Astronomy and Orbits � Analyze Orbital Motion Models (HSF-TF.B.5)

1) A binary star’s separation projected on the x-axis is s(t) = 14 cos(π t/5) solar radii. What is the time between consecutive maxima?

5 units

10 units

π units

14 units

ω = π/5 ⇒ T = 2π/ω = 2π / (π/5) = 10.

Hence, 10 units.

Explanation

ω = π/5 ⇒ T = 2π/ω = 2π / (π/5) = 10.

Hence, 10 units.

2) The Earth–Sun distance is approximated as R(t) = 1 + 0.0167 cos(2π t), AU, t in years with t = 0 at perihelion. What is R(0)?

0.9833 AU

1.0000 AU

1.0167 AU

2.0000 AU

R(0) = 1 + 0.0167·cos(0) = 1 + 0.0167 = 1.0167 AU.

Hence, 1.0167 AU.

Explanation

R(0) = 1 + 0.0167·cos(0) = 1 + 0.0167 = 1.0167 AU.

Hence, 1.0167 AU.

3) A small eccentricity orbit along x can be approximated: x(t) = a + e a cos(2π t/T). If a = 10, e = 0.1, T = 20, what are xmax and xmin?

10.1 and 9.9

11 and 9

12 and 8

10.2 and 9.8

ea = 10·0.1 = 1.

xmax = 10 + 1 = 11; xmin = 10 − 1 = 9.

Hence, 11 and 9.

Explanation

ea = 10·0.1 = 1.

xmax = 10 + 1 = 11; xmin = 10 − 1 = 9.

Hence, 11 and 9.

4) A planet’s x-position (in AU) is modeled by x(t) = 2 cos(π t), where t is in years and the y-position is y(t) = 2 sin(π t). What is the orbital period?

1/2 year

1 year

2 years

4 years

Angle θ = πt.

One full revolution needs Δθ = 2π ⇒ πT = 2π ⇒ T = 2.

Hence, 2 years.

Explanation

Angle θ = πt.

One full revolution needs Δθ = 2π ⇒ πT = 2π ⇒ T = 2.

Hence, 2 years.

5) A moon’s orbital radius is 420,000 km and its angular speed is ω = 2π/27.3 per day. Which equation gives the moon’s x-position (km) if at t = 0 it is at maximum x?

X(t) = 420000 sin(ω t)

X(t) = 420000 cos(ω t)

X(t) = 420000 cos(ω t + π/2)

X(t) = 420000 sin(ω t + π/2)

Maximum at t = 0 ⇒ use cosine with zero phase.

Hence, x(t) = 420000 cos(ω t).

Explanation

Maximum at t = 0 ⇒ use cosine with zero phase.

Hence, x(t) = 420000 cos(ω t).

6) A satellite’s radial distance r(t) from Earth is modeled as r(t) = 6800 + 50 sin(4π t), in km, t in hours. What is the period of its radial oscillation?

0.25 h

0.5 h

1 h

2 h

Angular speed ω = 4π rad/h.

T = 2π/ω = 2π/(4π) = 1/2 h.

Hence, 0.5 h.

Explanation

Angular speed ω = 4π rad/h.

T = 2π/ω = 2π/(4π) = 1/2 h.

Hence, 0.5 h.

7) A planet’s y-position is y(t) = 1.5 sin(2π t + π/3) AU. Which statement is true?

The amplitude is 2π.

The period is 1/2 year.

The phase shift is −π/3 to the right.

The amplitude is 1.5 AU.

Amplitude = coefficient of sine = 1.5.

Period = 1 (since 2πt), phase +π/3 is a left shift, not right.

Hence, amplitude is 1.5 AU.

Explanation

Amplitude = coefficient of sine = 1.5.

Period = 1 (since 2πt), phase +π/3 is a left shift, not right.

Hence, amplitude is 1.5 AU.

8) A spacecraft circles an asteroid: x(t) = 3 cos(0.1 t), y(t) = 3 sin(0.1 t), with t in minutes, distances in km. What is the orbital speed (km/min)?

$0

$1

$3

R = 3 km, ω = 0.1 rad/min.

Orbital speed v = Rω = 3·0.1 = 0.3 km/min.

Hence, 0.3 km/min.

Explanation

R = 3 km, ω = 0.1 rad/min.

Orbital speed v = Rω = 3·0.1 = 0.3 km/min.

Hence, 0.3 km/min.

9) A planet’s x-position is x(t) = 5 cos(ω t), and at t = 6 h the planet has completed exactly 3/8 of a revolution. What is ω (rad/h)?

π/8

π/4

π/2

3π/4

3/8 of a full turn ⇒ angle = (3/8)·2π = 3π/4.

ω·6 = 3π/4 ⇒ ω = (3π/4)/6 = π/8.

Hence, π/8 rad/h.

Explanation

3/8 of a full turn ⇒ angle = (3/8)·2π = 3π/4.

ω·6 = 3π/4 ⇒ ω = (3π/4)/6 = π/8.

Hence, π/8 rad/h.

10) A moon’s position obeys x(t) = R cos(ω t), y(t) = R sin(ω t). At t = 0, the moon is at (0, R). What equation matches this initial condition?

X(t) = R cos(ω t), y(t) = R sin(ω t)

X(t) = R sin(ω t), y(t) = R cos(ω t)

X(t) = R cos(ω t − π/2), y(t) = R sin(ω t − π/2)

X(t) = R cos(ω t + π/2), y(t) = R sin(ω t + π/2)

At t = 0: x(0) = 0, y(0) = R ⇒ x = R sin(ωt), y = R cos(ωt).

