Physics 9702/41 — May/June 2021

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

questions

100

marks

120

minutes

Topics Medical Physics · Quantum Physics · Magnetic Fields · Gravitational Fields · Ideal Gases · Thermodynamics · +7 more

Q1Gravitational FieldsFree sample

The Earth may be assumed to be an isolated uniform sphere with its mass of $6.0 \times 10^{24}\ \text{kg}$ concentrated at its centre.

A satellite of mass $1200\ \text{kg}$ is in a circular orbit about the Earth in the Earth’s gravitational field. The period of the orbit is $94$ minutes.

(a)

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 at that point.

Final answer

Gravitational force per unit mass at a point.

Detailed explanation

Background Concept

A gravitational field is a region where a mass experiences a gravitational force. To describe the field without having to specify a particular test mass each time, we define the gravitational field strength $g$ .

By definition,

g = \frac{F}{m}

where $F$ is the gravitational force on a small mass $m$ placed at the point. The word “small” means the test mass does not significantly change the field.

Understanding the Question

The question asks for the definition of gravitational field strength (not a formula like $g = GM/r^2$ ). So the answer should be in words, and must involve “force per unit mass” at a point.

Approach

Write the standard definition in a single sentence, making clear:

it is defined at a point in the field,
it is force per unit mass acting on a test mass.

Step-by-Step Reasoning

Take a small test mass $m$ at some point in a gravitational field.
The gravitational force on it is $F$ .
The field strength at that point is the force per unit mass:

g = \frac{F}{m}

Put this into words as the definition.

Key Takeaways

$g$ is a field quantity: depends on position, not on the chosen test mass.
Definition: force per unit mass at a point.

Common Mistakes

Defining $g$ as “ $GM/r^2$ ” instead of the definition (that is a derived expression for a spherical mass).
Missing “per unit mass”.
Saying “force on a unit mass” but not stating it is at a point.

Things to Be Careful About

Use “gravitational force” (not electric force, weight without context, etc.).
If direction is required in other contexts, $g$ is directed towards the mass causing the field, but here the 1-mark definition is usually just force per unit mass.

Techniques used

state a physics definition in terms of force per unit massspecify the point in the field at which the quantity is defined

(b)

Calculate the radius of the orbit of the satellite.

radius = ______ $\text{m}$

DifficultyMedium

Worked solution

Working

For a circular orbit,

\frac{GMm}{r^2} = m\frac{v^2}{r}

with $v = \frac{2\pi r}{T}$ , so

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

r^3 = \frac{GMT^2}{4\pi^2}

$T = 94\ \text{min} = 5640\ \text{s}$ , $M = 6.0\times 10^{24}\ \text{kg}$ .

r^3 = \frac{(6.67\times 10^{-11})(6.0\times 10^{24})(5640)^2}{4\pi^2}

r = 6.9\times 10^6\ \text{m}

Answer

$6.9 \times 10^6\ \text{m}$

Final answer

6.9 × 10^6 m

Detailed explanation

Background Concept

A satellite in a circular orbit requires a centripetal force towards the centre. In an orbit around Earth, this centripetal force is provided by gravity.

Gravitational force on satellite mass $m$ at distance $r$ from Earth’s centre:

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

Centripetal force needed for speed $v$ in a circle of radius $r$ :

F_c = m\frac{v^2}{r}

Orbital speed related to period $T$ :

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

Understanding the Question

You are given:

Earth mass $M = 6.0\times 10^{24}\ \text{kg}$ (model Earth as mass at its centre)
Satellite period $T = 94\ \text{min}$

You must find the orbital radius $r$ (distance from Earth’s centre to the satellite), not the height above Earth’s surface.

Approach

Convert the period to seconds.
Set gravitational force equal to centripetal force.
Replace $v$ using $v = 2\pi r/T$ .
Rearrange to the standard orbit relation:

T^2 \propto r^3

then solve for $r$ .

Step-by-Step Reasoning

Convert period:

T = 94\ \text{min} = 94\times 60 = 5640\ \text{s}

For a circular orbit, gravity provides the centripetal force:

\frac{GMm}{r^2} = m\frac{v^2}{r}

Cancel $m$ (orbit size does not depend on satellite mass):

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

Substitute $v = 2\pi r/T$ :

\frac{GM}{r^2} = \frac{1}{r}\left(\frac{2\pi r}{T}\right)^2 = \frac{4\pi^2 r}{T^2}

Rearrange for $r$ :

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

r^3 = \frac{GMT^2}{4\pi^2}

Substitute values:

r^3 = \frac{(6.67\times 10^{-11})(6.0\times 10^{24})(5640)^2}{4\pi^2}

Evaluating gives $r \approx 6.9\times 10^6\ \text{m}$ .

Key Takeaways

Circular orbit condition: gravitational force = centripetal force.
Useful derived result:

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

Always convert minutes to seconds before substitution.

