Calculate, for the surface of $\text{Io}$, the gravitational field strength g_Io due to the mass of $\text{Io}$. State an appropriate unit for your answer.

Show that the $\frac{\mathrm{gravitational} \mathrm{potential} \mathrm{due} \mathrm{to} \mathrm{Jupiter} \mathrm{at} \mathrm{the} \mathrm{orbit} \mathrm{of} \mathrm{Io}}{ \mathrm{gravitational} \mathrm{potential} \mathrm{due} \mathrm{to} \mathrm{Io} \mathrm{at} \mathrm{the} \mathrm{surface} \mathrm{of} \mathrm{Io}}$ is about 80.

Outline, using (b)(i), why it is not correct to use the equation $\sqrt{\frac{2G\times \text{mass of Io}}{\text{radius of Io}}}$ to calculate the speed required for the spacecraft to reach infinity from the surface of $\text{Io}$.

An engineer needs to move a space probe of mass 3600 kg from Ganymede to Callisto. Calculate the energy required to move the probe from the orbital radius of Ganymede to the orbital radius of Callisto. Ignore the mass of the moons in your calculation. 

$«\frac{GM}{r^2}=\frac{6.67\times {10}^{-11}\times 8.9\times {10}^{22}}{{\left(1.8\times {10}^6\right)}^2}=»1.8$ ✓

N kg⁻¹  OR  m s⁻²  ✓

$\frac{1.9\times {10}^{27}}{4.9\times {10}^8}$  AND  $\frac{8.9\times {10}^{22}}{1.8\times {10}^6}$ seen ✓

$«\frac{1.9\times {10}^{27}\times 1.8\times {10}^6}{4.9\times {10}^8\times 8.9\times {10}^{22}}=»78$  ✓

For MP1, potentials can be seen individually or as a ratio.

«this is the escape speed for $\text{Io}$ alone but» gravitational potential / field of Jupiter must be taken into account  ✓

OWTTE

$-G{M}_{\text{Jupiter}}\left(\frac{1}{1.88\times {10}^9}-\frac{1}{1.06\times {10}^9}\right)=«5.21\times {10}^7 {\text{J\hspace{0.05em}kg}}^{-1}»$  ✓

« multiplies by 3600 kg to get » 1.9 × 10¹¹«J» ✓

Award [2] marks if factor of ½ used, taking into account orbital kinetic energies, leading to a final answer of 9.4 x 10¹⁰«J».

Allow ECF from MP1

Award [2] marks for a bald correct answer.

Bohr modified the Rutherford model by introducing the condition mvr = n$\frac{h}{2\pi }$. Outline the reason for this modification.

Show that the speed v of an electron in the hydrogen atom is related to the radius r of the orbit by the expression

$$v=\sqrt{\frac{k{e}^2}{m_{\text{e}}r}}$$

where k is the Coulomb constant.

Using the answer in (b) and (c)(i), deduce that the radius r of the electron’s orbit in the ground state of hydrogen is given by the following expression.

$$r=\frac{h^2}{4{\pi }^{2}k{m}_{\text{e}}{e}^2}$$

Calculate the electron’s orbital radius in (c)(ii).

Explain what may be deduced about the energy of the electron in the β^– decay.

Suggest why the β^– decay is followed by the emission of a gamma ray photon.

Calculate the wavelength of the gamma ray photon in (d)(ii).

the electrons accelerate and so radiate energy

they would therefore spiral into the nucleus/atoms would be unstable

electrons have discrete/only certain energy levels

the only orbits where electrons do not radiate are those that satisfy the Bohr condition «mvr = n$\frac{h}{2\pi }$»

[3 marks]

$\frac{m_{\text{e}}{v}^2}{r}=\frac{k{e}^2}{r^2}$

OR

KE = $\frac{1}{2}$PE hence $\frac{1}{2}$m_ev² = $\frac{1}{2}\frac{k{e}^2}{r}$

«solving for v to get answer»

Answer given – look for correct working

[1 mark]

combining v = $\sqrt{\frac{k{e}^2}{m_{\text{e}}r}}$ with m_evr = $\frac{h}{2\pi }$ using correct substitution

«eg ${m_e}^2\frac{k{e}^2}{m_{\text{e}}r}{r}^2=\frac{h^2}{4{\pi }^2}$»

correct algebraic manipulation to gain the answer

Answer given – look for correct working

Do not allow a bald statement of the answer for MP2. Some further working eg cancellation of m or r must be shown

