Astronomy and Orbits � Build and Modify Trigonometric Orbit Equations (HSF-TF.B.5)

1) A satellite orbits Earth with altitude modeled by h(t) = 420 + 35 cos(2πt/90), where h is in km and t is in minutes. What is the maximum altitude?

420 km

385 km

455 km

35 km

Maximum occurs when cos = 1.

So, h = 420 + 35(1) = 455 km.

Hence, 455 km.

Explanation

Maximum occurs when cos = 1.

So, h = 420 + 35(1) = 455 km.

Hence, 455 km.

2) A moon's distance from its planet varies according to d(t) = 12,000 + 800 sin(πt/14), where d is in km and t is in days. What is the period of the orbit?

28 days

14 days

7 days

56 days

ω = π/14 ⇒ T = 2π/ω = 2π / (π/14) = 28 days.

Hence, 28 days.

Explanation

ω = π/14 ⇒ T = 2π/ω = 2π / (π/14) = 28 days.

Hence, 28 days.

3) An exoplanet's radial velocity is v(t) = 45 sin(2πt/4.2) m/s, where t is in days. What is the amplitude of the velocity variation?

4.2 m/s

45 m/s

2.1 m/s

90 m/s

Amplitude = coefficient of sine = 45 m/s.

Hence, 45 m/s.

Explanation

Amplitude = coefficient of sine = 45 m/s.

Hence, 45 m/s.

4) A comet's distance from the sun is r(θ) = 3.5 − 1.5 cos(θ) AU, where θ is the orbital angle. What is the perihelion distance (closest to sun)?

3.5 AU

5.0 AU

1.5 AU

2.0 AU

Minimum occurs when cosθ = 1.

r = 3.5 − 1.5 = 2.0 AU.

Hence, 2.0 AU.

Explanation

Minimum occurs when cosθ = 1.

r = 3.5 − 1.5 = 2.0 AU.

Hence, 2.0 AU.

5) A space station's position is given by x(t) = 8000 cos(πt/45) km and y(t) = 8000 sin(πt/45) km. What is the orbital radius?

8000 km

16000 km

11314 km

4000 km

x² + y² = 8000² ⇒ radius = 8000 km.

Hence, 8000 km.

Explanation

x² + y² = 8000² ⇒ radius = 8000 km.

Hence, 8000 km.

6) A planet's ecliptic latitude is β(t) = 7 sin(2πt/12) degrees, where t is in years. How often does the planet cross the ecliptic plane (β = 0)?

Every 12 years

Every 6 years

Every 3 years

Every 24 years

Crossings occur twice per cycle ⇒ every T/2.

T = 12 ⇒ T/2 = 6 years.

Hence, every 6 years.

Explanation

Crossings occur twice per cycle ⇒ every T/2.

T = 12 ⇒ T/2 = 6 years.

Hence, every 6 years.

7) A binary star system has one star's brightness varying as B(t) = 100 + 20 cos(2πt/8), where t is in days. At t = 2 days, what is the brightness?

120

80

100

110

B(2) = 100 + 20 cos(2π·2/8) = 100 + 20 cos(π/2) = 100.

Hence, 100.

Explanation

B(2) = 100 + 20 cos(2π·2/8) = 100 + 20 cos(π/2) = 100.

Hence, 100.

8) To model a satellite with altitude varying between 350 km and 450 km with a 2-hour period, which function is correct (t in hours)?

400 + 100 cos(πt)

400 + 50 cos(2πt)

350 + 100 cos(πt)

400 + 50 cos(πt)

Midline = (450 + 350)/2 = 400, amplitude = 50.

Period = 2 ⇒ ω = π.

Hence, 400 + 50 cos(πt).

Explanation

Midline = (450 + 350)/2 = 400, amplitude = 50.

Period = 2 ⇒ ω = π.

Hence, 400 + 50 cos(πt).

9) A moon's orbital speed varies as v(t) = 1.2 + 0.3 sin(πt/6) km/s. What is the average orbital speed?

1.2 km/s

1.5 km/s

0.9 km/s

0.3 km/s

Average speed = midline = 1.2 km/s.

Hence, 1.2 km/s.

Explanation

Average speed = midline = 1.2 km/s.

Hence, 1.2 km/s.

10) An asteroid's distance from the sun is d(t) = 2.8 + 1.2 cos(2πt/5), where t is in years. What is the aphelion distance (farthest from sun)?

2.8 AU

4.0 AU

1.6 AU

1.2 AU

Aphelion = midline + amplitude = 2.8 + 1.2 = 4.0 AU.

Hence, 4.0 AU.

Explanation

Aphelion = midline + amplitude = 2.8 + 1.2 = 4.0 AU.

Hence, 4.0 AU.

11) A satellite's ground track latitude is L(t) = 51.6 sin(2πt/90) degrees, where t is in minutes. What is the maximum northern latitude reached?

