Physics 9702/42 — May/June 2020

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

questions

100

marks

120

minutes

Topics Electric Fields · Magnetic Fields · Gravitational Fields · Ideal Gases · Thermodynamics · Oscillations · +7 more

Q1Gravitational FieldsFree sample

Answer all the questions in the spaces provided.

(a)

Define gravitational potential at a point.

DifficultyEasy

Worked solution

Answer

Gravitational potential at a point is the work done per unit mass by an external agent in bringing a small test mass from infinity to that point (with no change in kinetic energy).

Final answer

Work done per unit mass to bring a test mass from infinity to the point (no change in KE).

Detailed explanation

Background Concept

Gravitational potential $V$ (or $\phi$ ) is a scalar field quantity defined at a point. It is closely related to gravitational potential energy (GPE) $U$ :

U = mV

A key idea is that we define the potential relative to a reference point, usually infinity. For isolated masses, we take $V=0$ at infinity.

Understanding the Question

You are asked to define gravitational potential at a point. For full credit you must include:

it is work done per unit mass
the mass is brought from infinity
it is brought slowly / with no change in kinetic energy (so the work done is not “wasted” as kinetic energy)

Approach

Write the standard Cambridge definition using “work done per unit mass” and “from infinity”, and add the condition about no change in kinetic energy.

Step-by-Step Reasoning

Gravitational potential energy change is the work done by (or against) gravity when moving a mass in a gravitational field.
Dividing by the mass gives a definition that depends only on the field, not on the particular mass used.
Choosing infinity as the reference gives a universal zero for isolated masses.

So the definition is: work done per unit mass by an external agent in bringing a test mass from infinity to the point, with no change in kinetic energy.

Key Takeaways

Gravitational potential is energy per unit mass.
For isolated masses, the reference is infinity where $V=0$ .

Common Mistakes

Defining field strength ( $g$ , force per unit mass) instead of potential.
Missing “per unit mass”.
Not stating the reference point “from infinity”.
Not stating “no change in kinetic energy” / “moved slowly”.

Things to Be Careful About

Potential is a scalar (no direction).
In gravitational fields around a mass, potential values are typically negative (because $V=0$ at infinity).

Techniques used

state the definition of gravitational potentialexpress the definition as work done per unit mass from infinity

(b)

An isolated solid sphere of radius $r$ may be assumed to have its mass $M$ concentrated at its centre. The magnitude of the gravitational potential at the surface of the sphere is $\phi$ .

On Fig. 1.1, show the variation of the gravitational potential with distance $d$ from the centre of the sphere for values of $d$ from $d = r$ to $d = 4r$ .

Fig. 1.1

DifficultyMedium-Easy

Worked solution

Working

For $d \ge r$ ,

V(d) = -\frac{GM}{d}

Given magnitude at surface is $\phi$ , so $V(r)=-\phi$ .

\Rightarrow V(d) = -\phi\frac{r}{d}

Hence points: $(r,-\phi)$ , $(2r,-0.5\phi)$ , $(3r,-\tfrac{1}{3}\phi)$ , $(4r,-0.25\phi)$ , with a smooth curve approaching $0$ from below.

graph-b

Answer

Curve $V(d)=-\phi r/d$ from $d=r$ to $4r$ (negative, increasing towards $0$ ).

Final answer

Graph of V = -φ r/d: at r is -φ, at 2r is -0.5φ, at 3r is -φ/3, at 4r is -0.25φ, smooth curve tending to 0.

Detailed explanation

Background Concept

Outside a spherically symmetric mass, the gravitational field is the same as if all the mass were concentrated at the centre. The gravitational potential (taking $V=0$ at infinity) is

V(d) = -\frac{GM}{d}

This is an inverse relationship: as $d$ increases, $V$ becomes less negative and tends to $0$ .

Understanding the Question

You are told that at the surface ( $d=r$ ) the magnitude of the potential is $\phi$ . Since gravitational potential is negative (with zero at infinity), this means

V(r) = -\phi

You must sketch $V$ against $d$ from $r$ to $4r$ .

Approach

Start from $V(d)=-GM/d$ .
Use the given surface value to rewrite the expression in terms of $\phi$ .
Calculate the values at $d=r,2r,3r,4r$ .
Plot these points and draw a smooth $1/d$ curve approaching $0$ .