[2 marks]

« r = $\frac{{(6.63\times {10}^{-34})}^2}{4{\pi }^2\times 8.99\times {10}^9\times 9.11\times {10}^{-31}\times {(1.6\times {10}^{-19})}^2}$»

r = 5.3 × 10^–11 «m»

[1 mark]

the energy released is 3.54 – 0.48 = 3.06 «MeV»

this is shared by the electron and the antineutrino

so the electron’s energy varies from 0 to 3.06 «MeV»

[3 marks]

the palladium nucleus emits the photon when it decays into the ground state «from the excited state»

[1 mark]

Photon energy

E = 0.48 × 10⁶ × 1.6 × 10^–19 = «7.68 × 10^–14 J»

λ = «$\frac{hc}{E}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{7.68\times {10}^{-14}}$ =» 2.6 × 10^–12 «m»

Award [2] for a bald correct answer

Allow ECF from incorrect energy

[2 marks]

Outline the origin of the force that acts on Phobos.

Outline why this force does no work on Phobos.

The orbital period T of a moon orbiting a planet of mass M is given by

$$\frac{R^3}{T^2}=kM$$

where R is the average distance between the centre of the planet and the centre of the moon.

Show that $k=\frac{G}{4{\pi }^2}$

The following data for the Mars–Phobos system and the Earth–Moon system are available:

Mass of Earth = 5.97 × 10²⁴ kg

The Earth–Moon distance is 41 times the Mars–Phobos distance.

The orbital period of the Moon is 86 times the orbital period of Phobos.

Calculate, in kg, the mass of Mars.

The graph shows the variation of the gravitational potential between the Earth and Moon with distance from the centre of the Earth. The distance from the Earth is expressed as a fraction of the total distance between the centre of the Earth and the centre of the Moon.

Determine, using the graph, the mass of the Moon.

gravitational attraction/force/field «of the planet/Mars» ✔

Do not accept “gravity”.

the force/field and the velocity/displacement are at 90° to each other OR

there is no change in GPE of the moon/Phobos ✔

ALTERNATE 1

«using fundamental equations»

use of Universal gravitational force/acceleration/orbital velocity equations ✔

equating to centripetal force or acceleration. ✔

rearranges to get $k=\frac{G}{4{\pi }^2}$  ✔

ALTERNATE 2

«starting with $\frac{R^3}{T^2}=kM$»

substitution of proper equation for T from orbital motion equations ✔

substitution of proper equation for M OR R from orbital motion equations ✔

rearranges to get $k=\frac{G}{4{\pi }^2}$  ✔

$m_{\text{Mars}}={\left(\frac{R_{\text{Mars}}}{R_{\text{Earth}}}\right)}^3{\left(\frac{T_{\text{Earth}}}{T_{Mars}}\right)}^2{m}_{\text{Earth}}$ or other consistent re-arrangement ✔

6.4 × 10²³ «kg» ✔

read off separation at maximum potential 0.9 ✔

equating of gravitational field strength of earth and moon at that location OR ✔

7.4 × 10²² «kg» ✔

Allow ECF from MP1

This was generally well answered, although some candidates simply used the vague term “gravity” rather than specifying that it is a gravitational force or a gravitational field. Candidates need to be reminded about using proper physics terms and not more general, “every day” terms on the exam.

Some candidates connected the idea that the gravitational force is perpendicular to the velocity (and hence the displacement) for the mark. It was also allowed to discuss that there is no change in gravitational potential energy, so therefore no work was being done. It was not acceptable to simply state that the net displacement over one full orbit is zero. Unfortunately, some candidates suggested that there is no net force on the moon so there is no work done, or that the moon is so much smaller so no work could be done on it.

This was another “show that” derivation. Many candidates attempted to work with universal gravitation equations, either from memory or the data booklet, to perform this derivation. The variety of correct solution paths was quite impressive, and many candidates who attempted this question were able to receive some marks. Candidates should be reminded on “show that” questions that it is never allowed to work backwards from the given answer. Some candidates also made up equations (such as T = 2𝝿r) to force the derivation to work out.

This question was challenging for candidates. The candidates who started down the correct path of using the given derived value from 5bi often simply forgot that the multiplication factors had to be squared and cubed as well as the variables.

This question was left blank by many candidates, and very few who attempted it were able to successfully recognize that the gravitational fields of the Earth and Moon balance at 0.9r and then use the proper equation to calculate the mass of the Moon.