Hence, option B.

Explanation

At t = 0: x(0) = 0, y(0) = R ⇒ x = R sin(ωt), y = R cos(ωt).

Hence, option B.

11) An exoplanet’s brightness varies sinusoidally with period 12 days due to phase. A model is B(t) = 0.3 + 0.05 cos(ω t). What is ω?

π/6 rad/day

π/12 rad/day

2π/12 rad/day

2π rad/day

T = 12 ⇒ ω = 2π/T = 2π/12.

Hence, 2π/12 rad/day.

Explanation

T = 12 ⇒ ω = 2π/T = 2π/12.

Hence, 2π/12 rad/day.

12) A satellite in a circular equatorial orbit has position x(t) = 7000 cos(kt), y(t) = 7000 sin(kt) km. If its period is 90 minutes, what is k in rad/min?

2π/45

2π/60

2π/75

2π/90

k = 2π/T = 2π/90 rad/min.

Hence, 2π/90.

Explanation

k = 2π/T = 2π/90 rad/min.

Hence, 2π/90.

13) A ring particle oscillates vertically as z(t) = 20 sin(6t) meters. What is the maximum vertical speed?

20 m/s

6 m/s

120 m/s

3.33 m/s

v(t) = z′(t) = 20·6 cos(6t).

v_max = 20·6 = 120 m/s.

Hence, 120 m/s.

Explanation

v(t) = z′(t) = 20·6 cos(6t).

v_max = 20·6 = 120 m/s.

Hence, 120 m/s.

14) A planet’s x-projection is x(t) = 2 + 0.4 cos(2π t − π/2) AU. Which is true?

Mean distance is 0.4 AU.

Amplitude is 2 AU.

Phase shift corresponds to a quarter-period delay.

Minimum x-value is 2.4 AU.

Phase −π/2 ⇒ right shift of T/4 (quarter-period delay).

Mean = 2, amplitude = 0.4, min = 1.6.

Hence, quarter-period delay.

Explanation

Phase −π/2 ⇒ right shift of T/4 (quarter-period delay).

Mean = 2, amplitude = 0.4, min = 1.6.

Hence, quarter-period delay.

15) A moon’s angular position is θ(t) = θ0 + ω t. If θ increases by 5π/6 radians in 10 hours, what is ω?

π/12 rad/h

π/6 rad/h

π/3 rad/h

5π/6 rad/h

ω = Δθ/Δt = (5π/6)/10 = π/12 rad/h.

Hence, π/12 rad/h.

Explanation

ω = Δθ/Δt = (5π/6)/10 = π/12 rad/h.

Hence, π/12 rad/h.

16) A probe’s motion in the orbital plane is x(t) = 4 cos(t) km, y(t) = 2 sin(t) km. What is the shape of the trajectory and its semi-axes?

Circle radius 4

Circle radius 2

Ellipse with semi-axes 4 and 2

Ellipse with semi-axes 2 and 1

x/4 = cos t, y/2 = sin t ⇒ (x/4)² + (y/2)² = 1.

Ellipse with semi-axes 4 (x) and 2 (y).

Hence, ellipse, 4 & 2.

Explanation

x/4 = cos t, y/2 = sin t ⇒ (x/4)² + (y/2)² = 1.

Ellipse with semi-axes 4 (x) and 2 (y).

Hence, ellipse, 4 & 2.

17) A sinusoidal model for Earth’s heliocentric x-position is x(t) = cos(2π t), t in years. At what t does x(t) first reach zero after t = 0?

T = 1/4 year

T = 1/2 year

T = 3/4 year

T = 1 year

cos(2πt) = 0 ⇒ 2πt = π/2 ⇒ t = 1/4.

Hence, 1/4 year.

Explanation

cos(2πt) = 0 ⇒ 2πt = π/2 ⇒ t = 1/4.

Hence, 1/4 year.

18) A star’s radial velocity is v(t) = 12 sin(2π t/5) m/s. What is its maximum acceleration a_max?

12 m/s²

(24π/5) m/s²

(12·2π/5) m/s²

(12·(2π/5)²) m/s²

a(t) = v′(t) = 12·(2π/5) cos(2πt/5).

Maximum magnitude a_max = 12·(2π/5).

Hence, (12·2π/5) m/s².

Explanation

a(t) = v′(t) = 12·(2π/5) cos(2πt/5).

Maximum magnitude a_max = 12·(2π/5).

Hence, (12·2π/5) m/s².

19) A satellite altitude oscillates as h(t) = 400 + 15 cos(π t) km, t in hours. What is the time from maximum altitude to minimum altitude?

1 h

2 h

3 h

4 h

T = 2π/π = 2 h.

Max → min takes T/2 = 1 h.

Hence, 1 h.

Explanation

T = 2π/π = 2 h.

Max → min takes T/2 = 1 h.

Hence, 1 h.

20) A comet’s x-position is modeled by x(t) = 8 cos(π t/6 + π/3) AU. What is the time between successive x = 8 AU events?

6 units

8 units

10 units

12 units

x = 8 when phase = 2πn.

Change in phase Δ(πt/6) = 2π ⇒ Δt = 12.

Hence, 12 units.

Explanation

x = 8 when phase = 2πn.

Change in phase Δ(πt/6) = 2π ⇒ Δt = 12.

Hence, 12 units.

Understanding Trigonometric Orbit Models