Common Mistakes

Using Earth radius instead of orbital radius (or giving altitude when asked for radius).
Forgetting to convert minutes to seconds.
Using $v = r\omega$ but then using the wrong expression for $\omega$ .
Algebra error: writing $r^2$ instead of $r^3$ when rearranging.

Things to Be Careful About

$r$ is measured from Earth’s centre.
Keep enough significant figures during working; round at the end.
Check reasonableness: $r$ should be slightly larger than Earth’s radius ( $\approx 6.4\times 10^6\ \text{m}$ ).

Techniques used

equate gravitational force to centripetal force for a circular orbituse the relationship between orbital period and angular speedconvert time units and solve for orbital radius using algebraevaluate a cube root and express the result to appropriate significant figures

(c)

Rockets on the satellite are fired so that the satellite enters a different circular orbit that has a period of $150$ minutes. The change in the mass of the satellite may be assumed to be negligible.

(i)

Show that the radius of the new orbit is $9.4 \times 10^6\ \text{m}$ .

DifficultyMedium-Easy

Worked solution

Working

Using

r^3 = \frac{GMT^2}{4\pi^2}

$T = 150\ \text{min} = 9000\ \text{s}$ .

r^3 = \frac{(6.67\times 10^{-11})(6.0\times 10^{24})(9000)^2}{4\pi^2}

r = 9.4\times 10^6\ \text{m}

Answer

$9.4\times 10^6\ \text{m}$

Final answer

9.4 × 10^6 m

Detailed explanation

Background Concept

For objects orbiting a central mass $M$ in a circular orbit, equating gravity to centripetal force leads to

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

This is essentially Kepler’s third law for circular orbits about the same central body.

Understanding the Question

After firing rockets, the satellite is in a new circular orbit with period $150\ \text{min}$ . You must show the radius is $9.4\times 10^6\ \text{m}$ . Earth mass is still $6.0\times 10^{24}\ \text{kg}$ .

Approach

Use the same orbit relation as in part (b):

Convert $T$ to seconds.
Rearrange to $r^3 = GMT^2/(4\pi^2)$ .
Substitute and evaluate; then take the cube root.

Step-by-Step Reasoning

Convert time:

T = 150\times 60 = 9000\ \text{s}

r^3 = \frac{GMT^2}{4\pi^2}

Substitute $G = 6.67\times 10^{-11}$ and $M = 6.0\times 10^{24}$ :

r^3 = \frac{(6.67\times 10^{-11})(6.0\times 10^{24})(9000)^2}{4\pi^2}

Evaluate $r^3$ numerically and take cube root to obtain

r \approx 9.4\times 10^6\ \text{m}

which matches the value to be shown.

Key Takeaways

Longer orbital period implies larger orbital radius.
Relationship $T^2 \propto r^3$ is the fastest route for these questions.

Common Mistakes

Not converting minutes to seconds.
Accidentally using $T \propto r^3$ instead of $T^2 \propto r^3$ .
Using the satellite mass (it cancels out).

Things to Be Careful About

Quote the radius from Earth’s centre.
Keep standard form and appropriate significant figures consistent with given data.

Techniques used

use the circular orbit period-radius relationshipconvert time units consistentlysubstitute into an expression and take a cube rootquote the result in standard form

(ii)

State, with a reason, whether the gravitational potential energy of the satellite increases or decreases.

DifficultyMedium-Easy

Worked solution

Answer

The gravitational potential energy increases because

U = -\frac{GMm}{r}

and the new orbit has larger $r$ , so $U$ becomes less negative (greater).

Final answer

Increases (becomes less negative) because r increases and U = −GMm/r.

Detailed explanation

Background Concept

For a mass $m$ in the gravitational field of a spherical mass $M$ , the gravitational potential energy (taking zero at infinity) is

U = -\frac{GMm}{r}

This is negative because work must be done to move the mass from distance $r$ out to infinity against the attractive force.

As $r$ increases, $1/r$ decreases, so $-GMm/r$ becomes less negative. That means $U$ increases (moves upwards towards 0).

Understanding the Question

The satellite is moved to a different circular orbit with a longer period (150 min), which corresponds to a larger orbital radius than before. The question asks whether the satellite’s gravitational potential energy increases or decreases, with a reason.

Approach

Use how $U$ depends on $r$ :

write $U = -GMm/r$ ,
compare $r$ before and after,
decide how the negative sign affects the conclusion.

Step-by-Step Reasoning

Initial orbit radius $r_1$ (from part b) is smaller than new orbit radius $r_2$ (from part c(i)). So $r_2 > r_1$ .
Use

U = -\frac{GMm}{r}

If $r$ increases, the magnitude $GMm/r$ decreases.
Because of the minus sign, $U$ becomes less negative, e.g. from $-2\times 10^{10}\ \text{J}$ to $-1\times 10^{10}\ \text{J}$ .
A less negative number is a larger number, so gravitational potential energy increases.