Explain why a centripetal force is needed for the planet to be in a circular orbit.

Calculate the value of the centripetal force.

Show that the gravitational potential due to the planet and the star at the surface of the planet is about −5 × 10⁹J kg⁻¹.

Estimate the escape speed of the spacecraft from the planet–star system.

«circular motion» involves a changing velocity ✓

«Tangential velocity» is «always» perpendicular to centripetal force/acceleration ✓

there must be a force/acceleration towards centre/star ✓

without a centripetal force the planet will move in a straight line ✓

$F=\frac{(6.67\times {10}^{-11})(8\times {10}^{24})(3.2\times {10}^{30})}{(4.4\times {10}^{10}{)}^2}=8.8\times {10}^{23}$ «N» ✓

V_planet = «−»$\frac{(6.67\times {10}^{-11})(8\times {10}^{24})}{9.1\times {10}^6}=$«−» 5.9 × 10⁷ «J kg⁻¹» ✓

V_star = «−»$\frac{(6.67\times {10}^{-11})(3.2\times {10}^{30})}{4.4\times {10}^{10}}=$«−» 4.9 × 10⁹ «J kg⁻¹» ✓

V_planet + V_star = «−» 4.9 «09» × 10⁹ «J kg⁻¹» ✓

Must see substitutions and not just equations.

use of v_esc = $\sqrt{2V}$ ✓

v = 9.91 × 10⁴ «m s⁻¹» ✓

State what is meant by gravitational field strength.

Show that V = –g(R + h).

Draw a graph, on the axes, to show the variation of the gravitational potential V of the planet with height h above the surface of the planet.

A planet has a radius of 3.1 × 10⁶ m. At a point P a distance 2.4 × 10⁷ m above the surface of the planet the gravitational field strength is 2.2 N kg^–1. Calculate the gravitational potential at point P, include an appropriate unit for your answer.

The diagram shows the path of an asteroid as it moves past the planet.

When the asteroid was far away from the planet it had negligible speed. Estimate the speed of the asteroid at point P as defined in (b).

The mass of the asteroid is 6.2 × 10¹² kg. Calculate the gravitational force experienced by the planet when the asteroid is at point P.

the «gravitational» force per unit mass exerted on a point/small/test mass

[1 mark]

at height h potential is V = –$\frac{GM}{(R+h)}$

field is g = $\frac{GM}{{(R+h)}^2}$

«dividing gives answer»

Do not allow an answer that starts with g = –$\frac{\mathrm{\Delta }V}{\mathrm{\Delta }r}$ and then cancels the deltas and substitutes R + h

[2 marks]

correct shape and sign

non-zero negative vertical intercept

[2 marks]

V = «–2.2 × (3.1 × 10⁶ + 2.4 × 10⁷) =» «–» 6.0 × 10⁷ J kg^–1

Unit is essential

Allow eg MJ kg^–1 if power of 10 is correct

Allow other correct SI units eg m²s^–2, N m kg^–1

[1 mark]

total energy at P = 0 / KE gained = GPE lost

«$\frac{1}{2}$mv² + mV = 0 ⇒» v = $\sqrt{-2V}$

v = «$\sqrt{2\times 6.0\times {10}^7}$ =» 1.1 × 10⁴ «ms^–1»

Award [3] for a bald correct answer

Ignore negative sign errors in the workings

Allow ECF from 6(b)

[3 marks]

ALTERNATIVE 1

force on asteroid is «6.2 × 10¹² × 2.2 =» 1.4 × 10¹³ «N»

«by Newton’s third law» this is also the force on the planet

ALTERNATIVE 2

mass of planet = 2.4 x 10²⁵ «kg» «from V = –$\frac{GM}{(R+h)}$»

force on planet «$\frac{GMm}{{(R+h)}^2}$» = 1.4 × 10¹³ «N»

MP2 must be explicit

[2 marks]

Outline what is meant by gravitational field strength at a point.

Newton’s law of gravitation applies to point masses. Suggest why the law can be applied to a satellite orbiting Mars.

Mars has a mass of 6.4 × 10²³ kg. Show that, for Mars, k is about 9 × 10^–13s²m^–3.

The time taken for Mars to revolve on its axis is 8.9 × 10⁴ s. Calculate, in m s^–1, the orbital speed of the satellite.

Show that the intensity of solar radiation at the orbit of Mars is about 600 W m^–2.