Step-by-Step Reasoning

From the formula,

V(d) = -\frac{GM}{d}

At $d=r$ :

V(r) = -\frac{GM}{r} = -\phi

So $GM = \phi r$ .
Substitute back:

V(d) = -\frac{\phi r}{d} = -\phi\frac{r}{d}

Now evaluate key points:

$d=r$ : $V=-\phi$
$d=2r$ : $V=-\phi\,\frac{1}{2}=-0.5\phi$
$d=3r$ : $V=-\phi\,\frac{1}{3}$
$d=4r$ : $V=-\phi\,\frac{1}{4}=-0.25\phi$

Plot these and join with a smooth curve that is always negative and flattens as $d$ increases (approaching $0$ asymptotically).

graph-b

Key Takeaways

For $d \ge r$ , gravitational potential varies as $-1/d$ .
The potential becomes less negative with distance and tends to $0$ at infinity.

Common Mistakes

Drawing a straight line instead of a curved $1/d$ shape.
Plotting positive values (forgetting potential is negative).
Making the curve cross $V=0$ between $r$ and $4r$ (it should not).
Using $-1/d^2$ (confusing potential with field strength $g$ ).

Things to Be Careful About

The question says magnitude is $\phi$ ; you must use $V(r)=-\phi$ .
Ensure correct fractional values at $2r,3r,4r$ and a smooth curve through them.

Techniques used

use the gravitational potential of a point massscale the potential using the given surface valueplot an inverse relationship and mark key points

(c)

The sphere in (b) is a planet with radius $r$ of $6.4 \times 10^6\ \text{m}$ and mass $M$ of $6.0 \times 10^{24}\ \text{kg}$ . The planet has no atmosphere.

A rock of mass $3.4 \times 10^3\ \text{kg}$ moves directly towards the planet. Its distance from the centre of the planet changes from $4r$ to $3r$ .

(i)

Calculate the change in gravitational potential energy of the rock.

change = ______ $\text{J}$

DifficultyMedium-Easy

Worked solution

Working

U = -\frac{GMm}{d}

\Delta U = U_{3r}-U_{4r}=-\frac{GMm}{3r}-\left(-\frac{GMm}{4r}\right)

\Delta U = -\frac{GMm}{r}\left(\frac{1}{3}-\frac{1}{4}\right)=-\frac{GMm}{12r}

With $G=6.67\times 10^{-11}$ , $M=6.0\times 10^{24}\ \text{kg}$ , $m=3.4\times 10^{3}\ \text{kg}$ , $r=6.4\times 10^{6}\ \text{m}$ :

\Delta U = -\frac{(6.67\times 10^{-11})(6.0\times 10^{24})(3.4\times 10^{3})}{12(6.4\times 10^{6})} = -1.77\times 10^{10}\ \text{J}

Answer

$-1.8\times 10^{10}\ \text{J}$

Final answer

-1.8 × 10^10 J

Detailed explanation

Background Concept

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

U = -\frac{GMm}{d}

where $d$ is the distance from the planet’s centre. The negative sign reflects that energy must be supplied to take the mass from near the planet out to infinity.

A change in GPE is always

\Delta U = U_{\text{final}} - U_{\text{initial}}

Understanding the Question

A rock moves directly towards a planet, changing its distance from the centre from $4r$ to $3r$ . You are asked for the change in gravitational potential energy of the rock.

$r = 6.4\times 10^{6}\ \text{m}$
$M = 6.0\times 10^{24}\ \text{kg}$
$m = 3.4\times 10^{3}\ \text{kg}$
initial distance $d_i=4r$
final distance $d_f=3r$

Because it moves closer, you should expect $U$ to become more negative, so $\Delta U$ should be negative.

Approach

Use $U=-GMm/d$ at $d=4r$ and $d=3r$ , then subtract:

Write $U_{3r}$ and $U_{4r}$ .
Compute $\Delta U = U_{3r}-U_{4r}$ .
Substitute the numbers and keep the correct sign.