Key Takeaways

With zero at infinity, gravitational potential energy is negative.
Increasing orbital radius increases $U$ (towards 0).

Common Mistakes

Saying “decreases because gravity is weaker” (force decreases, but potential energy increases).
Confusing “increase in magnitude” with “increase in value” when quantities can be negative.

Things to Be Careful About

Always state what reference is implied (for CIE, $U=0$ at infinity is standard).
Use “increases (becomes less negative)” to make the sign clear and avoid ambiguity.

Techniques used

use the expression for gravitational potential energy in a central fieldcompare values based on how a quantity varies with distancereason using sign (negative potential energy) and monotonic change with radius

(iii)

Determine the magnitude of the change in the gravitational potential energy of the satellite.

change in potential energy = ______ $\text{J}$

DifficultyMedium

Worked solution

Working

U = -\frac{GMm}{r}

So the change is

\Delta U = U_2 - U_1 = GMm\left(\frac{1}{r_1}-\frac{1}{r_2}\right)

Using $G = 6.67\times 10^{-11}$ , $M = 6.0\times 10^{24}\ \text{kg}$ , $m = 1200\ \text{kg}$ ,
$r_1 = 6.9\times 10^6\ \text{m}$ and $r_2 = 9.4\times 10^6\ \text{m}$ :

\Delta U = (6.67\times 10^{-11})(6.0\times 10^{24})(1200)\left(\frac{1}{6.9\times 10^6}-\frac{1}{9.4\times 10^6}\right)

\Delta U = 1.9\times 10^{10}\ \text{J}

Answer

$1.9\times 10^{10}\ \text{J}$

Final answer

1.9 × 10^10 J

Detailed explanation

Background Concept

With gravitational potential energy defined to be zero at infinity, a mass $m$ at distance $r$ from the centre of a spherical mass $M$ has

U = -\frac{GMm}{r}

If the object moves from radius $r_1$ to $r_2$ , the change is

\Delta U = U_2 - U_1 = -\frac{GMm}{r_2} - \left(-\frac{GMm}{r_1}\right)=GMm\left(\frac{1}{r_1}-\frac{1}{r_2}\right)

If $r_2>r_1$ , then $1/r_1 - 1/r_2$ is positive, so $\Delta U$ is positive (energy increases).

Understanding the Question

You have two circular orbit radii:

initial radius $r_1$ from part (b)
new radius $r_2 = 9.4\times 10^6\ \text{m}$ from part (c)(i)

The satellite mass is $m = 1200\ \text{kg}$ . You must find the magnitude of the change in gravitational potential energy in joules.

Approach

Use $U = -GMm/r$ .
Compute $\Delta U = U_2 - U_1 = GMm(1/r_1 - 1/r_2)$ .
Substitute values carefully in standard form.

Step-by-Step Reasoning

Write the change formula:

\Delta U = GMm\left(\frac{1}{r_1}-\frac{1}{r_2}\right)

Calculate the constant factor:

GMm = (6.67\times 10^{-11})(6.0\times 10^{24})(1200) \approx 4.8\times 10^{17}

Calculate the bracket:

\frac{1}{r_1}-\frac{1}{r_2} = \frac{1}{6.9\times 10^6}-\frac{1}{9.4\times 10^6}

Numerically this is about $3.9\times 10^{-8}\ \text{m}^{-1}$ .
4. Multiply:

\Delta U \approx (4.8\times 10^{17})(3.9\times 10^{-8}) \approx 1.9\times 10^{10}\ \text{J}

The value is positive because the satellite moved to a higher orbit; the question asks for magnitude, so quote $1.9\times 10^{10}\ \text{J}$ .

Key Takeaways

Use $U=-GMm/r$ (zero at infinity).
For a move from $r_1$ to $r_2$ :

\Delta U = GMm\left(\frac{1}{r_1}-\frac{1}{r_2}\right)

Higher orbit means larger $U$ (less negative).

Common Mistakes

Forgetting the negative sign in $U=-GMm/r$ and getting the wrong sign for $\Delta U$ .
Using $\Delta U = GMm(1/r_2 - 1/r_1)$ (sign error).
Mixing radii (using altitude above Earth instead of distance from centre).
Rounding $r_1$ too aggressively before calculating $1/r$ .

Things to Be Careful About

Use consistent units: $r$ in $\text{m}$ , masses in $\text{kg}$ .
Carry sufficient significant figures in intermediate steps; round final answer appropriately.
“Magnitude of the change” means give a positive number even though potential energies themselves are negative.

Techniques used

use gravitational potential energy expression U = −GMm/rcompute a difference in potential energy between two radiihandle standard form and significant figures correctly

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