Determine, in K, the mean surface temperature of Mars. Assume that Mars acts as a black body.

The atmosphere of Mars is composed mainly of carbon dioxide and has a pressure less than 1 % of that on the Earth. Outline why the mean temperature of Earth is strongly affected by gases in its atmosphere but that of Mars is not.

force per unit mass ✔

acting on a small/test/point mass «placed at the point in the field» ✔

Mars is spherical/a sphere «and of uniform density so behaves as a point mass» ✔

satellite has a much smaller mass/diameter/size than Mars «so approximates to a point mass» ✔

«$\frac{m{v}^2}{r}=\frac{GMm}{r^2}$ hence» $v=\sqrt{\frac{GM}{R}}$. Also $v=\frac{2\pi R}{T}$

OR

$m{\omega }^{2}r=\frac{GMm}{r^2}$ hence ${\omega }^2=\frac{GM}{R^3}$ ✔

uses either of the above to get $T^2=\frac{4{\pi }^2}{GM}{R}^3$

OR

uses $k=\frac{4{\pi }^2}{GM}$ ✔

k = 9.2 × 10⁻¹³ / 9.3 × 10⁻¹³

Unit not required

$R^3=\frac{T^2}{k}=\frac{{\left(8.9\times {10}^4\right)}^2}{9.25\times {10}^{-13}}$  R = 2.04 × 10⁷ «m» ✔

v = «$\omega r=\frac{2\pi \times 2.04\times {10}^7}{89000}=$» 1.4 × 10³ «m s^–1»

OR

v = «$\sqrt{\frac{GM}{R}}=\sqrt{\frac{6.67\times {10}^{-11}\times 6.4\times {10}^{23}}{2.04\times {10}^7}}=$» 1.4 × 10³ «m s^–1» ✔

use of $I\propto \frac{1}{r^2}$ «1.36 × 10³ × $\frac{1}{{1.5}^2}$» ✔

604 «W m^–2» ✔

use of $\frac{600}{4}$ for mean intensity ✔

temperature/K = «$\sqrt[4]{\frac{600}{4\times 5.67\times {10}^{-8}}}=$» 230 ✔

reference to greenhouse gas/effect ✔

recognize the link between molecular density/concentration and pressure ✔

low pressure means too few molecules to produce a significant heating effect

OR

low pressure means too little radiation re-radiated back to Mars ✔

The greenhouse effect can be described, it doesn’t have to be named

Show that the intensity of the solar radiation at the location of Titan is 16 W m⁻².

Titan has an atmosphere of nitrogen. The albedo of the atmosphere is 0.22. The surface of Titan may be assumed to be a black body. Explain why the average intensity of solar radiation absorbed by the whole surface of Titan is 3.1 W m⁻².

Show that the equilibrium surface temperature of Titan is about 90 K.

The mass of Titan is 0.025 times the mass of the Earth and its radius is 0.404 times the radius of the Earth. The escape speed from Earth is 11.2 km s⁻¹. Show that the escape speed from Titan is 2.8 km s⁻¹.

The orbital radius of Titan around Saturn is $R$ and the period of revolution is $T$.

Show that $T^2=\frac{4{\mathrm{\pi }}^2{R}^{ 3}}{GM}$ where $M$ is the mass of Saturn.

The orbital radius of Titan around Saturn is 1.2 × 10⁹m and the orbital period is 15.9 days. Estimate the mass of Saturn.

Show that the mass of a nitrogen molecule is 4.7 × 10⁻²⁶ kg.

Estimate the root mean square speed of nitrogen molecules in the Titan atmosphere. Assume an atmosphere temperature of 90 K.

Discuss, by reference to the answer in (b), whether it is likely that Titan will lose its atmosphere of nitrogen.

incident intensity $\frac{1360}{9.3^2}$ OR $15.7\approx 16$ «W m⁻²» ✓

Allow the use of 1400 for the solar constant.

exposed surface is ¼ of the total surface ✓

absorbed intensity = (1−0.22) × incident intensity ✓

0.78 × 0.25 × 15.7  OR  3.07 «W m⁻²» ✓

Allow 3.06 from rounding and 3.12 if they use 16 W m⁻².

