Physics 9702/42 — October/November 2018

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

questions

100

marks

120

minutes

Topics Magnetic Fields · Gravitational Fields · Nuclear Physics · Temperature · Ideal Gases · Oscillations · +6 more

Q1Gravitational FieldsFree sample

(a)

(i)

State what is meant by gravitational field strength.

DifficultyEasy

Worked solution

Answer

Gravitational field strength $g$ at a point is the gravitational force per unit mass on a small test mass placed at that point:

g = \frac{F}{m}

Final answer

Gravitational field strength is the gravitational force per unit mass at a point (g = F/m).

Detailed explanation

Background Concept

A gravitational field is a region where a mass experiences a gravitational force. The gravitational field strength $g$ at a point is defined by

g = \frac{F}{m}

where:

$F$ is the gravitational force on a small test mass,
$m$ is the test mass.

So $g$ tells you how many newtons of gravitational force act on each kilogram of mass at that point. Its units are $\text{N kg}^{-1}$ , which is equivalent to $\text{m s}^{-2}$ .

Understanding the Question

You are asked to state what gravitational field strength means. For a 1-mark “state” question, the examiner expects the precise definition (force per unit mass) and/or the equation.

Approach

Recall the standard definition used throughout A-level gravitation: field strength equals force per unit mass on a small test mass.

Step-by-Step Reasoning

Take a small test mass $m$ at the point in the field.
It experiences a gravitational force $F$ .
Define the field strength as the force per unit mass:

g = \frac{F}{m}

Key Takeaways

$g$ is defined by force per unit mass.
Units: $\text{N kg}^{-1}$ (same as $\text{m s}^{-2}$ ).

Common Mistakes

Defining $g$ as “force due to gravity” (missing “per unit mass”).
Saying “acceleration due to gravity” as the definition without mentioning force per unit mass (that is an explanation/implication, not the definition).

Things to Be Careful About

The test mass should be “small” so it does not significantly disturb the gravitational field you are measuring.

Techniques used

state the definition of gravitational field strengthexpress the definition as an equation

(ii)

Explain why, at the surface of a planet, gravitational field strength is numerically equal to the acceleration of free fall.

DifficultyMedium-Easy

Worked solution

Answer

At the surface, the gravitational force on mass $m$ is its weight $F = mg$ .
By Newton’s second law, $F = ma$ .
So $mg = ma \Rightarrow g = a$ (acceleration of free fall).

Final answer

Because for a freely falling mass, F = mg and also F = ma, so mg = ma and hence g = a.

Detailed explanation

Background Concept

Two key ideas are being connected:

Gravitational field strength definition:

g = \frac{F}{m}

Newton’s second law for a net force $F$ on mass $m$ producing acceleration $a$ :

F = ma

Near the surface of a planet, the gravitational force on an object is its weight:

W = mg

Understanding the Question

The question asks why, at the planet’s surface, the value of $g$ (field strength) is numerically the same as the acceleration a freely falling object has (often called “acceleration of free fall”).

Approach

Use the fact that for free fall (ignoring air resistance) the only force is weight. Then apply Newton’s second law and compare with the expression $W = mg$ .

Step-by-Step Reasoning

For an object of mass $m$ in free fall at the surface:

The gravitational force on it is

F = W = mg

The same force causes an acceleration $a$ , so by Newton’s second law:

F = ma

Equate the two expressions for the same force:

mg = ma

Cancel $m$ (non-zero):

g = a

So the gravitational field strength is numerically equal to the free-fall acceleration.

Key Takeaways

$g$ is “force per unit mass”, but in free fall that force produces an acceleration.
If gravity is the only force: $g$ and the free-fall acceleration are the same.

Common Mistakes

Writing $g = 9.81$ without any explanation.
Forgetting to mention Newton’s second law ( $F = ma$ ), which is the required link.
Not specifying free fall / ignoring air resistance.

Things to Be Careful About

If air resistance is significant, the resultant force is less than $mg$ , so $a < g$ (e.g. at terminal velocity). The equality holds when gravity is the only significant force.

Techniques used

apply Newton's second lawuse the definition of weight in a gravitational fieldequate expressions for force per unit mass

(b)

An isolated uniform spherical planet has radius $R$ .
The acceleration of free fall at the surface of the planet is $g$ .

On Fig. 1.1, sketch a graph to show the variation of the acceleration of free fall with distance $x$ from the centre of the planet for values of $x$ in the range $x = R$ to $x = 4R$ .

Fig. 1.1

DifficultyMedium-Easy

Worked solution

graph-b

Answer

For $x \ge R$ ,

g(x) = \frac{GM}{x^2} \propto \frac{1}{x^2}

So the curve passes through $(R,\,g)$ and then falls as an inverse-square curve, e.g.

g(2R)=\frac{g}{4},\quad g(3R)=\frac{g}{9},\quad g(4R)=\frac{g}{16}

Final answer

An inverse-square decrease from g at x = R: points (2R, g/4), (3R, g/9), (4R, g/16) joined by a smooth curve.

Detailed explanation

Background Concept

For a spherically symmetric planet, the gravitational field outside the planet behaves as if all its mass were concentrated at the centre. Hence for distance $x$ from the centre (with $x \ge R$ ), the gravitational field strength is

g(x) = \frac{GM}{x^2}

This is an inverse-square relationship: doubling $x$ makes $g$ four times smaller.

Understanding the Question

You are given that the acceleration of free fall (i.e. $g$ ) at the surface ( $x = R$ ) equals $g$ . You must sketch how $g$ changes from $x=R$ out to $x=4R$ .

Important clues:

The planet is uniform spherical and isolated.
The range begins at $x=R$ , so we are only considering points at and outside the surface.

