Solutions to Odd Exercises

Solutions to all Odd Problems of BASIC CALCULUS OF PLANETARY ORBITS AND INTERPLANETARY FLIGHT Chapter 1. Solutions of Odd Problems Problem 1.1. In Figure 1.33, let V represent Mercury instead of Venus. Let α = \V ES be the maximum angle that Copernicus observed after measuring it many times. As in the case of Venus, α is at its maximum when the line of sight EV is tangent to the orbit of Mercury (assumed to be a circle). This means that when α is at its maximum, the triangle EV S is a right triangle VS with right angle at V . Since Copernicus concluded from his observations that ES ≈ 0:38, we VS know that sin α = ES ≈ 0:38. Working backward (today we would push the inverse sine button 1 ◦ of a calculator) provides the conclusion that α ≈ 22 3 . So the maximum angle that Copernicus 1 ◦ observed was approximately 22 3 . From this he could conclude that EV ≈ 0:38. Problem 1.3. Since the lines of site BA and B0A of Figure 1.36a are parallel, Figure 1.36b tells us that (α + β) + (α0 + β0) = π as asserted. By a look at the triangle ∆BB0M, β + β0 + θ = π. So β + β0 = π − θ, and hence α0 + α + (π − θ) = π. So θ = α + α0. Problem 1.5. Let R the radius of the circle of Figure 1.40b. Turn to Figure 1.40a, and notice that cos ' = R . The law of cosines applied to the triangle of Figure 1.40b tells us that (BB0)2 = rE R2 + R2 − 2R · R cos θ = 2R2(1 − cos θ). So BB0 = Rp2(1 − cos θ): Since the point B rotates on 1 the circle of Figure 1.40b at a constant rate, the ratio of 6 6 hours to 24 hours is the same as the 1 ◦ 6 6 ◦ ratio of θ expressed in degrees to 360 . So θ = 24 360 = 92:5 , and 0 p ◦ p ◦ BB = (rE cos ') 2(1 − cos θ) = (6364:57 cos 52:92 ) 2(1 − cos 92:5 ) ≈ 5543:98 km. a3 Problem 1.7. Newton's computations of T 2 for Saturn's five moons, were 1:953 2:53 3:53 83 243 45:312 ≈ 0:00361; 65:692 ≈ 0:00362; 108:422 ≈ 0:00365; 372:692 ≈ 0:00369, and 1903:82 ≈ 0:00381, where for each of the five moons, a is the radius of the circular orbit (and hence also the semimajor axis of the orbit) measured in outer ring radii, and T is the corresponding period of the orbit in hours. Given the difficulty of determining the radii and the periods of the orbits of these five moons with precision with the instruments that existed in Newton's time, these results seem close enough to confirm Kepler's third law for Saturn and these moons. Figure Sol 1.1 shows a planet (comet or asteroid) at its aphelion position A and a short stretch of its orbit QA before its arrival at A. Its time of travel from Q to A is ∆t. It has already been Q R Δs A S Sol 1.1 established that a planet (or comet or asteroid) attains its minimum speed vmin at aphelion. 1 Problem 1.9. i. Let arc QA be the length of the arc between Q and A. Since ∆t is the time of arc QA travel of the body, its average velocity vav during its trip from Q to A is equal to ∆t . For a very arc QA ∆s small ∆t, Q is close to A so that arc QA is close to ∆s, so that vav = ∆t ≈ ∆t . ii. For a small ∆s, the area ∆A of the elliptical sector SAQ is approximately equal to the area of 1 1 ∆A 2 (SA·∆s) the triangle ∆SAQ. Since the area of this triangle is 2 (SA · ∆s) it follows that κ = ∆t ≈ ∆t . 1 2 (a+c)·∆s abπ ∆s 2abπ From Figure 1.9, SA = a + c, so that κ ≈ ∆t . Since κ = T It follows that ∆t ≈ (a+c)T : ∆s iii. When ∆t is pushed to zero, Q gets pushed to A, so that the average speed vav ≈ ∆t , gets pushed to the speed vmin at aphelion. Simultaneously, the approximation of the area ∆A of the 1 1 elliptical sector SAQ by the area 2 (SA · ∆s) = 2 (a + c) · ∆s of ∆SAQ becomes tighter and tighter. 