Physics 9702/41 — October/November 2020

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

questions

100

marks

120

minutes

Topics Gravitational Fields · Medical Physics · Magnetic Fields · Thermodynamics · Oscillations · Electric Fields · +5 more

Q1Gravitational FieldsFree sample

Answer all the questions in the spaces provided.

(a)

(i)

State what is meant by a field of force.

DifficultyEasy

Worked solution

Answer

A field of force is a region of space in which a suitable test object (e.g. a mass) experiences a force.

The field may be represented by field lines showing the direction of the force, with the spacing indicating the strength.

Final answer

A region of space where a suitable test object experiences a force (represented by field lines giving direction and relative strength).

Detailed explanation

Background Concept

A field describes how one object can exert an influence on another without contact. Instead of thinking of the force as acting only when two objects touch, we say that one object creates a field in the space around it. If a suitable test object is placed in that region, it experiences a force.

A field of force is therefore linked to both:

the source (e.g. a mass producing a gravitational field, or a charge producing an electric field), and
the effect (a force on a test object placed at a point in space).

Field lines are a common way to represent fields. The direction of the field line at a point shows the direction of the force on a suitable test object, and the density (spacing) of lines represents relative strength.

Understanding the Question

You are asked to state what is meant by a field of force. This is a definition-style question, so the mark-worthy points are:

“region of space”
“test object experiences a force”
Optionally, to gain the second mark, you normally add how it is represented (field lines: direction/strength).

Approach

Give a concise definition that includes the idea of a region in space and the force on a test object. Then add one extra sentence about field lines to secure the second mark.

Step-by-Step Reasoning

A field must exist in space, not only at the source, so include “region of space”.
It must cause an observable effect: a force on something placed there. So state that a suitable test object experiences a force.
For a second marking point, explain the representation: field lines show the direction of the force, and closer lines indicate a stronger field.

Key Takeaways

A field is a description of force pervading space.
A field is detected by placing a test object and observing a force.
Field lines indicate direction and relative strength.

Common Mistakes

Saying only “a force acts at a distance” without mentioning a region of space.
Forgetting the role of a test object (how the field is detected).
Confusing “field” with “force” (the field is the property of space; the force is what happens to an object in the field).

Things to Be Careful About

Use general wording (“suitable test object”), not only “mass”, because the term “field of force” can apply to other fields too.
If you mention field lines, ensure you link them correctly: tangent gives direction, spacing gives strength.

Techniques used

define a field as a region in which a test object experiences a forcestate how a field can be represented by field lines indicating direction and strength

(ii)

Define 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:

g = \frac{F}{m}

(unit $\text{N kg}^{-1}$ ).

Final answer

Force per unit mass on a small test mass at the point (g = F/m).

Detailed explanation

Background Concept

In a gravitational field, a mass experiences a gravitational force. To describe the field itself (independent of what test mass you choose), we define the gravitational field strength $g$ .

It is defined as:

g = \frac{F}{m}

where:

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

Because $F$ is proportional to $m$ , the ratio $F/m$ is the same for any small test mass at that point.

The unit is $\text{N kg}^{-1}$ , which is equivalent to $\text{m s}^{-2}$ (since $1\ \text{N} = 1\ \text{kg m s}^{-2}$ ).

Understanding the Question

The question asks for a definition, so you need the phrase “force per unit mass” and ideally the equation $g = F/m$ .

Approach

State the definition in words, and support it with the defining equation and unit.

Step-by-Step Reasoning

Identify that the field quantity should not depend on the test mass. So use the ratio force / mass.
Write $g = F/m$ .
Quote the unit $\text{N kg}^{-1}$ .

Key Takeaways

$g$ is a property of the field at a point, not of the object.
$g = F/m$ and units are $\text{N kg}^{-1}$ .

Common Mistakes

Defining $g$ as “force” (missing “per unit mass”).
Using $g = GM/r$ (wrong power of $r$ ; that is not field strength).
Forgetting “small test mass” (to avoid the test mass significantly affecting the field).

Things to Be Careful About

Do not confuse gravitational field strength $g$ with gravitational potential $\phi$ (which has units $\text{J kg}^{-1}$ ).
If you give units, use $\text{N kg}^{-1}$ (or $\text{m s}^{-2}$ ), not just “N”.

Techniques used

use the definition of gravitational field strength as force per unit massstate the defining equation g = F/m with correct units

(b)

An isolated planet may be assumed to be a uniform sphere of radius $3.39 \times 10^6\ \text{m}$ with its mass of $6.42 \times 10^{23}\ \text{kg}$ concentrated at its centre.

Calculate the gravitational field strength at the surface of the planet.

field strength = ______ $\text{N kg}^{-1}$

DifficultyMedium-Easy

Worked solution

Working

For a spherical planet,

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

g = \frac{(6.67 \times 10^{-11})(6.42 \times 10^{23})}{(3.39 \times 10^{6})^2}

g \approx 3.7\ \text{N kg}^{-1}

Answer

$3.7\ \text{N kg}^{-1}$

Final answer

3.7 N kg^-1

Detailed explanation

Background Concept

For a spherically symmetric mass distribution (such as a uniform sphere), the gravitational field outside the sphere is the same as if all the mass were concentrated at its centre. This allows us to treat the planet like a point mass when calculating $g$ at or above its surface.

Newton's law of gravitation gives the force on a mass $m$ at distance $r$ from the planet's centre:

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

Gravitational field strength is force per unit mass:

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

Understanding the Question

You are given:

planet radius $r = 3.39 \times 10^6\ \text{m}$ ,
planet mass $M = 6.42 \times 10^{23}\ \text{kg}$ ,
$G = 6.67 \times 10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$ .

