Oxford Revise AQA A Level Physics | Chapter P14 answers

P14: Gravitational fields

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Extra information

Mark

AO

Spec reference

01.1

g is the (gravitational) force per unit mass

Allow \(\frac{F}{m}\) if F and m are explained

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3.7.2.2

AO1

01.2

\(
\begin{aligned}
& \rho = \frac{M}{V}, V = \frac{4}{3} \pi r^3\\
& \text{g } = \frac{GM}{r^2} = \frac{G\rho \frac{4}{3}\pi r^3}{r^2} = G\rho \frac{4}{3}\pi r\\
& \text{If density constant}, g \alpha r\\
& \text{If } g \text{ less, then } r \text{ must be less}\\
\end{aligned}
\)

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1

1

3.7.2.2

AO2

01.3

Area under the existing curve shaded in

from 2.4 (× 106) to the right/infinity

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1

3.7.2.3

AO1

01.4

Either by estimating area under curve:

\(
\begin{aligned}
& 220 \text{ squares }\pm 5\\
& \text{Each square } = 0.1 \times 0.4 \times 10^{6}\text{ J kg}^{-1}\\
& V_g = 220 \times 0.1 \times 0.4 \times 10^{6}\text{ J kg}^{-1}\\
& = 8.8 \times 10^6\text{ J kg}^{-1}\\
\end{aligned}
\)

OR

\(
\begin{aligned}
& \text{Use of surface data to gain } GM\\
& g = \frac{GM}{r^2} \text{ and }gr^2 = GM\\
& V_g = \frac{GM}{r} = \frac{gr^2}{r} = gr = 3.7 \times 2.4 \times 10^{6} = 8.9 \times 10^{6} (\text{J kg}^{-1})\\
\end{aligned}
\)

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1

3.7.2.3

AO2

01.5

\(
\begin{aligned}
& \frac{GMm}{r^2} = \frac{1}{2}m{v^2}\\
& \frac{2GM}{r} = v^2\\
& v^2 = 2 \times 9 \times 10^{6}\\
& v = 4200\text{ m s}^{-1}\\
\end{aligned}
\)

All values of Vg yield 4200 m s–1 to 2 s.f.

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1

3.7.2.4

AO2

01.6

Straight line drawn from (0, 0) to (2.4, 3.7)

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3.7.2.2

AO1

02.1

\(
\begin{aligned}
& g = \frac{GM}{r^2},\ {V_g} = \frac{GM}{r}\\
& V_g = \left( \frac{GM}{R^2} \right),\ V_g = gR\\
\end{aligned}
\)

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3.7.2.2

3.7.2.3

AO1

02.2

\(
\begin{aligned}
& \frac{GMm}{r} = \frac{1}{2}\ mv^2\\
& \frac{GM}{r} = \frac{1}{2}\ v^2\\
& gR = \frac{1}{2}\ v^2\\
& v = \sqrt{2gR}\\
\end{aligned}
\)

Algebra must be clear

\(
\begin{aligned}
& \text{Alternative }mV_{\mathrm{g}} = \frac{1}{2}\ mv^2\\
& \therefore v = \sqrt{2V_{\mathrm{g}}} = \sqrt{2gR}\\
\end{aligned}
\)

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1

3.7.2.4

AO2

02.3

\(
\begin{aligned}
& v = \sqrt{2gR}\\
& v = \sqrt{2 \times 9.81 \times 6.37 \times {10}^6} = 11\,000\mathrm{\ m\ s}^{-1} (11\,200)\\
\end{aligned}
\)

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3.7.2.4

AO1

02.4

\(
\begin{aligned}
& \text{Mass of hydrogen } = \frac{0.002}{6.02 \times 10^{23}} = 3.32 \times 10^{-27}\mathrm{\ kg}\\
& \frac{1}{2} m(c_\mathrm{rms})^{2} = \frac{3}{2} kT\\
& \left( \frac{m}{3k} \right)(c_\mathrm{rms})^{2} = T\\
& T = \frac{3.32 \times 10^{-27}\mathrm{\ kg}}{3 \times 1.38 \times 10^{-23}} \times 11\ 000^2\\
& T = 9700\text{ K (using all unrounded numbers gives 10 000 K)}\\
\end{aligned}
\)

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1

1

3.6.2.3

AO3

02.5

Value used in 02.4 uses the mean speed of the molecules

At 650 K there will be a range of molecular speeds and some will have enough speed to escape the atmosphere