1 ∆A 2 (a+c)·∆s 1 abπ Therefore, the approximations κ = ∆t ≈ ∆t ≈ 2 (a + c)vmin snap to equalities. Since κ = T 2abπ this leads to the conclusion vmin = (a+c)T : p p 2 2 2 iv. Since c = "a and b = a − c , we get a + c = a(1 + ") and b = a 1 − " . So vmin = p p p 2 2 2a( 1−")( 1+") q 2a 1−" π p π 2aπ 1−" a(1+") T = ( 1+")2 T = T 1+" : Recall that 149,597,870.7 km 1 year 1 au/year = × ≈ 4:74 km/sec: 1 year 31,557,600 sec Problem 1.11. With Halley's semimajor axis a = 17:83 au, eccentricity " = 0:967, and period T = 75.32 years, we get 2aπ q 1+" 2(17:83)π q 1+0:967 vmax = T 1−" = 75:32 1−0:967 ≈ 11:483 au/year ≈ 54:43 km/sec and 2aπ q 1−" 2(17:83)π q 1−0:967 vmin = T 1+" = 75:32 1+0:967 ≈ 0:193 au/year ≈ 0:91 km/sec: 2 2 MME 4π M ME d 4π M ME d MME Problem 1.13. Using both F = G 2 and F = 2 , we get 2 = G 2 , hence d T ME +M T ME +M d 2 2 3 2 3 2 3 4π d G 4π d 4π d 4π d ME +M M 2 = 2 , and therefore 2 · = G: So · 2 = ME +M and · 2 = = 1+ : T ME +M d T ME +M G T GME T ME ME Problem 1.15. i. Let F be the force of gravity of Earth on the Moon, ME the mass of Earth, and M that of the Moon. By Newton's law of universal gravitation, 14 22 GME M (3:98600×10 )(7:3461×10 ) 20 F ≈ 2 = 8 2 = 1:98165 × 10 newtons. aM (3:844×10 ) ii. Let F be the force of gravity of the Sun on the Moon, MS the mass of the Sun, and M that of the Moon. By Newton's law of universal gravitation, 20 22 GMS M (1:32712×10 )(7:3461×10 ) 20 F ≈ 2 = 11 2 = 4:35628 × 10 newtons. aE (1:49598×10 ) iii. So the gravitational force of the Sun on the Moon is more than twice than that of Earth on the Moon. So why is the Moon orbiting Earth? The fact is that both Earth and Moon are orbiting the Sun. As these orbiting motions occur, the Earth pulls the Moon into orbit around it. −11 m3 Problem 1.17. With G = 6:67384 × 10 kg·sec2 in MKS, we get −11 m3 −11 m3 (100 cm)3 1 kg 1 −8 cm3 G = 6:67384 × 10 kg·sec2 = 6:67384 × 10 kg·sec2 · 1 m3 · 1000 gr · sec = 6:67384 × 10 gr·sec2 : 2 Problem 1.19. We know from section 1G that 1 foot = 30.48 cm and 1 inch = 2.54 cm. With the 1 2:54 2:54 spheres of Figure 1.47 \distant but by 4 inch", 2d = 4 so that d = 8 = 0:3175 cm. Since the 1 radius of the spheres is 2 foot, c = 15:24 + d = 15:24 + 0:3175 = 15:5575 cm. The volume of each 4 3 3 3 of the two spheres is 3 π(15:24) cm ≈ 14827 cm . Under the assumption that their densities are the same as that of the Earth, the mass of each sphere is (14827)(5:54) gr = 82;142 gr. Problem 1.21. If can determine the time t it takes for the mutual force of attraction to move each of the spheres through a distance d, then the two spheres will touch after time t and the problem q 1 F 2 2 2mx(t) 2mx(t) is solved. The equality x(t) = 2 m t informs us that t = F and hence that t = F . With x(t) = d = 0:3175 cm and F equal to the magnitude of the maximal force of 0:485 dynes throughout the motion of the sphere, we get q 2mx(t) q 2(82;142)(0:3175) t = F = 0:485 = 327:94 sec. q 2mx(t) Under the assumption that the minimal force acts throughout the motion, we get t = F = q 2(82;142)(0:3175) 0:465 = 334:92 sec and it takes a little longer for the two spheres to meet. Since the force of gravity between the two sphere varies from the smaller of these two values to the larger, 1 the time it takes for the two spheres to meet lies between the two calculated values, thus about 5 2 minutes. This conclusion is of course at odds with Newton's assertion about the matter. Why? While Newton could compute GM with M the Earth's mass (from his version of Kepler's third law), he had no clue about G (as he had asserted himself). Problem 1.23. i. Let a and " be the semimajor axis and eccentricity of Sputnik 1's elliptical orbit.

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