You need the gravitational field strength at the surface, so the distance from the centre is exactly the radius $r$ .

Approach

Use the surface-field formula:

Write $g = GM/r^2$ .
Substitute $G$ , $M$ , and $r$ .
Calculate and give the answer in $\text{N kg}^{-1}$ .

Step-by-Step Reasoning

Start with

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

Substitute values:

g = \frac{(6.67 \times 10^{-11})(6.42 \times 10^{23})}{(3.39 \times 10^{6})^2}

Handle powers of ten and the square carefully:

Numerator: $6.67 \times 6.42 \approx 42.8$ and $10^{-11} \times 10^{23} = 10^{12}$ , so numerator $\approx 4.28 \times 10^{13}$ .
Denominator: $(3.39 \times 10^6)^2 = 3.39^2 \times 10^{12} \approx 11.5 \times 10^{12} = 1.15 \times 10^{13}$ .

g \approx \frac{4.28 \times 10^{13}}{1.15 \times 10^{13}} \approx 3.7\ \text{N kg}^{-1}

Key Takeaways

For points on/above a spherical planet: $g = GM/r^2$ .
At the surface, $r$ is the planet's radius.
Squaring a number in standard form squares both the coefficient and the power of ten.

Common Mistakes

Using $r = 3.39 \times 10^6\ \text{m}$ but forgetting to square it.
Squaring only the coefficient (e.g. $3.39^2$ ) and forgetting $10^{6}$ becomes $10^{12}$ .
Giving the unit as $\text{N}$ or $\text{N m}^{-1}$ instead of $\text{N kg}^{-1}$ .

Things to Be Careful About

The formula uses distance from the centre, not height above surface.
Keep sufficient significant figures through the working, then round at the end (typically to 2 or 3 s.f.).

Techniques used

apply g = GM/r^2 at the surface of a spherical planetsubstitute given values with correct powers of tencalculate and quote the result with correct unit and significant figures

(c)

Calculate the height above the surface of the planet in (b) at which the gravitational field strength is $1.0\%$ less than its value at the surface of the planet.

height = ______ $\text{m}$

DifficultyMedium

Worked solution

Working

At surface:

g_0 = \frac{GM}{r^2}

At height $h$ :

g = \frac{GM}{(r+h)^2}

Given $g = 0.99g_0$ :

\frac{GM}{(r+h)^2} = 0.99\frac{GM}{r^2}

(r+h)^2 = \frac{r^2}{0.99}

r+h = \frac{r}{\sqrt{0.99}}

h = r\left(\frac{1}{\sqrt{0.99}} - 1\right)

With $r = 3.39 \times 10^6\ \text{m}$ :

h \approx 1.7 \times 10^{4}\ \text{m}

Answer

$1.7 \times 10^{4}\ \text{m}$

Final answer

1.7 × 10^4 m

Detailed explanation

Background Concept

The gravitational field strength due to a spherical mass is an inverse-square function of distance from the centre:

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

where $R$ is the distance from the planet's centre.

If you move away from the planet, $R$ increases and $g$ decreases. A small percentage change in $g$ corresponds to a related change in $R$ because of the square.

Understanding the Question

From part (b), the planet radius is

r = 3.39 \times 10^6\ \text{m}

The question asks: at what height above the surface is the field strength 1.0% less than at the surface?

So if $g_0$ is the surface field, we want $g = 0.99g_0$ at distance $r+h$ from the centre.

Approach

Write expressions for $g_0$ (at $r$ ) and $g$ (at $r+h$ ).
Use the condition $g = 0.99g_0$ to form an equation.
Cancel common factors ( $GM$ ), solve for $h$ .

Step-by-Step Reasoning

Surface:

g_0 = \frac{GM}{r^2}

At height $h$ above the surface, the distance from the centre is $r+h$ , so:

g = \frac{GM}{(r+h)^2}

Given $g$ is 1.0% less:

g = 0.99g_0

Substitute the expressions:

\frac{GM}{(r+h)^2} = 0.99\frac{GM}{r^2}

Cancel $GM$ (same planet, so same $G$ and $M$ ):

\frac{1}{(r+h)^2} = \frac{0.99}{r^2}

Rearrange:

(r+h)^2 = \frac{r^2}{0.99}

Take square roots (use the positive root since distances are positive):

r+h = \frac{r}{\sqrt{0.99}}

h = r\left(\frac{1}{\sqrt{0.99}} - 1\right)

Now calculate:

$\sqrt{0.99} \approx 0.995$
$1/\sqrt{0.99} \approx 1.005$

Hence

h \approx 3.39 \times 10^6 \times 0.005 \approx 1.7 \times 10^4\ \text{m}

So the height is about $17\ \text{km}$ .

Key Takeaways

Use $g \propto 1/R^2$ to compare field strengths at two distances.
A percentage change in $g$ produces about half that fractional change in $R$ (because of the square), for small changes.
Height above surface is found from $R - r$ .

Common Mistakes

Using $r$ instead of $r+h$ in the inverse-square formula.
Treating 1.0% less as $0.01g_0$ instead of $0.99g_0$ .
Forgetting to take the square root when solving for $r+h$ .
Giving $r+h$ as the final answer instead of the height $h$ .

Things to Be Careful About

Keep $r$ in metres; do not switch to km unless you convert back.
When taking square roots, ensure you use $\sqrt{0.99}$ not $0.99$ .
Final answer should be the height above the surface, not distance from the centre.

Techniques used

set up a ratio using the inverse-square dependence g ∝ 1/r^2solve an equation involving square roots to find the new radiusconvert from distance from centre to height above the surface

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