{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Physics Solution Manual for 1100 and 2101

6 kms 586 103 ms equation 3 now gives the orbital

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . However we do know that the speed v of a satellite is equal to the circumference (2π r) of its orbit divided by the period T, so v = 2 π r / T . Chapter 5 Problems 263 SOLUTION Solving the relation v = 2 π r / T for r gives r = vT / 2π . Substituting this value for r into Equation (1) yields 3/2 v T / ( 2π ) TA rA = 3/2 = A A = 3/ 2 TB rB vB TB / ( 2π ) 3/ 2 ( vA TA ) 3/ 2 ( vB TB ) 3/ 2 Squaring both sides of this equation, algebraically solving for the ratio TA/TB, and using the fact that vA = 3vB gives 3 3 TA vB vB 1 = 3= = 3 TB vA ( 3vB ) 27 ______________________________________________________________________________ 36. REASONING AND SOLUTION The period of the moon's motion (approximately the length of a month) is given by T= 4π 2 r 3 = GM E 4π 2 ( 3.85 × 108 m ) ( 6.67 ×10−11 N ⋅ m 2 / kg 2 )( 5.98 ×1024 kg ) 3 = 2.38 × 106 s = 27.5 days 37. SSM REASONING Equation 5.2 for the centripetal acceleration applies to both the plane and the satellite, and the centripetal acceleration is the same for each. Thus, we have ac = 2 v plane rplane = 2 v satellite rsatellite or v plane = Fr Gr G H plane satellite I v J J K satellite The speed of the satellite can be obtained directly from Equation 5.5. SOLUTION Using Equation 5.5, we can express the speed of the satellite as v satellite = Gm E rsatellite Substituting this expression into the expression obtained in the reasoning for the speed of the plane gives 264 DYNAMICS OF UNIFORM CIRCULAR MOTION Fr I Fr v =G Gr J GJ H K Gr H 15 c b mg6.67 × 10 N ⋅ m = v plane = plane plane satellite satellite satellite −11 v plane 2 I Gm r Jr = r J K / kg h 5.98 c × 10 plane E satellite GmE satellite 2 24 6.7 × 10 6 m kg h = 12 m / s GM S r − 11 2 2 (Equation 5.5), where MS is the mass of the star, G = 6.674 × 10 N ⋅ m / kg is the universal gravitational constant, and r is the orbital radius. This expression can be solved for 2π r MS. However, the orbital radius r is not known, so we will use the relation v = T (Equation 5.1) to eliminate r in favor of the known quantities v and T (the period). Returning to Equation 5.5, we square both sides and solve for the mass of the star: 38. REASONING The speed v of a planet orbiting a star is given by v = GM S = v2 r Then, solving v = or MS = rv 2 G (1) 2π r vT for r yields r = , which we substitute into Equation (1): T 2π vT 2 v 3 rv 2π = v T MS = = G G 2π G 2 (2) We will use Equation (2) to calculate the mass of the star in part a. In part b, we will solve Equation (2) for the period T of the faster planet, which should be shorter than that of the slower planet. SOLUTION a. The speed of the slower planet is v = 43.3 km/s = 43.3×103 m/s. Its orbital period in seconds is T = (7.60 yr)[(3.156 × 107 s)/(1 yr)] = 2.40×108 s. Substituting these values into Equation (2) yields the mass of the star: ( )( 3 ) ) 43.3 × 103 m/s 2.40 × 108 s v 3T MS = = = 4.65 × 1031 kg −11 2 2 2π G 2π 6.674 × 10 N ⋅ m / kg ( This is roughly 23 times the mass of the sun. Chapter 5 Problems 265 b. Solving Equation (2) for the orbital period T, we obtain v 3T = MS 2π G or T= 2π GM S (3) v3 The speed of the faster planet is v = 58.6 km/s = 58.6 × 103 m/s. Equation (3) now gives the orbital period of the faster planet in seconds: T= ( )( 3 (58.6 ×103 m/s ) 2π 6.674 ×10−11 N ⋅ m2 / kg 2 4.65 ×1031 kg ) = 9.69 ×107 s Lastly, we convert the period from seconds to years: ( ) 1 yr T = 9.69 ×107 s 7 3.156 ×10 s = 3.07 yr 39. REASONING The satellite’s true weight W when at rest on the surface of the planet is the gravitational force the planet exerts on it. This force is given by W = GM P m r 2 (Equation 4.4), where G is the universal gravitational constant, MP is the mass of the planet, m is the mass of the satellite, and r is the distance between the satellite and the center of the planet. When the satellite is at rest on the planet’s surface, its distance from the planet’s 2 center is RP, the radius of the planet, so we have W = GM P m RP . The satellite’s mass m is given, as is the planet’s radius RP. But we must use the relation T = 2π r 3 2 (Equation 5.6) GM P to determine the planet’s mass MP in terms of the satellite’s orbital period T and orbital radius r. Squaring both sides of Equation 5.6 and solving for MP, we obtain T2 = () 2 ( GM P ) 22 π 2 r 3 2 Substituting Equation (1) into W = W= 2 = 4π 2 r 3 GM P GM P m 2 RP or MP = 4π 2 r 3 GT 2 (1) (Equation 4.4), we find that Gm G m 4π 2 r 3 4π 2r 3m MP ) = 2 = 2 2 2( RP RP G T 2 RPT (2) 266 DYNAMICS OF UNIFORM CIRCULAR MOTION SOLUTION All of the quantities in Equation (2), except for the period T, are given in SI base units, so we must convert the period from hours to seconds, the SI base unit for time: T = (2.00 h)[(3600 s)/(1 h)] = 7.20×103 s. The satellite’s orbital radius r in Equation (2) is the distance between the satellite and the center of the orbit, which is the planet’s center. Therefore, the orbital radius is the sum of the planet’s radius RP and the satellite’s height h...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online