A satellite is in a circular orbit around a planet of mass M , as shown in the diagram. a . ii . Explain why the satellite has acceleration even though its speed is constant.

The satellite is moving in circular path with uniform speed, but at every point on the circular path the direction of the satellite changes corresponding to linear velocity. Thus, the velocity is not constant instead keeps changing because of the direction. Hence, satellite has acceleration even though its speed is constant.

A satellite is in a circular orbit around a planet of mass M , as shown in the diagram. b . Show that the angular speed ω is related to the orbit radius r by r 3 ω 2 = G M .

The Gravitational force on the satellite of mass m due to planet of mass M is: G M m r 2 The Gravitational force on satellite is equal to centripetal force on satellite that is m v 2 r . Thus, G M m r 2 = m v 2 r So, we get, G M = v 2 r Since, the relation between linear velocity v and angular velocity ω is v = ω r . Substituting, this in above equation. We get, G M = ω r 2 r Hence ω 2 r 3 = G M .

A satellite is in a circular orbit around a planet of mass M , as shown in the diagram. d . Titan and Enceladus are two of Saturn’s moons. Data about these moons are given in the table. Moon Orbit radius / m Angular speed / rad s − 1 Titan 1 . 22 × 10 9 Enceladus 2 . 38 × 10 8 5 . 31 × 10 − 5 ii . Determine the period of revolution of Titan in days.

Since, the centripetal force on Titan is equal to Gravitational force of Saturn and the centripetal force on Enceladus is equal to Gravitational force of Saturn. That is, m T v 2 r T = G M m T r T 2 and m E v 2 r E = G M m E r E 2 Then, v T 2 r T = G M and v E 2 r E = G M So we get, ω T 2 r T 3 = G M = ω E 2 r E 3 ⇒ ω T = ω E r E 3 r T 3 Substituting the corresponding value. ω T = 5 . 31 × 10 - 5 2 . 38 × 10 8 1 . 22 × 10 9 rad s - 1 ω T = 4 . 58 × 10 - 6 rad s - 1 The relation between time period T and angular speed ω is T = 2 π ω . So, the time period of revolution of Titan is: T T = 2 π ω T = 2 π 4 . 58 × 10 - 6 s = 1 . 37 × 10 6 s So, for revolution in terms of days: 1 . 37 × 10 6 24 × 3600 days = 15 . 856 days ≈ 15 . 9 days Hence, the period of revolution of Titan is 15 . 9 days .

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A satellite is in a circular orbit around a planet of mass $M$ , as shown in the diagram.

$a . i .$ On a copy of the diagram draw arrows to represent the velocity and acceleration of the satellite.

Important Questions on Circular Motion and Gravitation

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Physics for the IB Diploma 6th Edition>Chapter 6 - Circular Motion and Gravitation>Exam-style questions>Q 16

A satellite is in a circular orbit around a planet of mass $M$ , as shown in the diagram.

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$a . ii .$ Explain why the satellite has acceleration even though its speed is constant.

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Physics for the IB Diploma 6th Edition>Chapter 6 - Circular Motion and Gravitation>Exam-style questions>Q 16

A satellite is in a circular orbit around a planet of mass $M$ , as shown in the diagram.

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$b .$ Show that the angular speed $ω$ is related to the orbit radius $r$ by $r^{3} ω^{2} = G M$ .

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Physics for the IB Diploma 6th Edition>Chapter 6 - Circular Motion and Gravitation>Exam-style questions>Q 16

A satellite is in a circular orbit around a planet of mass $M$ , as shown in the diagram.

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$c .$ Because of friction with the upper atmosphere, the satellite slowly moves into another circular orbit with a smaller radius than the answer in $(b)$ that is $G M = ω^{2} r^{3}$ . Suggest the effect of this on the satellite’s:

$i .$ angular speed

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Physics for the IB Diploma 6th Edition>Chapter 6 - Circular Motion and Gravitation>Exam-style questions>Q 16

A satellite is in a circular orbit around a planet of mass $M$ , as shown in the diagram.

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$ii .$ linear speed.

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Physics for the IB Diploma 6th Edition>Chapter 6 - Circular Motion and Gravitation>Exam-style questions>Q 16

A satellite is in a circular orbit around a planet of mass $M$ , as shown in the diagram.

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$d .$ Titan and Enceladus are two of Saturn’s moons. Data about these moons are given in the table.

Moon	Orbit radius / $m$	Angular speed / $rad s^{- 1}$
Titan	$1.22 \times 10^{9}$
Enceladus	$2.38 \times 10^{8}$	$5.31 \times 10^{- 5}$

$i .$ Determine the mass of Saturn.

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Physics for the IB Diploma 6th Edition>Chapter 6 - Circular Motion and Gravitation>Exam-style questions>Q 16

A satellite is in a circular orbit around a planet of mass $M$ , as shown in the diagram.

Question Image

$d .$ Titan and Enceladus are two of Saturn’s moons. Data about these moons are given in the table.

Moon	Orbit radius / $m$	Angular speed / $rad s^{- 1}$
Titan	$1.22 \times 10^{9}$
Enceladus	$2.38 \times 10^{8}$	$5.31 \times 10^{- 5}$

$ii .$ Determine the period of revolution of Titan in days.

A satellite is in a circular orbit around a planet of mass M, as shown in the diagram. a.i. On a copy of the diagram draw arrows to represent the velocity and acceleration of the satellite.

Important Questions on Circular Motion and Gravitation

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A satellite is in a circular orbit around a planet of mass $M$ , as shown in the diagram.

$a . i .$ On a copy of the diagram draw arrows to represent the velocity and acceleration of the satellite.