Astronomy and Orbits � Real-World Modeling and Interpretation (HSF-TF.B.5)

1) A satellite's distance from Earth's center is modeled by d(t) = 6800 + 400 cos(0.25 t) km, where t is in hours. What is the maximum distance?

6400 km

6800 km

7200 km

7600 km

Maximum occurs when cos = 1.

So, d = 6800 + 400(1) = 7200 km.

Hence, 7200 km.

Explanation

Maximum occurs when cos = 1.

So, d = 6800 + 400(1) = 7200 km.

Hence, 7200 km.

2) A moon’s height above a planet’s surface is h(t) = 900 + 250 sin(π t / 6) km. What is the period?

6 hours

12 hours

18 hours

24 hours

ω = π/6 ⇒ T = 2π/ω = 2π ÷ (π/6) = 12 h.

Hence, 12 hours.

Explanation

ω = π/6 ⇒ T = 2π/ω = 2π ÷ (π/6) = 12 h.

Hence, 12 hours.

3) An asteroid’s distance from the sun varies as r(θ) = 3.5 − 1.1 cos θ AU. What is the minimum distance?

2.4 AU

3.5 AU

4.6 AU

1.1 AU

Minimum occurs when cos θ = 1.

r = 3.5 − 1.1 = 2.4 AU.

Hence, 2.4 AU.

Explanation

Minimum occurs when cos θ = 1.

r = 3.5 − 1.1 = 2.4 AU.

Hence, 2.4 AU.

4) A spacecraft’s vertical oscillation is z(t) = 150 sin(2π t / 96) km, t in minutes. What is the frequency in cycles per minute?

1/96

2/96

96

150

Frequency f = 1/T, and T = 96.

So, f = 1/96 cycles per minute.

Hence, 1/96.

Explanation

Frequency f = 1/T, and T = 96.

So, f = 1/96 cycles per minute.

Hence, 1/96.

5) A probe’s radial distance is d(t) = 10 + 3 sin(0.5 t) thousand km. Between what distances does it travel?

5 to 15 thousand km

10 to 13 thousand km

3 to 17 thousand km

7 to 13 thousand km

Amplitude = 3 ⇒ range = 10 ± 3 ⇒ [7, 13].

Hence, 7 to 13 thousand km.

Explanation

Amplitude = 3 ⇒ range = 10 ± 3 ⇒ [7, 13].

Hence, 7 to 13 thousand km.

6) A planet’s radial distance is r(t) = 1.8 + 0.2 cos(2π t / 365) AU, t in days. What is the amplitude and midline?

Amplitude 0.2 AU, midline 1.8 AU

Amplitude 1.8 AU, midline 0.2 AU

Amplitude 365 AU, midline 2π AU

Amplitude 0.1 AU, midline 2.0 AU

Amplitude = 0.2 (coefficient of cosine).

Midline = 1.8 (constant term).

Hence, amplitude 0.2 AU, midline 1.8 AU.

Explanation

Amplitude = 0.2 (coefficient of cosine).

Midline = 1.8 (constant term).

Hence, amplitude 0.2 AU, midline 1.8 AU.

7) A satellite’s x and y coordinates are x(t) = 5000 cos(π t / 60), y(t) = 5000 sin(π t / 60) km. What is the orbital period?

60 minutes

90 minutes

120 minutes

240 minutes

ω = π/60 ⇒ T = 2π / (π/60) = 120 min.

Hence, 120 minutes.

Explanation

ω = π/60 ⇒ T = 2π / (π/60) = 120 min.

Hence, 120 minutes.

8) A comet’s distance is modeled by d(t) = 6 − 2 cos(0.4 t) AU. If we want the same midline but double the amplitude, which model is correct?

D(t) = 6 − 4 cos(0.4 t)

D(t) = 6 − 2 cos(0.8 t)

D(t) = 6 + 4 cos(0.4 t)

D(t) = 8 − 4 cos(0.4 t)

Double amplitude from 2 → 4, keep midline 6, same phase.

Hence, d(t) = 6 − 4 cos(0.4 t).

Explanation

Double amplitude from 2 → 4, keep midline 6, same phase.

Hence, d(t) = 6 − 4 cos(0.4 t).

9) A moon’s orbital radius (in thousand km) is r(t) = 15 + 5 sin(0.2 t). How long (in hours) between successive maxima?

2π hours

5π hours

10π hours

20π hours

ω = 0.2 ⇒ T = 2π / 0.2 = 10π h.

Hence, 10π hours.

Explanation

ω = 0.2 ⇒ T = 2π / 0.2 = 10π h.

Hence, 10π hours.

10) A satellite’s altitude is h(t) = 420 + 60 sin(π t / 45) km. What is h(22.5)?

420 km

480 km

360 km

450 km

Substitute t = 22.5:

h = 420 + 60 sin(π·22.5/45) = 420 + 60 sin(π/2) = 420 + 60(1) = 480.

Hence, 480 km.

Explanation

Substitute t = 22.5:

h = 420 + 60 sin(π·22.5/45) = 420 + 60 sin(π/2) = 420 + 60(1) = 480.

Hence, 480 km.

11) An exoplanet’s radial velocity is v(t) = 35 sin(2π t / 3.6) m/s, t in days. What is the period?

1.8 days

3.6 days

7.2 days

35 days

ω = 2π / 3.6 ⇒ T = 3.6 days.

