Physics 9702/43 — October/November 2010

Cambridge A-Level · A Level Structured Questions · worked solutions for every part, with the mark scheme

questions

100

marks

105

minutes

Topics Electric Fields · Magnetic Fields · Medical Physics · Motion in a Circle · Gravitational Fields · Ideal Gases · +7 more

Q1Motion in a CircleGravitational FieldsFree sample

A planet of mass $m$ is in a circular orbit of radius $r$ about the Sun of mass $M$ , as illustrated in Fig. 1.1.

Fig. 1.1

The magnitude of the angular velocity and the period of revolution of the planet about the Sun are $\omega$ and $T$ respectively.

(a)

State

(i)

what is meant by angular velocity,

DifficultyEasy

Worked solution

Answer

Angular velocity $\omega$ is the rate of change of angular displacement $\theta$ with time:

\omega = \frac{\mathrm{d}\theta}{\mathrm{d}t}

(unit: $\text{rad s}^{-1}$ ).

Final answer

Angular velocity is the rate of change of angular displacement with time, ω = dθ/dt (unit rad s⁻¹).

Detailed explanation

Background Concept

When an object moves in a circle, its position can be described by an angle $\theta$ (in radians) measured about the centre. The angular velocity $\omega$ tells you how quickly this angle is changing.

By definition,

\omega = \frac{\mathrm{d}\theta}{\mathrm{d}t}

The SI unit is $\text{rad s}^{-1}$ . (Radians are technically dimensionless, but we write rad to show it is an angular quantity.)

Understanding the Question

The planet is moving in a circular orbit, so it continually sweeps out angular displacement about the Sun. The question asks you to state what “angular velocity” means.

Approach

Give the definition in words and/or as an equation: “rate of change of angular displacement with time”. Including the equation and unit secures full credit.

Step-by-Step Reasoning

In circular motion, the angular position is $\theta$ .
“How fast the angle changes” means change in angle per unit time.
Therefore $\omega = \mathrm{d}\theta/\mathrm{d}t$ , with unit $\text{rad s}^{-1}$ .

Key Takeaways

Angular velocity is a rate: angle per time.
Standard definition: $\omega = \mathrm{d}\theta/\mathrm{d}t$ .

Common Mistakes

Defining it as “speed” without mentioning it is angular (not linear) or without reference to angular displacement.
Giving units as $\text{s}^{-1}$ only; examiners usually expect $\text{rad s}^{-1}$ .

Things to Be Careful About

Use $\theta$ in radians when using circular-motion formulae such as $v = r\omega$ .
Do not confuse angular velocity $\omega$ with angular displacement $\theta$ .

Techniques used

use the definition of angular velocity as rate of change of angular displacementstate the unit of angular velocity in SI units

(ii)

the relation between $\omega$ and $T$ .

DifficultyEasy

Worked solution

Answer

In one period $T$ , the angular displacement is $2\pi$ rad, so

\omega = \frac{2\pi}{T}

Final answer

ω = 2π / T

Detailed explanation

Background Concept

For uniform circular motion, the object repeats its motion every period $T$ . In that time it completes one full revolution, which corresponds to an angular displacement of $2\pi$ radians.

Angular velocity is the rate of change of angular displacement:

\omega = \frac{\Delta \theta}{\Delta t}

Understanding the Question

You are told the planet’s angular velocity is $\omega$ and its period of revolution is $T$ . The question asks for the relationship between these two quantities.

Approach

Use $\Delta\theta = 2\pi$ for one full orbit and $\Delta t = T$ .

Step-by-Step Reasoning

For one revolution:

\Delta \theta = 2\pi

Time taken:

\Delta t = T

\omega = \frac{\Delta \theta}{\Delta t} = \frac{2\pi}{T}

Key Takeaways

One full revolution corresponds to $2\pi$ rad.
$\omega$ and $T$ are inversely related: faster orbiting means smaller period.

Common Mistakes

Writing $\omega = 2\pi T$ instead of dividing by $T$ .
Confusing period $T$ with frequency $f$ ; remember $f = 1/T$ and $\omega = 2\pi f$ .

Things to Be Careful About

The formula assumes uniform angular speed (true for a circular orbit at constant speed).
Keep $2\pi$ exact rather than using a decimal too early.

Techniques used

use one complete revolution as an angular displacement of 2 pi radiansrelate angular velocity to period via angular displacement per revolution

(b)

Show that, for a planet in a circular orbit of radius $r$ , the period $T$ of the orbit is given by the expression

T^2 = cr^3

where $c$ is a constant. Explain your working.

