Physics 9702/43 — 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 Oscillations · Medical Physics · Quantum Physics · Magnetic Fields · Gravitational Fields · Motion in a Circle · +8 more

Q1Gravitational FieldsMotion in a CircleFree 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 force per unit mass on a small test mass at that point:

g = \frac{F}{m}

Final answer

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 $g$ is defined so that it tells you “how much force each kilogram would feel”.

By definition,

g = \frac{F}{m}

where:

$F$ is the gravitational force on a small test mass,
$m$ is the mass of that test object,
$g$ has units of $\text{N kg}^{-1}$ (equivalently $\text{m s}^{-2}$ ).

Understanding the Question

The question asks for the definition (not a calculation). You must state what gravitational field strength means in words, and it is best to include the defining equation $g = F/m$ .

Approach

Write the standard definition: “force per unit mass on a small test mass at a point” and (optionally but helpfully) give $g = F/m$ .

Step-by-Step Reasoning

Identify the physical meaning: field strength measures gravitational effect independent of the chosen test mass.
State it precisely: force per unit mass.
Express it mathematically: $g = F/m$ .

Key Takeaways

$g$ is defined as force per unit mass.
A good definition includes both words and the equation $g = F/m$ .

Common Mistakes

Defining $g$ as “force” without dividing by mass.
Confusing gravitational field strength $g$ with gravitational force $F$ .

Things to Be Careful About

Mention “at a point” and “on a small test mass” (so the test mass does not significantly alter the field).
Units: $\text{N kg}^{-1}$ (or $\text{m s}^{-2}$ ).

Techniques used

state a definition using force per unit massexpress the definition as an equation

(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} = \frac{mv^2}{r}

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

Also $v = \dfrac{2\pi r}{T}$ so

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

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

$T = 94 \times 60 = 5640\ \text{s}$ .

r = \left(\frac{(6.67\times10^{-11})(6.0\times10^{24})(5640)^2}{4\pi^2}\right)^{1/3} = 6.85\times10^6\ \text{m}

Answer

$6.85\times10^6\ \text{m}$

Final answer

6.85 × 10^6 m

Detailed explanation

Background Concept

A satellite in a circular orbit must have a centripetal acceleration towards the centre. The only significant force on the satellite is Earth’s gravitational force, so gravity provides the centripetal force.

Key equations:

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

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

Required centripetal force for circular motion at speed $v$ :

F_c = \frac{mv^2}{r}

Orbital speed in terms of period $T$ :

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

Combining gives the standard “Kepler-type” result for circular orbits:

T^2 \propto r^3

Understanding the Question

You are told:

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

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

Approach

Convert $T$ to seconds.
Set gravitational force equal to centripetal force.
Use $v = 2\pi r/T$ to remove $v$ .
Rearrange to get $r^3 = \dfrac{GMT^2}{4\pi^2}$ and take the cube root.

Step-by-Step Reasoning

Convert the period:

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

Equate forces (gravity provides centripetal force):

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

Cancel $m$ (this is why satellite mass is irrelevant for the radius-period relation):

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

Use $v = 2\pi r/T$ and square it:

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

Rearrange:

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

Substitute values ( $G = 6.67\times10^{-11}$ ):

r = \left(\frac{(6.67\times10^{-11})(6.0\times10^{24})(5640)^2}{4\pi^2}\right)^{1/3} = 6.85\times10^6\ \text{m}

Key Takeaways

For a circular orbit: gravitational force = centripetal force.
Using $v = 2\pi r/T$ leads to $r^3 = GMT^2/(4\pi^2)$ .
The satellite mass cancels.

Common Mistakes

Forgetting to convert minutes to seconds.
Using Earth radius or subtracting it (the question asks for orbital radius from the centre).
Algebra slip: using $r^2$ instead of $r^3$ in the final relation.

Things to Be Careful About

Keep $G$ in SI units so that $r$ comes out in metres.
Cube roots are sensitive: keep sufficient significant figures during intermediate steps, then round the final answer appropriately.

Techniques used

equate gravitational force to centripetal forceuse v = 2πr/T to eliminate speedrearrange to an r^3 proportionality and take a cube rootconvert period into SI units

(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

For a circular orbit,

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

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

r = \left(\frac{(6.67\times10^{-11})(6.0\times10^{24})(9000)^2}{4\pi^2}\right)^{1/3} = 9.35\times10^6\ \text{m} \approx 9.4\times10^6\ \text{m}

Answer

$9.4\times10^6\ \text{m}$

Final answer

9.4 × 10^6 m

Detailed explanation

Background Concept

For circular orbits around a central mass $M$ , gravity supplies the centripetal force. This leads to the relation:

T^2 = \frac{4\pi^2}{GM} r^3 \quad\text{or}\quad r^3 = \frac{GMT^2}{4\pi^2}

So if the period increases, the orbital radius must also increase.

Understanding the Question

The satellite changes to a new circular orbit with period $150\ \text{min}$ . You must use the same orbit relation to calculate the new radius and show it is $9.4\times10^6\ \text{m}$ .

Approach

Convert $150\ \text{min}$ to seconds.
Substitute $G$ , $M$ , and $T$ into $r^3 = GMT^2/(4\pi^2)$ .
Take the cube root and round to match the stated value.

Step-by-Step Reasoning

Convert the period:

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

Substitute into the formula:

r = \left(\frac{(6.67\times10^{-11})(6.0\times10^{24})(9000)^2}{4\pi^2}\right)^{1/3}

Evaluate:

r = 9.35\times10^6\ \text{m} \approx 9.4\times10^6\ \text{m}

Key Takeaways

For circular orbits: larger period means larger orbital radius.
The relationship is $r^3 \propto T^2$ .