Hence, 3.6 days.

Explanation

ω = 2π / 3.6 ⇒ T = 3.6 days.

Hence, 3.6 days.

12) A spacecraft’s distance is d(t) = 12 + 4 cos(π t / 10) thousand km. To shift the first maximum from t = 0 to t = 5, which phase shift is correct?

D(t) = 12 + 4 cos(π t / 10 − π/2)

D(t) = 12 + 4 cos(π (t + 5) / 10)

D(t) = 12 + 4 cos(π t / 10 + π/2)

D(t) = 12 + 4 cos(π (t − 5) / 10)

To delay maximum by 5, replace t with (t − 5).

Hence, 12 + 4 cos(π (t − 5) / 10).

Explanation

To delay maximum by 5, replace t with (t − 5).

Hence, 12 + 4 cos(π (t − 5) / 10).

13) A moon orbits with distance d(t) = 22 + 6 cos(0.3 t) thousand km. When does it first reach its closest distance (t ≥ 0)?

T = (π)/0.6

T = 2π/0.3

T = π/0.3

T = (2π)/0.6

Minimum when cos = −1 ⇒ 0.3t = π ⇒ t = π / 0.3.

Hence, π / 0.3.

Explanation

Minimum when cos = −1 ⇒ 0.3t = π ⇒ t = π / 0.3.

Hence, π / 0.3.

14) A satellite’s north-south position is y(t) = 250 sin(2π t / T) km. If T is halved, what happens to the frequency?

Unchanged

Halves

Doubles

Quadruples

f = 1 / T ⇒ if T halves, f doubles.

Hence, frequency doubles.

Explanation

f = 1 / T ⇒ if T halves, f doubles.

Hence, frequency doubles.

15) An asteroid’s polar orbit is modeled by r(θ) = 5 / (1 + 0.25 cos θ) AU. What is r(0)?

4 AU

5 AU

6.25 AU

20/5 AU

r(0) = 5 / (1 + 0.25·1) = 5 / 1.25 = 4 AU.

Hence, 4 AU.

Explanation

r(0) = 5 / (1 + 0.25·1) = 5 / 1.25 = 4 AU.

Hence, 4 AU.

16) A planet’s star-planet distance varies from 0.9 AU (min) to 1.1 AU (max). Which model fits?

D(t) = 1.0 + 0.1 cos(ω t)

D(t) = 1.0 − 0.2 cos(ω t)

D(t) = 1.1 + 0.1 cos(ω t)

D(t) = 0.9 + 0.2 cos(ω t)

Midline = (1.1 + 0.9)/2 = 1.0; amplitude = 0.1.

Starts at maximum ⇒ +cos.

Hence, 1.0 + 0.1 cos(ω t).

Explanation

Midline = (1.1 + 0.9)/2 = 1.0; amplitude = 0.1.

Starts at maximum ⇒ +cos.

Hence, 1.0 + 0.1 cos(ω t).

17) A probe’s altitude is h(t) = 1000 + 200 sin(0.6 t) km. What is the minimum altitude?

800 km

1000 km

1200 km

600 km

Minimum = midline − amplitude = 1000 − 200 = 800 km.

Hence, 800 km.

Explanation

Minimum = midline − amplitude = 1000 − 200 = 800 km.

Hence, 800 km.

18) A planet’s x-position is x(t) = 3 cos(2π t / 2) AU and y-position is y(t) = 2 sin(2π t / 2) AU. What is the orbit shape and period?

Circle, period 2

Ellipse, period 2

Circle, period 1

Ellipse, period 1

x/3 = cos(πt), y/2 = sin(πt) ⇒ ellipse with semi-axes 3, 2.

Period = 2π / (π) = 2.

Hence, ellipse, period 2.

Explanation

x/3 = cos(πt), y/2 = sin(πt) ⇒ ellipse with semi-axes 3, 2.

Period = 2π / (π) = 2.

Hence, ellipse, period 2.

19) A moon’s distance is d(t) = M + A cos(ω t), with max 28 and min 12 (thousand km). Which pair (M, A) works?

(20, 8)

(28, 12)

(12, 8)

(20, 16)

M = (max + min)/2 = (28 + 12)/2 = 20,

A = (max − min)/2 = (28 − 12)/2 = 8.

Hence, (20, 8).

Explanation

M = (max + min)/2 = (28 + 12)/2 = 20,

A = (max − min)/2 = (28 − 12)/2 = 8.

Hence, (20, 8).

20) A satellite model h(t) = 600 + 180 sin(π t / 30) km needs its maximum increased to 840 km while keeping its minimum unchanged. Which new model works?

H(t) = 630 + 210 sin(π t / 30)

H(t) = 600 + 240 sin(π t / 30)

H(t) = 660 + 180 sin(π t / 30)

H(t) = 540 + 240 sin(π t / 30)

Old max = 600 + 180 = 780; want 840 ⇒ increase midline & amplitude equally by 30.

Midline = 630; amplitude = 210.

Hence, h(t) = 630 + 210 sin(π t / 30).

Explanation

Old max = 600 + 180 = 780; want 840 ⇒ increase midline & amplitude equally by 30.

Midline = 630; amplitude = 210.

Hence, h(t) = 630 + 210 sin(π t / 30).

Applying Trigonometric Models to Real Orbits