Step-by-Step Reasoning

Start with

U(d) = -\frac{GMm}{d}

U_{3r} = -\frac{GMm}{3r},\quad U_{4r}=-\frac{GMm}{4r}

Change in GPE:

\Delta U = U_{3r}-U_{4r}=-\frac{GMm}{3r}+\frac{GMm}{4r}

Factor out $GMm/r$ :

\Delta U = -\frac{GMm}{r}\left(\frac{1}{3}-\frac{1}{4}\right)

Compute the bracket:

\frac{1}{3}-\frac{1}{4}=\frac{4-3}{12}=\frac{1}{12}

\Delta U = -\frac{GMm}{12r}

Substitute values:

\Delta U = -\frac{(6.67\times 10^{-11})(6.0\times 10^{24})(3.4\times 10^{3})}{12(6.4\times 10^{6})}

This gives

\Delta U \approx -1.77\times 10^{10}\ \text{J}

To 2 s.f. (limited by given data),

\Delta U \approx -1.8\times 10^{10}\ \text{J}

Key Takeaways

Use $U=-GMm/d$ for gravitational potential energy relative to infinity.
Always compute $\Delta U = U_{\text{final}}-U_{\text{initial}}$ .
Moving closer to the planet makes $U$ more negative, so $\Delta U<0$ .

Common Mistakes

Getting the sign wrong (writing a positive change when moving closer).
Using $d=r$ instead of $d=3r$ or $4r$ .
Confusing $r$ (planet radius) with $d$ (distance from centre).
Using $g=GM/r^2$ and then incorrectly multiplying by distance without integrating.

Things to Be Careful About

Distances must be measured from the centre of the planet.
Keep powers of ten under control; using standard form helps.
Quote an appropriate number of significant figures and include the unit $\text{J}$ .

Techniques used

use gravitational potential energy U = -GMm/rcompute a change as final minus initialuse standard form and correct significant figures

(ii)

Explain whether the rock’s speed increases, decreases or stays the same.

DifficultyMedium-Easy

Worked solution

Answer

Speed increases.

As the rock moves closer, its gravitational potential energy decreases (becomes more negative). With no atmosphere, energy is conserved, so the lost GPE is converted to kinetic energy, increasing the speed.

Final answer

Increases.

Detailed explanation

Background Concept

When only conservative forces act (such as gravity), the total mechanical energy is conserved:

E = U + K = \text{constant}

Gravity does work on an incoming object, transferring energy from gravitational potential energy $U$ to kinetic energy $K=\tfrac{1}{2}mv^2$ .

If there is an atmosphere, drag would remove mechanical energy as thermal energy, but the question states there is no atmosphere.

Understanding the Question

The rock moves towards the planet (from $4r$ to $3r$ ). You must decide whether its speed increases, decreases, or stays the same, and explain why.

Key clues:

Motion is directly towards the planet, so gravity acts along the motion.
“No atmosphere” implies negligible resistive forces, so mechanical energy is conserved.

Approach

Use energy conservation:

Moving closer makes gravitational potential energy more negative (decreases).
With no losses, that decrease must appear as an increase in kinetic energy.
Increasing kinetic energy implies increasing speed.

Step-by-Step Reasoning

At a smaller distance $d$ ,

U=-\frac{GMm}{d}

becomes more negative because $d$ is smaller.

Therefore $\Delta U<0$ for motion inwards.
With no atmosphere, there is no significant drag, so the decrease in $U$ becomes an increase in $K$ :

\Delta K = -\Delta U > 0

Since

K=\frac{1}{2}mv^2,

an increase in $K$ means $v$ increases.

Equivalently: the gravitational force is towards the planet, same direction as the motion, so the rock accelerates and its speed rises.

Key Takeaways

In a vacuum, gravity converts GPE into KE.
Moving towards a planet increases speed because the object accelerates towards the centre.

Common Mistakes

Saying speed is constant because “gravity is constant” (it is not constant with distance, and anyway a non-zero force causes acceleration).
Forgetting the significance of “no atmosphere” (drag would reduce speed gain).
Confusing “potential increases” with “potential energy increases” (here, both $V$ and $U$ become more negative when closer).

Things to Be Careful About

Use correct sign language: “GPE decreases / becomes more negative” when moving inward.
Your explanation should mention either conservation of energy or the force/acceleration argument explicitly.

Techniques used

apply conservation of energyrelate decrease in gravitational potential energy to increase in kinetic energylink direction of gravitational force to acceleration

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