σT ⁴ = 3.07

OR

T = 86 «K» ✓

$v=«\sqrt{\frac{2GM}{R}}=»\sqrt{\frac{0.025}{0.404}}\times 11.2$

OR

$2.79$ «km s⁻¹» ✓

correct equating of gravitational force / acceleration to centripetal force / acceleration ✓

correct rearrangement to reach the expression given ✓

Allow use of $\sqrt{\frac{GM}{R}}=\frac{2\mathrm{\pi }R}{T}$ for MP1.

$T=15.9\times 24\times 3600$ «s» ✓

$M=\frac{4{\mathrm{\pi }}^2{\left(1.2\times {10}^9\right)}^3}{6.67\times {10}^{-11}\times {\left(15.9\times 24\times 3600\right)}^2}=5.4\times {10}^{26}$«kg» ✓

Award [2] marks for a bald correct answer.

Allow ECF from MP1.

$m=\frac{28\times {10}^{-3}}{6.02\times {10}^{23}}$

OR

$4.65\times {10}^{-26}$ «kg» ✓

$«\frac{1}{2}m{v}^2=\frac{3}{2}kT\Rightarrow »v=\sqrt{\frac{3kT}{m}}$ ✓

$v=«\sqrt{\frac{3\times 1.38\times {10}^{-23}\times 90}{4.651\times {10}^{-26}}}=»283\approx 300$ «ms⁻¹» ✓

Award [2] marks for a bald correct answer.

Allow 282 from a rounded mass.

no, molecular speeds much less than escape speed ✓

Allow ECF from incorrect (d)(ii).

Label with arrows on the diagram the magnetic force F on the proton.

Label with arrows on the diagram the velocity vector v of the proton.

For this proton, determine, in m, the radius of the circular path. Give your answer to an appropriate number of significant figures.

For this proton, calculate, in s, the time for one full revolution.

F towards centre ✔

v tangent to circle and in the direction shown in the diagram ✔

«$qvB=\frac{m{v}^2}{R}\Rightarrow »R=\frac{mv}{qB}/\frac{1.673\times {10}^{-27}\times 2.16\times {10}^6}{1.60\times {10}^{-19}\times 0.042}$  ✔

R = 0.538«m»  ✔

R = 0.54«m»   ✔

$T=\frac{2\pi R}{v}/\frac{2\pi \times 0.54}{2.16\times {10}^6}$  ✔

$T=1.6\times {10}^{-6}$«s»   ✔

Examiners were requested to be lenient here and as a result most candidates scored both marks. Had we insisted on e.g. straight lines drawn with a ruler or a force arrow passing exactly through the centre of the circle very few marks would have been scored. For those who didn’t know which way the arrows were supposed to be the common guesses were to the left and up the page. Some candidates neglected to label the arrows.

Examiners were requested to be lenient here and as a result most candidates scored both marks. Had we insisted on e.g. straight lines drawn with a ruler or a force arrow passing exactly through the centre of the circle very few marks would have been scored. For those who didn’t know which way the arrows were supposed to be the common guesses were to the left and up the page. Some candidates neglected to label the arrows.

This was generally well answered although usually to 3 sf. Common mistakes were to substitute 0.042 for F and 1 for q. Also some candidates tried to answer in terms of electric fields.

This was well answered with many candidates scoring ECF from the previous part.

State the direction of the resultant force on the ball.

On the diagram, construct an arrow of the correct length to represent the weight of the ball.

Show that the magnitude of the net force F on the ball is given by the following equation.

                                          $$F=\frac{mg}{\tan \theta }$$

The radius of the bowl is 8.0 m and θ = 22°. Determine the speed of the ball.

Outline whether this ball can move on a horizontal circular path of radius equal to the radius of the bowl.

Outline why the ball will perform simple harmonic oscillations about the equilibrium position.

Show that the period of oscillation of the ball is about 6 s.

The amplitude of oscillation is 0.12 m. On the axes, draw a graph to show the variation with time t of the velocity v of the ball during one period.

A second identical ball is placed at the bottom of the bowl and the first ball is displaced so that its height from the horizontal is equal to 8.0 m.