90 degrees

25.8 degrees

51.6 degrees

103.2 degrees

Amplitude = 51.6 ⇒ maximum latitude = 51.6°.

Hence, 51.6 degrees.

Explanation

Amplitude = 51.6 ⇒ maximum latitude = 51.6°.

Hence, 51.6 degrees.

12) If a planet's x-coordinate is x(t) = 5 cos(ωt) and y-coordinate is y(t) = 5 sin(ωt), and the period is 12 years, what is ω?

2π/12 rad/year

12/2π rad/year

6/π rad/year

π/6 rad/year

ω = 2π / T = 2π / 12.

Hence, 2π/12 rad/year.

Explanation

ω = 2π / T = 2π / 12.

Hence, 2π/12 rad/year.

13) A star's apparent magnitude varies as m(t) = 4.5 + 0.8 cos(2πt/3 − π/2), where t is in days. At t = 0, what is the magnitude?

4.5

5.3

3.7

0.8

m(0) = 4.5 + 0.8 cos(−π/2) = 4.5 + 0 = 4.5.

Hence, 4.5.

Explanation

m(0) = 4.5 + 0.8 cos(−π/2) = 4.5 + 0 = 4.5.

Hence, 4.5.

14) To increase the amplitude of an orbit model h(t) = 500 + 100 sin(2πt/T) from 100 to 150 while keeping the same period and midline, the new model should be:

500 + 150 sin(πt/T)

500 + 150 sin(2πt/T)

650 + 100 sin(2πt/T)

500 + 100 sin(3πt/T)

Only amplitude changes ⇒ replace 100 with 150.

Hence, 500 + 150 sin(2πt/T).

Explanation

Only amplitude changes ⇒ replace 100 with 150.

Hence, 500 + 150 sin(2πt/T).

15) A moon orbits with distance d(t) = 15 + 3 cos(πt/8) thousand km. At what time t (0 ≤ t ≤ 16 hours) does it first reach maximum distance?

4 hours

8 hours

0 hours

16 hours

Maximum occurs when cos = 1 ⇒ at t = 0.

Hence, 0 hours.

Explanation

Maximum occurs when cos = 1 ⇒ at t = 0.

Hence, 0 hours.

16) An exoplanet transits its star with light intensity I(t) = 1.0 − 0.02 sin²(πt/6), where t is in hours. What is the minimum intensity during transit?

1

0.02

0.96

0.98

Min occurs when sin² = 1 ⇒ I = 1.0 − 0.02 = 0.98.

Hence, 0.98.

Explanation

Min occurs when sin² = 1 ⇒ I = 1.0 − 0.02 = 0.98.

Hence, 0.98.

17) A satellite's altitude is h(t) = 600 + 200 cos(2πt/120) km. To double the period while keeping amplitude and midline the same, the new model is:

600 + 200 cos(2πt/240)

600 + 200 cos(2πt/60)

600 + 200 cos(4πt/120)

600 + 400 cos(2πt/120)

Doubling period ⇒ replace 120 with 240.

Hence, 600 + 200 cos(2πt/240).

Explanation

Doubling period ⇒ replace 120 with 240.

Hence, 600 + 200 cos(2πt/240).

18) A planet's heliocentric distance varies from 0.7 AU to 1.5 AU. Which model correctly represents this with a cosine function starting at maximum?

1.1 + 0.8 cos(ωt)

1.1 + 0.4 cos(ωt)

0.7 + 0.8 cos(ωt)

1.5 − 0.4 cos(ωt)

Midline = (1.5 + 0.7)/2 = 1.1, amplitude = 0.4.

Starts at maximum ⇒ +cos.

Hence, 1.1 + 0.4 cos(ωt).

Explanation

Midline = (1.5 + 0.7)/2 = 1.1, amplitude = 0.4.

Starts at maximum ⇒ +cos.

Hence, 1.1 + 0.4 cos(ωt).

19) A moon's vertical position is z(t) = 250 sin(2πt/18 + π/3) km. At t = 0, what is the position (nearest km)?

0 km

250 km

217 km

125 km

z(0) = 250 sin(π/3) = 250·(√3/2) ≈ 216.5 ≈ 217 km.

Hence, 217 km.

Explanation

z(0) = 250 sin(π/3) = 250·(√3/2) ≈ 216.5 ≈ 217 km.

Hence, 217 km.

20) For circular orbital motion x(t) = R cos(ωt), y(t) = R sin(ωt), if the period is 96 minutes and R = 6800 km, what is the orbital speed (nearest km/min)?

70.8 km/min

222.1 km/min

141.6 km/min

444 km/min

v = Rω = R(2π/T) = 6800·(2π/96) ≈ 444 km/min.

Hence, 444 km/min.

Explanation

v = Rω = R(2π/T) = 6800·(2π/96) ≈ 444 km/min.

Hence, 444 km/min.