DifficultyMedium

Worked solution

Working

Gravitational force provides centripetal force:

\frac{GMm}{r^2} = mr\omega^2

\omega^2 = \frac{GM}{r^3}

Using $\omega = \frac{2\pi}{T}$ ,

\left(\frac{2\pi}{T}\right)^2 = \frac{GM}{r^3}

T^2 = \frac{4\pi^2}{GM} r^3

So $T^2 = cr^3$ where

c = \frac{4\pi^2}{GM}

Answer

$T^2 = cr^3$ with $c = \dfrac{4\pi^2}{GM}$ .

Final answer

T^2 = (4π^2/GM) r^3, so c = 4π^2/(GM).

Detailed explanation

Background Concept

A planet in a circular orbit must have a centripetal (inward) acceleration to continuously change the direction of its velocity.

For circular motion of radius $r$ and angular speed $\omega$ :

a = r\omega^2

The required centripetal force is therefore:

F_c = ma = mr\omega^2

For a planet orbiting the Sun, the only significant force on the planet is the gravitational attraction between the Sun (mass $M$ ) and the planet (mass $m$ ):

F_g = \frac{GMm}{r^2}

A stable circular orbit occurs when gravity supplies exactly the centripetal force.

Understanding the Question

You are asked to show that the period $T$ satisfies

T^2 = cr^3

for a circular orbit of radius $r$ , where $c$ is a constant (for a given central mass $M$ ). This is essentially Kepler’s third law for circular orbits.

Approach

Write the gravitational force on the planet.
Write the centripetal force needed for circular motion.
Set them equal (because gravity provides the centripetal force).
Use $\omega = 2\pi/T$ to replace $\omega$ with $T$ .
Rearrange into the required form $T^2 = cr^3$ and identify $c$ .

Step-by-Step Reasoning

orbit-fbd

Gravitational force on the planet due to the Sun is

F_g = \frac{GMm}{r^2}

For circular motion, the required centripetal force is

F_c = mr\omega^2

In a circular orbit, gravity is the centripetal force:

\frac{GMm}{r^2} = mr\omega^2

Cancel $m$ (important insight: the planet mass does not affect the period):

\frac{GM}{r^2} = r\omega^2

\omega^2 = \frac{GM}{r^3}

Relate $\omega$ to period:

\omega = \frac{2\pi}{T}

Square both sides and substitute:

\left(\frac{2\pi}{T}\right)^2 = \frac{GM}{r^3}

Rearrange to make $T^2$ the subject:

\frac{4\pi^2}{T^2} = \frac{GM}{r^3}

T^2 = \frac{4\pi^2}{GM} r^3

Hence it matches $T^2 = cr^3$ with

c = \frac{4\pi^2}{GM}

So for planets orbiting the same Sun (same $M$ ), $T^2/r^3$ is constant.

Key Takeaways

Circular orbit condition: gravitational force = centripetal force.
Derivation leads to Kepler’s third-law form $T^2 \propto r^3$ .
The orbit period depends on $M$ and $r$ , not on the planet’s own mass $m$ .

Common Mistakes

Using the wrong centripetal force form, e.g. $mv^2r$ instead of $mv^2/r$ .
Forgetting to square $\omega = 2\pi/T$ .
Dropping a power of $r$ when rearranging, leading to $T^2 \propto r^2$ or $r^4$ .

Things to Be Careful About

$r$ is the orbital radius (distance from the Sun’s centre), not the planet’s radius.
This result assumes a circular orbit and that gravity is the only significant force.
When identifying $c$ , treat $G$ and $M$ as constants for all planets orbiting the same Sun.

Techniques used

equate gravitational force to centripetal force for a circular orbitsubstitute omega = 2 pi / Trearrange algebraically to obtain a power-law relationshipidentify the constant of proportionality

(c)

Data for the planets Venus and Neptune are given in Fig. 1.2.

planet	$r / 10^8\ \text{km}$	$T / \text{years}$
Venus	1.08	0.615
Neptune	45.0

Fig. 1.2

Assume that the orbits of both planets are circular.

(i)

Use the expression in (b) to calculate the value of $T$ for Neptune.

$T$ = ______ $\text{years}$

DifficultyMedium-Easy

Worked solution

Working

For planets orbiting the Sun,

\frac{T^2}{r^3} = \text{constant}

T_N = T_V\left(\frac{r_N}{r_V}\right)^{3/2}

T_N = 0.615\left(\frac{45.0}{1.08}\right)^{3/2} = 1.65 \times 10^2

Answer

T \approx 165\ \text{years}

Final answer

165 years

Detailed explanation

Background Concept

From part (b), for circular orbits around the same central mass $M$ (the Sun),

T^2 = cr^3

where $c = 4\pi^2/(GM)$ is the same for all planets orbiting the Sun. Therefore,

\frac{T^2}{r^3} = c = \text{constant}

This means you can compare two planets using ratios without ever calculating $c$ .

