relativity predicts that the elliptical orbits of planets precess around the sun. • In Section 4, you will confirm Einstein’s theory of general relativity by predicting the precession in the orbits of Mercury and Venus. In the figure below, a planet moves about the Sun (located at the origin). The planet’s position is marked by polar coordinates (r, θ) as well as Cartesian coordinates (x, y). For convenience, the angle θ = 0 is chosen to be the planet at perihelion, or the position in the orbit closest to the sun. In other words, when θ = 0, r achieves its minimum value of P . Recall from Precalculus that x = r cos θ and y = r sin θ. (1) We will use (without proof) the following facts from physics. " d2 r dθ Newton’s Second Law, written in polar coordinates: F = ma = m −r dt2 dt GM m Newton’s Universal Law of Gravitation: F =− r2 dθ Conservation of Angular Momentum, written in polar coordinates: mr2 = `, dt  2 # (2) (3) (4) where G is the gravitational constant of the universe, m is the mass of the planet, M is the mass of the sun, and ` is the constant angular momentum of the planet. y r θ x Perihelion P c a 1 Governing Differential Equation under Newtonian Mechanics Problem 1.1 Please write “Yes” if you’ve read the entire project (so far). Problem 1.2 Use (4) to solve for dθ . (Yes, this is easy.) dt Problem 1.3 According to the Chain Rule, dr dr dθ = . Use the result of Problem 1.2 to show that dt dθ dt dr ` d =− dt m dθ 1 . r   (5) Hint: This can be confirmed by differentiating the right-hand side of (5) using the Chain Rule (or, if you prefer, implicit differentiation). Problem 1.4 According to the Chain Rule, d2 r dr0 dr0 dθ d = = = 2 dt dt dθ dt dθ  dr dt  · dθ . dt (6) Use (5), (6) and the result of Problem 1.2 to show that d2 r `2 d2 = − dt2 m2 r2 dθ2 1 . r   (7) Problem 1.5 Combine (2), (3), (7) and the result of Problem 1.2 to obtain the governing differential equation   d2 1 1 GM m2 + = , (8) dθ2 r r `2 which may be rewritten as 1 α (9) f 0 (0) = 0. (10) f 00 (θ) + f (θ) = where f (θ) = 1 `2 and α = . r(θ) GM m2 Problem 1.6 Explain why the initial conditions are f (0) = 1 , P 2 Mercury’s Orbit under Newtonian Mechanics In this problem, you will show that the solution of (9) and (10) is an ellipse, expressed in polar coordinates. In other words, Newton’s Law of Gravitation predict that planets move in ellipses. Problem 2.1 Find the general solution of (9). Then use the initial conditions (10) to show that f (θ) = 1 + ε cos θ , α (11) where α−P . P The physical meaning of ε will become clear in the next few problems. ε= (12) Problem 2.2 In light of (11), f (θ) is maximized (i.e., the planet is at perihelion) when θ is a multiple of what angle? Problem 2.3 Recall that f (θ) = 1/r(θ). Use this to rewrite (11) as α − εr cos θ = r. (13) Problem 2.4 Square both sides of (13), use (1) to convert to Cartesian coordinates, and complete the square to show that (13) can be rewritten as (x − h)2 y 2 + 2 = 1, a2 b (14) where α , 1 − ε2 α b = √ , 1 − ε2 αε h = − . 1 − ε2 a = (15) (16) (17) Therefore, if 0 < ε < 1 so that b is defined, the orbit of a planet is an ellipse!∗ ∗ Recall from Precalculus that the distance of the focus from the center of the ellipse is given by p αε c = a2 − b2 = 1 − ε2 and that the eccentricity of the ellipse is given by e = c/a = ε. Therefore, the undetermined constant ε of (12) represents the eccentricity of the planet’s orbit. Also, since c = −h, the sun is located at a focus of the ellipse. Though we do not use this fact here, other possible orbits are circles (ε = 0), parabolas (ε = 1), and hyperbolas (ε > 1). 3 Einstein’s General Relativity If the universe consisted of only Mercury and the sun, Mercury’s trajectory would trace the same ellipse over and over again. However, there are eight other planets in the solar system, and these planets tug and nudge the orbit of Mercury ever so slightly.† The practical effect of these nudges is that the orbit of Mercury precesses, or rotates like a spiral. Since the planets are much smaller than the sun and are further away from the Sun than Mercury, this precession is very small. However, this effect can be measured. Every century, the perihelion of Mercury precesses by 57400 of arc (roughly a sixth of a degree‡ ). Newton’s Law of Gravitation can be used to calculate the amount of the precession of Mercury; however, they predict a precession of only 53100 of arc per century. This discrepancy between observation and prediction was first observed in 1845 and was, for a long time, the outstanding unresolved difficulty in Newton’s Law. Einstein’s general theory of relativity, which was published seventy years later in 1915, exactly accounts for the missing 4300 per century (within the tolerances of observational error). This was the first physical confirmation of general relativity. Furthermore, general relativity predicted that the orbit of Venus also precesses, but by only about 900 of arc per century. This small discrepancy was unobservable in 1915 but was confirmed in 1960. The figure below shows the (greatly exaggerated) effect of precession on a planet’s otherwise elliptical orbit. In the figure, each perihelion is precessed by an angle of φ = 40◦ . After nine orbits, the planet returns to its original position. According to general relativity — which we accept here without proof — the governing differential equation for a planet’s motion is not (9) but should be corrected as f 00 (θ) + f (θ) = 1 + δ[f (θ)]2 , α where (18) 3GM (19) c2 