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1

3.6.2.3

AO3

03.1

Gravitational potential Vg at a point is defined as the work done/energy required to bring 1 kg/unit mass from infinity to that point in space

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3.7.2.3

AO1

03.2

If \(V \propto \frac{1}{r}\) then Vr should equal a constant

Take pairs of data (at least 2) and see if this is correct

Allow plot of a graph of V vs \(\frac{1}{r}\)

Should be a straight line through the origin

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1

3.7.2.3

MS0.3

AO2

03.3

Tangent drawn at 14 × 106 m

Gradient calculated, e.g. \(\frac{58 \times 10^6}{27 \times 10^6}\)

g = 2.1 ± 0.2

Allow for 1 mark value calculated using

\(g = \frac{GM}{r^2}\), which gives value of 2.0

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1

3.7.2.3

AO2

03.4

Graph rising as it moves towards the Moon and then decreasing closer to the Moon

Starts at –63 and Earth’s surface, ends at a value smaller at Moon’s surface

Does not go to zero

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1

1

3.7.2.3

AO3

04.1

The potential difference between the lines is constant but the distance is not

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3.7.2.3

AO2

04.2

Lines drawn towards the centre of the Earth perpendicular to surface (by eye) and potential lines

Arrow pointing to the centre

Should stop at the surface

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1

3.7.2.2

AO1

04.3

\(
\begin{aligned}
& {V_g} = \frac{GM}{r}\\
& r = \frac{GM}{V_g}\\
& r = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}\mathrm{\ kg}}{40 \times 10^6} = 1 \times 10^7\mathrm{\ m}\\
\end{aligned}
\)

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1

3.7.2.3

AO1

04.4

gravitation potential remains constant / ΔVg = 0

Since \({V_g} = \frac{GM}{r}\) and (the mass of the Earth is constant and) the height of orbit is constant

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1

3.7.2.3

AO1

05.1

Arrow down labelled W / mg / weight

Arrow along string labelled Tension (pointing away from bob)

Arrow to the left labelled Force/gravitational force of attraction

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1

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3.4.1.5

AO1

05.2

The force of attraction between two masses is proportional to the product of the masses and inversely proportional to the distance between them squared

Allow equation but terms must be defined

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3.7.2.1

AO1

05.3

\(\mathrm{T\ } \cos \theta = \frac{GmM_{\mathrm{E}}}{R^2}\ \text{ or } T\sin \theta = \frac{GMm}{d^2}\)

Divide one equation by the other (or substitute for T)

\(
\begin{aligned}
& \frac{T \sin \theta}{T\cos \theta} = \frac{\frac{GMm}{d^2}}{\frac{GmM_{\mathrm{E}}}{R^2}}\\
& \tan \theta = \frac{MR^2}{M_{\mathrm{E}}{d^2}}\\
\end{aligned}
\)

Allow force triangle from 05.2 and use of \(\tan = \frac{opp}{adj}\)

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3.4.1.1

AO2

05.4

\(
\begin{aligned}
& \%\text{ difference } = \frac{\text{measured}-\text{actual}}{\text{actual}}\\
& = \frac{4560-5510}{5510} \times 100\% = \left( – \right)17\%\\
\end{aligned}
\)

ignore minus sign

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3.1.2

AO2

06.1

\(
\begin{aligned}
& F = \frac{GMm}{r^2}\text{ and } F = \frac{mv^2}{r}\text{ or } g = \frac{GM}{r^2}\text{ and } a = \frac{v^2}{r}\\
& \frac{GMm}{r^2} = \frac{mv^2}{r}\\
& \frac{GM}{r} = {v^2}\\
& v = \frac{2\pi r}{T}\\
& \frac{GM}{r} = \frac{4\pi^2r^2}{T^2}\\
& T^{2} = \frac{4\pi^2r^3}{GM}\\
& \text{Since others constant }{T^2} \propto r^3\\
\end{aligned}
\)

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3.7.2.4

3.6.1.1

AO1

06.2

Appropriate test proposed \(\frac{T^2}{r^3}\) = constant

Data tested at least three times

Relationship holds for the moons

Moon

\(\frac{T^2}{r^3}I\)

× 10-8 days2 Mm3

\(\frac{r^2}{T^2}I\)

× 106 Mm3 days2

Io

4.164

24.02

Europa

4.174

23.96

Ganymede

4.179

23.93

Callisto

4.172

23.97

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1

1

3.7.2.4

AO2

06.3

\(\frac{T^2}{r^3} = \frac{4\pi^2}{GM}\) use of constant in appropriate units or pair of data from the table