Understanding the Question

You are given:

Venus: $r_V = 1.08 \times 10^8\ \text{km}$ , $T_V = 0.615\ \text{years}$
Neptune: $r_N = 45.0 \times 10^8\ \text{km}$ , $T_N$ unknown

You must find $T_N$ assuming circular orbits about the same Sun.

Approach

Use the ratio form:

\frac{T_N^2}{r_N^3} = \frac{T_V^2}{r_V^3}

Then solve for $T_N$ . The factors of $10^8\ \text{km}$ cancel automatically if you use the tabulated values consistently.

Step-by-Step Reasoning

Start with

\frac{T_N^2}{r_N^3} = \frac{T_V^2}{r_V^3}

Rearrange:

T_N^2 = T_V^2\left(\frac{r_N}{r_V}\right)^3

Take square root:

T_N = T_V\left(\frac{r_N}{r_V}\right)^{3/2}

Now substitute (using the numbers from the table; common units cancel):

\frac{r_N}{r_V} = \frac{45.0}{1.08} = 41.7

T_N = 0.615 \times (41.7)^{3/2}

Compute $(41.7)^{3/2} = 41.7\sqrt{41.7} \approx 41.7 \times 6.46 \approx 269$ .

Thus

T_N \approx 0.615 \times 269 \approx 165\ \text{years}

Key Takeaways

For orbits around the same central mass: $T^2/r^3$ is constant.
Ratio methods avoid calculating $G$ or $M$ and avoid unit conversion here.

Common Mistakes

Forgetting the exponent and using $T \propto r^3$ instead of $T \propto r^{3/2}$ .
Mixing units between planets (e.g. one $r$ in km and one in m) if not using the table consistently.
Rounding too early in the ratio, causing noticeable numerical drift.

Things to Be Careful About

The relationship applies because both orbit the Sun (same $M$ ). It would not be valid if one orbited a different star.
Keep at least 3 significant figures during intermediate steps; round at the end to a sensible number (typically 3 s.f. or to match the data).

Techniques used

use proportionality T^2 proportional to r^3 for bodies orbiting the same central massuse ratio methods to eliminate the constant of proportionalityapply power-law scaling to calculate an unknown period

(ii)

Determine the linear speed of Venus in its orbit.

speed = ______ $\text{km s}^{-1}$

DifficultyMedium-Easy

Worked solution

Working

v = \frac{2\pi r}{T}

r = 1.08\times 10^8\ \text{km}

T = 0.615\ \text{years} = 0.615\times 3.16\times 10^7\ \text{s} = 1.94\times 10^7\ \text{s}

v = \frac{2\pi(1.08\times 10^8)}{1.94\times 10^7} = 3.49\times 10^1\ \text{km s}^{-1}

Answer

v \approx 35\ \text{km s}^{-1}

Final answer

35 km s⁻¹

Detailed explanation

Background Concept

For uniform circular motion, an object travels one circumference $2\pi r$ in one period $T$ . Therefore the linear (tangential) speed is

v = \frac{2\pi r}{T}

You must ensure $r$ and $T$ are in compatible units (e.g. km and s to get $\text{km s}^{-1}$ ).

Understanding the Question

You need the orbital speed of Venus.
Given:

$r = 1.08 \times 10^8\ \text{km}$
$T = 0.615\ \text{years}$

Required: speed in $\text{km s}^{-1}$ , so convert years to seconds.

Approach

Use $v = 2\pi r / T$ .
Convert $T$ from years to seconds.
Substitute and calculate, keeping units consistent.

Step-by-Step Reasoning

Start with

v = \frac{2\pi r}{T}

Convert the period to seconds. Using $1\ \text{year} \approx 3.16\times 10^7\ \text{s}$ ,

T = 0.615 \times 3.16\times 10^7 = 1.94\times 10^7\ \text{s}

Substitute $r$ in km and $T$ in s:

v = \frac{2\pi (1.08\times 10^8\ \text{km})}{1.94\times 10^7\ \text{s}}

Numerically this gives

v \approx 34.9\ \text{km s}^{-1} \approx 35\ \text{km s}^{-1}

Key Takeaways

Linear speed in a circular orbit: $v = 2\pi r/T$ .
Correct unit conversion is essential to match requested units.

Common Mistakes

Using $v = r/T$ (missing the $2\pi$ ).
Forgetting to convert years to seconds, producing a speed in $\text{km year}^{-1}$ .
Converting $r$ into metres but still reporting $\text{km s}^{-1}$ .

Things to Be Careful About

Use a consistent value for seconds per year (typically $3.16\times 10^7\ \text{s}$ ) and keep enough significant figures during calculation.
The speed found is tangential speed; direction is continually changing, but the magnitude is constant for a circular orbit.

Techniques used

use v = 2 pi r / T for uniform circular motionconvert period from years to secondscarry out unit-consistent substitution to obtain speed

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