Approach

Use the inverse-square law $g \propto 1/x^2$ for $x \ge R$ .
Compute a few anchor points at $2R, 3R, 4R$ in terms of $g$ .
Sketch a smooth curve decreasing and flattening (since inverse-square gets less steep as $x$ increases).

Step-by-Step Reasoning

At the surface:

g(R)=g

At $x = 2R$ :

g(2R)=\frac{GM}{(2R)^2}=\frac{GM}{4R^2}=\frac{1}{4}\frac{GM}{R^2}=\frac{g}{4}

At $x = 3R$ :

g(3R)=\frac{GM}{9R^2}=\frac{g}{9}\approx 0.11g

At $x = 4R$ :

g(4R)=\frac{GM}{16R^2}=\frac{g}{16}=0.0625g

Plot these points and draw a smooth inverse-square curve through them, decreasing and approaching zero as $x$ increases.

graph-b

Key Takeaways

Outside a spherical planet, $g(x)$ follows $1/x^2$ .
Use a few exact fractional points ( $g/4$ , $g/9$ , $g/16$ ) to get the sketch correct.

Common Mistakes

Drawing a straight line decrease (should be a curve).
Making $g$ drop to zero by $x=4R$ (it is small but not zero).
Using $g \propto 1/x$ instead of $1/x^2$ .

Things to Be Careful About

The question starts at $x = R$ , not $x = 0$ . (Inside the planet the relationship would be different, but that is not needed here.)
Ensure the curve passes exactly through $(R, g)$ and is consistent with the scale (e.g. $g/4$ is at $0.25g$ on the axis).

Techniques used

use the inverse-square dependence of gravitational field strength outside a sphereevaluate g at multiples of the radiussketch a smooth curve through key points

(c)

The planet in (b) has radius $R$ equal to $3.4 \times 10^3\ \text{km}$ and mean density $4.0 \times 10^3\ \text{kg m}^{-3}$ .

Calculate the acceleration of free fall at a height $R$ above its surface.

acceleration of free fall = ______ $\text{m s}^{-2}$

DifficultyMedium

Worked solution

Working

Radius:

R = 3.4 \times 10^3\ \text{km} = 3.4 \times 10^6\ \text{m}

Mass of planet:

M = \rho \left(\frac{4}{3}\pi R^3\right)

At height $R$ above surface, distance from centre $r = 2R$ , so

g = \frac{GM}{r^2} = \frac{G\rho \left(\frac{4}{3}\pi R^3\right)}{(2R)^2} = \frac{G\rho \pi R}{3}

Substitute:

g = \frac{(6.67 \times 10^{-11})\,\pi\,(4.0 \times 10^3)\,(3.4 \times 10^6)}{3} = 0.95\ \text{m s}^{-2}

Answer

$0.95\ \text{m s}^{-2}$

Final answer

0.95 m s^-2

Detailed explanation

Background Concept

For a spherical planet, the gravitational field strength at distance $r$ from its centre is

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

If the planet has mean density $\rho$ and radius $R$ , its mass is

M = \rho V = \rho \left(\frac{4}{3}\pi R^3\right)

Combining these allows $g$ to be found from $\rho$ and $R$ .

Understanding the Question

You are told:

$R = 3.4 \times 10^3\ \text{km}$ (so you must convert to metres),
$\rho = 4.0 \times 10^3\ \text{kg m}^{-3}$ ,
find the acceleration of free fall at a height $R$ above the surface.

A height $R$ above the surface means the distance from the centre is

r = R + R = 2R

Approach

Convert $R$ to SI units.
Find the planet’s mass using $M = \rho (4/3)\pi R^3$ .
Use $g = GM/r^2$ with $r = 2R$ .
Substitute values and give the answer in $\text{m s}^{-2}$ .

Step-by-Step Reasoning

Convert radius to metres:

R = 3.4 \times 10^3\ \text{km} = 3.4 \times 10^6\ \text{m}

Mass of the planet:

M = \rho\left(\frac{4}{3}\pi R^3\right)

Distance from centre at height $R$ :

r = 2R

Field strength there:

\begin{aligned} g &= \frac{GM}{r^2} = \frac{G\rho \left(\frac{4}{3}\pi R^3\right)}{(2R)^2} \\ &= \frac{G\rho \left(\frac{4}{3}\pi R^3\right)}{4R^2} = \frac{G\rho\pi R}{3} \end{aligned}

Substitute numbers:

\begin{aligned} g &= \frac{(6.67 \times 10^{-11})\,\pi\,(4.0 \times 10^3)\,(3.4 \times 10^6)}{3} \\ &\approx 0.95\ \text{m s}^{-2} \end{aligned}

(You can also note that outside the planet $g \propto 1/r^2$ , so at $r=2R$ the value is one quarter of the surface value.)

Key Takeaways

Use $M = \rho (4/3)\pi R^3$ to get mass from density.
Use $g = GM/r^2$ with $r$ measured from the centre.
At $r = 2R$ , the field strength is a factor of $1/4$ of its value at $r=R$ .

Common Mistakes

Using $r = R$ instead of $r = 2R$ (forgetting height is above the surface).
Not converting $\text{km}$ to $\text{m}$ , leading to errors by a factor of $10^3$ .
Forgetting the $\pi$ or $4/3$ in the sphere volume.

Things to Be Careful About

Keep everything in SI units: $R$ in $\text{m}$ , $\rho$ in $\text{kg m}^{-3}$ , $G$ in SI.
Use a sensible number of significant figures (here typically 2 s.f. from given data).

Techniques used

compute the planet’s mass from density and volumeapply g = GM/r^2 at a given distance from the centresubstitute numerical values with correct unit conversions

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