\(\frac{T^2}{r^3} = 3.1 \times 10^{-16} s^2\mathrm{\ m}^{-3}\) \(M = \frac{4 \pi^2}{G \times 3.1 \times {10}^{-16}} = 1.9 \times 10^{27}\mathrm{\ kg}\)

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1

3.7.2.4

AO3

06.4

T2r3

2 log T \(\propto\) 3 log r

log t \(\propto \frac{3}{2}\) log r

Straight-line graph with gradient = \(\frac{3}{2}\)

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1

3.7.2.4

MS3.11

AO3

07.1

Arrow pointing towards centre of Earth (judged by eye)

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3.7.2.1

AO1

07.2

To remain in orbit, there must be a force perpendicular to direction of motion

This satellite could not maintain this orbit without an engine/other force/energy input

owtte

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3.6.1.1

AO1

07.3

Use of r = (36 × 106 + 6.37 × 106)

T = 24 × 60 × 60 = 86 400 s

use of \(v = \frac{2\pi r}{T}\) = 3081 m s−1 ≈ 3 km s−1

\(
\begin{aligned}
& \frac{GMm}{r^2} = \frac{mv^2}{r}\\
& \frac{GMm}{r} = {v^2}\\
& v = \sqrt{\frac{GM}{r}}\\
& v = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{36 \times 10^6 + 6.37 \times 10^6}}\\
& v = 3100\mathrm{\ m\ s}^{-1}\\
\end{aligned}
\)

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3.6.1.1

3.7.2.1

AO2

07.4

\(
\begin{aligned}
& \text{Use of }E = E_{\mathrm{k}} + E_{\mathrm{p}}\\
& E_{k} = \frac{1}{2} m{v^2} = \frac{GMm}{2r}\\
& E_{p} = -\frac{GMm}{r}\\
& E = \frac{GMm}{2r} – \frac{GMm}{r} = -\frac{GMm}{2r}\\
& E = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 282}{2 \times \left( 36 \times 10^6 + 6.37 \times 10^6 \right)}\\
& E = -1.3 \times 10^{9}\mathrm{\ J}\\
\end{aligned}
\)

Students may also have used

\(\frac{1}{2}m{v^2}\) to yield same answer

Do not award final mark if minus sign not included

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1

1

3.7.2.4

AO2

08.1

Arrow drawn pointing to centre of the space station

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3.6.1.1

AO1

08.2

\(
\begin{aligned}
& a = \omega^{2} r\\
& \frac{9.81}{25} = \omega^{2}\\
& \omega = 0.63\text{ rad s}^{-1}\\
& \omega = \frac{2\pi}{T}\\
& T = \frac{2\pi}{\omega} = 10\text{ s}\\
\end{aligned}
\)

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3.6.1.1

3.7.2.2

AO2

08.3

Suggested height: 1.8 m (allow between 1.5 m and 2.0 m)

r = 25 – 1.8 = 23.2 m

a = ω2r

a = 0.632 × 23.2 = 9.2 m s–2

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3.1.3

3.6.1.1

AO3

08.4

Larger radius means the height of astronaut is a smaller fraction of the radius – so difference over body marginal (or wtte)

Difficulty/expense of taking such large amounts of material into space

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1

3.1.2

AO3

Skills box answers

Question

Answer

1

Plot a graph of ln(T/ days) against ln(r / 103 km). Obtain a straight-line graph of gradient 1.5 and intercept −7.9.

2

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

Substituting in values for \(G, M\) and \(r\) gives \(T^2 = \frac{4\pi^2 \left( 3.5 \times 10^8 \right)^3}{6.67 \times 10^{-11} \times 1.02 \times 10^{26}}\)

T 2 = 2.49 × 1011 s2. Therefore \(T = \sqrt{\left( 2.49 \times 10^{11}\mathrm{\ s}^2 \right)}\) = 4.99 × 105 s or 5.77 days.

3

Rearranging the equation for M gives \(M = \frac{4 \pi^2 r^2}{G T^2}\)

Converting the values of r and T into standard form:

r = 2.38 × 108 m and T = (1.37 × 24 × 60 × 60) = 1.18 × 105 s.

Substituting these into the rearranged equation gives \(M = \frac{4\pi^2 \left( 2.38 \times 10^8 \right)^3}{6.67 \times 10^{-11} \left( 1.18 \times 10^5 \right)^2} = 5.7 \times 10^{26}\mathrm{\ kg}\)

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