Common Mistakes

Not converting minutes to seconds.
Quoting $9.35\times10^6$ as $9.35\times10^5$ (power-of-ten error).

Things to Be Careful About

Keep consistent SI units.
Don’t round too early before the cube root; it can shift the final value.

Techniques used

apply the circular-orbit period-radius relationconvert period into SI unitsevaluate a cube root to obtain orbital radius

(ii)

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

DifficultyEasy

Worked solution

Answer

The gravitational potential energy increases because

U = -\frac{GMm}{r}

and $r$ increases, so $U$ becomes less negative (closer to zero).

Final answer

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

Detailed explanation

Background Concept

Gravitational potential energy (GPE) of a mass $m$ in the gravitational field of a spherical mass $M$ (taking $U=0$ at infinity) is

U = -\frac{GMm}{r}

The minus sign is important: it shows the system is bound. As you move further away (larger $r$ ), the value of $U$ increases towards zero.

Understanding the Question

The satellite changes from a smaller circular orbit to a larger one (since the period increases from $94$ min to $150$ min). You must state whether its GPE increases or decreases, and give a reason.

Approach

Use the formula $U=-GMm/r$ and compare what happens when $r$ increases.

Step-by-Step Reasoning

New period is larger, so the new orbital radius is larger.
Since

U = -\frac{GMm}{r}

increasing $r$ makes $1/r$ smaller, so $-GMm/r$ becomes less negative.
3. “Less negative” means the numerical value of $U$ has increased (e.g. from $-5$ to $-3$ is an increase).

Key Takeaways

With $U=0$ at infinity, gravitational potential energy is negative for bound orbits.
Moving to a higher orbit increases $U$ (and energy must be supplied).

Common Mistakes

Saying it decreases because the gravitational force is smaller. (Force decreases, but GPE increases.)
Forgetting the negative sign in $U=-GMm/r$ .

Things to Be Careful About

Always interpret “increase/decrease” using the signed value of $U$ , not just the magnitude.
Don’t confuse potential energy with gravitational potential ( $\phi = U/m$ ); both have the same sign behaviour with $r$ .

Techniques used

use U = -GMm/r for gravitational potential energycompare signs to decide increase or decrease

(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

Gravitational potential energy:

U = -\frac{GMm}{r}

With $r_1 = 6.85\times10^6\ \text{m}$ and $r_2 = 9.4\times10^6\ \text{m}$ ,

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

\Delta U = (6.67\times10^{-11})(6.0\times10^{24})(1200)\left(\frac{1}{6.85\times10^6}-\frac{1}{9.4\times10^6}\right)

\Delta U = 1.9\times10^{10}\ \text{J}

Answer

Change in gravitational potential energy (magnitude) $= 1.9\times10^{10}\ \text{J}$

Final answer

1.9 × 10^10 J

Detailed explanation

Background Concept

For a mass $m$ in the gravitational field of a spherical mass $M$ , the gravitational potential energy relative to infinity is

U = -\frac{GMm}{r}

A change in GPE between radii $r_1$ and $r_2$ is

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

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

Understanding the Question

You are asked for the magnitude of the change in the satellite’s gravitational potential energy when moving from the first orbit (period $94$ min) to the second orbit (period $150$ min).

Given/previous results:

Earth mass $M = 6.0\times10^{24}\ \text{kg}$ .
Satellite mass $m = 1200\ \text{kg}$ .
Initial orbital radius from (b): $r_1 = 6.85\times10^6\ \text{m}$ .
New orbital radius from (c)(i): $r_2 = 9.4\times10^6\ \text{m}$ .

Approach

Write $U=-GMm/r$ for each orbit.
Form $\Delta U = U_2 - U_1$ .
Since the question asks for magnitude, give $|\Delta U|$ (which here equals $\Delta U$ because it is positive).

Step-by-Step Reasoning

Write expressions:

U_1 = -\frac{GMm}{r_1},\qquad U_2 = -\frac{GMm}{r_2}

Subtract:

\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)

Calculate the constant factor:

GMm = (6.67\times10^{-11})(6.0\times10^{24})(1200) = 4.80\times10^{17}

Calculate the bracket:

\frac{1}{r_1} = \frac{1}{6.85\times10^6} = 1.46\times10^{-7}\ \text{m}^{-1}

\frac{1}{r_2} = \frac{1}{9.4\times10^6} = 1.06\times10^{-7}\ \text{m}^{-1}

Difference:

\left(\frac{1}{r_1}-\frac{1}{r_2}\right) \approx 3.96\times10^{-8}\ \text{m}^{-1}

Multiply:

\Delta U = (4.80\times10^{17})(3.96\times10^{-8}) \approx 1.9\times10^{10}\ \text{J}

Since the question asks for the magnitude, the answer is $1.9\times10^{10}\ \text{J}$ .

Key Takeaways

Use $U=-GMm/r$ (negative) with $U=0$ at infinity.
For moving to a higher orbit ( $r$ increases), GPE increases.
Changes are easiest using reciprocals: $\Delta U = GMm(1/r_1 - 1/r_2)$ .

Common Mistakes

Dropping the negative sign and concluding the wrong direction of change.
Using $\Delta U = GMm(1/r_2 - 1/r_1)$ without taking magnitude, giving a negative number when magnitude is requested.
Using orbital height above Earth’s surface instead of radius from Earth’s centre.

Things to Be Careful About

Use consistent radii values (carry forward from part (b) and part (c)(i)).
Keep powers of ten under control when calculating $1/r$ .
Final answer should be to 2 or 3 significant figures, consistent with the data given.

Techniques used

use U = -GMm/r for each orbitcompute a change using differences of reciprocalsuse values of orbital radius obtained from earlier parts

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