and c is the speed of light. Also, the initial conditions (10) still apply. The differential equation (18) cannot be solved exactly, but an accurate approximation can be found by instead solving 1 f 00 (θ) + f (θ) = + δ[f1 (θ)]2 , (20) α where we have replaced the f on the right-hand side with f1 , the Newtonian solution given by (11). δ= † ‡ Similar nudges in the orbit of Uranus led to the discovery of Neptune in 1846. Remember that one degree contains 3600 seconds of arc. Problem 3.1 Please write “Yes” if you have read the entire project (so far). Problem 3.2 Substitute (11) into the right-hand side of (20) and expand the right-hand side. Problem 3.3 Find the general solution of (20). • You can use either the method of undetermined coefficients or variation of parameters to find a particular solution. If you use the method of undetermined coefficients, be sure to use a trigonometric identity to replace cos2 θ with an expression involving cos 2θ. • Remember that α, δ, and ε are constants; the only variable on the right-hand side is θ. Problem 3.4 In Problem 3.3, you showed that the general solution of (20) was f (θ) = c1 cos θ + c2 sin θ + 1 δ ε2 δ εδ δε2 + 2 + 2 + 2 θ sin θ − 2 cos 2θ. α α 2α α 6α Now use the initial conditions (??) to solve for c1 and c2 and show that " δ ε2 ε2 cos 2θ (3 + ε2 ) cos θ 1 + ε cos θ + 2 1+ + εθ sin θ − − f (θ) = α α 2 6 3 # (21) Notice that the first term of (21) describes an elliptical orbit, as we saw in Problem 2.4, while the remaining terms are small perturbations — since the coefficient δ/α2 is a very, very small number — to that elliptical orbit. Problem 3.5 Let’s now pretend to be physicists and start estimating everything in sight to simplify the form of f (θ). For large θ (many orbits), explain why f (θ) may be safely approximated by 1 εδ 1 δθ f (θ) ≈ (1 + ε cos θ) + 2 θ sin θ = 1 + ε cos θ · 1 + sin θ · α α α α    . (22) Hint: Compare the magnitudes of the perturbation terms in (21) as θ gets large. Problem 3.6 Use a trig identity and the fact that δ/α is a small number to show that 1 δθ 1 + ε cos θ − α α  f (θ) ≈   . Hint: You will need to use the first terms in the Maclaurin expansions of sin x and cos x. (23) 4 Mercury’s Orbit under General Relativity According to Newtonian mechanics, a planet’s orbit traces the same ellipse over and over again. By contrast, as we will now show, according to general relativity, each perihelion is rotated by a small angle of φ. . . just as in the figure in Section 3. Are the effects of general relativity observable? As it turns out, for φ to be as large as possible, we need a (the semimajor axis of the orbit; see the first page of this project) to be small and ε (the eccentricity of the orbit) to be large. By a fortunate coincidence, Mercury — the closest planet to the sun — has the most elliptical orbit of the eight planets. Problem 4.1 Let’s now return to the solution from Problem 3.6 of orbits under general relativity. Use (15), (19) and (23) to show that perihelia occur when θ is a multiple of 2π + φ, where φ= 2π 2πδ 6πGM − 2π ≈ = 2 . 1 − (δ/α) α ac (1 − ε2 ) (24) (Notice that this is different than your answer to Problem 2.2, the elliptical orbit from Newtonian mechanics.) This angle φ is the amount of the precession — in other words, the angle past a full rotation of 2π radians where the next perihelion occurs. The units of φ are radians per orbit. Hint: For the second step, remember that δ/α is a very, very small number. Problem 4.2 Calculate the predicted value of φ for Mercury according to general relativity using the following numbers (in the SI system): gravitational constant mass of the Sun semimajor axis of Mercuty’s orbit speed of light eccentricity of Mercury’s orbit time for Mercury to complete one orbit G M a c ε T = = = = = = 6.6726 × 10−11 N-m2 /kg2 1.9929 × 1030 kg 5.7871 × 1010 m 2.9979 × 108 m/s 0.2056 0.2408 years Remember that the units for φ are radians per orbit; you’ll need to convert this to seconds of arc per century. The observed Newtonian discrepancy in the precession of Mercury’s orbit is 43.1100 ± 0.4500 of arc per century. If your answer lies in this range, you have verified the general theory of relativity! Problem 4.3 Now predict the precession of Venus’ orbit in seconds of arc per century. To do this, you need to know the following information about Venus: semimajor axis of Venus’s orbit eccentricity of Venus’s orbit time for Venus to complete one orbit a ε T = 1.0813 × 1011 m = 0.0068 = 0.6152 years As mentioned earlier, this prediction was made 45 years before telescopes were advanced enough to observe precession in the orbit of Venus.

Tutor Answer

(Top Tutor) Studypool Tutor
School: Carnegie Mellon University
Studypool has helped 1,244,100 students
flag Report DMCA
Similar Questions
Hot Questions
Related Tags
Study Guides

Brown University





1271 Tutors

California Institute of Technology




2131 Tutors

Carnegie Mellon University




982 Tutors

Columbia University





1256 Tutors

Dartmouth University





2113 Tutors

Emory University





2279 Tutors

Harvard University





599 Tutors

Massachusetts Institute of Technology



2319 Tutors

New York University





1645 Tutors

Notre Dam University





1911 Tutors

Oklahoma University





2122 Tutors

Pennsylvania State University





932 Tutors

Princeton University





1211 Tutors

Stanford University





983 Tutors

University of California





1282 Tutors

Oxford University





123 Tutors

Yale University





2325 Tutors