The first ball is released and eventually strikes the second ball. The two balls remain in contact. Determine, in m, the maximum height reached by the two balls.

towards the centre «of the circle» / horizontally to the right

Do not accept towards the centre of the bowl

[1 mark]

downward vertical arrow of any length

arrow of correct length

Judge the length of the vertical arrow by eye. The construction lines are not required. A label is not required

eg: 

[2 marks]

ALTERNATIVE 1

F = N cos θ

mg = N sin θ

dividing/substituting to get result

ALTERNATIVE 2

right angle triangle drawn with F, N and W/mg labelled

angle correctly labelled and arrows on forces in correct directions

correct use of trigonometry leading to the required relationship

tan θ = $\frac{\text{O}}{A}=\frac{mg}{F}$

[3 marks]

$\frac{mg}{\tan \theta }$ = m$\frac{v^2}{r}$

r = R cos θ

v = $\sqrt{\frac{gR{\cos }^2\theta }{\sin \theta }}/\sqrt{\frac{gR\cos \theta }{\tan \theta }}/\sqrt{\frac{9.81\times 8.0\cos 22}{\tan 22}}$

v = 13.4/13 «ms ^–1»

Award [4] for a bald correct answer 

Award [3] for an answer of 13.9/14 «ms ^–1». MP2 omitted

[4 marks]

there is no force to balance the weight/N is horizontal

so no / it is not possible

Must see correct justification to award MP2

[2 marks]

the «restoring» force/acceleration is proportional to displacement

Direction is not required

[1 mark]

ω = «$\sqrt{\frac{g}{R}}$» = $\sqrt{\frac{9.81}{8.0}}$ «= 1.107 s^–1»

T = «$\frac{2\pi }{\omega }$ = $\frac{2\pi }{1.107}$ =» 5.7 «s»

Allow use of or g = 9.8 or 10

Award [0] for a substitution into T = 2π$\sqrt{\frac{I}{g}}$

[2 marks]

sine graph

correct amplitude «0.13 m s^–1»

correct period and only 1 period shown

Accept ± sine for shape of the graph. Accept 5.7 s or 6.0 s for the correct period.

Amplitude should be correct to ±$\frac{1}{2}$ square for MP2

eg: v /m s^–1   

[3 marks]

speed before collision v = «$\sqrt{2gR}$ =» 12.5 «ms^–1»

«from conservation of momentum» common speed after collision is $\frac{1}{2}$ initial speed «v_c = $\frac{12.5}{2}$ = 6.25 ms^–1»

h = «$\frac{{v_c}^2}{2g}=\frac{{6.25}^2}{2\times 9.81}$» 2.0 «m»

Allow 12.5 from incorrect use of kinematics equations

Award [3] for a bald correct answer

Award [0] for mg(8) = 2mgh leading to h = 4 m if done in one step.

Allow ECF from MP1

Allow ECF from MP2

[3 marks]

Outline why the gravitational potential is negative.

The gravitational potential due to the Sun at a distance r from its centre is V_S. Show that

rV_S = constant.

Calculate the gravitational potential energy of the Earth in its orbit around the Sun. Give your answer to an appropriate number of significant figures.

Calculate the total energy of the Earth in its orbit.

An asteroid strikes the Earth and causes the orbital speed of the Earth to suddenly decrease. Suggest the ways in which the orbit of the Earth will change.

Outline, in terms of the force acting on it, why the Earth remains in a circular orbit around the Sun.

potential is defined to be zero at infinity

so a positive amount of work needs to be supplied for a mass to reach infinity

V_S = $-\frac{GM}{r}$ so r x V_S «= –GM» = constant because G and M are constants

GM = 1.33 x 10²⁰ «J m kg^–1»

GPE at Earth orbit «= –$\frac{1.33\times {10}^{20}\times 6.0\times {10}^{24}}{1.5\times {10}^{11}}$» = «–» 5.3 x 10³³ «J»

Award [1 max] unless answer is to 2 sf.

Ignore addition of Sun radius to radius of Earth orbit.

ALTERNATIVE 1
work leading to statement that kinetic energy $\frac{GMm}{2r}$, AND kinetic energy evaluated to be «+» 2.7 x 10³³ «J»

energy «= PE + KE = answer to (b)(ii) + 2.7 x 10³³» = «–» 2.7 x 10³³ «J»

ALTERNATIVE 2
statement that kinetic energy is $=-\frac{1}{2}$ gravitational potential energy in orbit

so energy «$=\frac{\text{answer to (b)(ii)}}{2}$» = «–» 2.7 x 10³³ «J»

Various approaches possible.

«KE will initially decrease so» total energy decreases
OR
«KE will initially decrease so» total energy becomes more negative

Earth moves closer to Sun

new orbit with greater speed «but lower total energy»

changes ellipticity of orbit

centripetal force is required

and is provided by gravitational force between Earth and Sun

Award [1 max] for statement that there is a “centripetal force of gravity” without further qualification.