On the diagram, draw the shape of the electric field between the plates.

Calculate the electric field strength between the plates.

Calculate the force on M.

Determine the change in the electric potential energy of M as it moves from the positive to the negative plate.

Construct the nuclear equation for the decay of radium-226.

Radium-226 has a half-life of 1600 years. Determine the time, in years, it takes for the activity of radium-226 to fall to 5% of its original activity.

minimum of three lines equally spaced and distributed, perpendicular to the plates and downwards; edge effect shown; } (condone lines that do not touch plates)

\(4.3 \times {10^5}{\text{ (N}}{{\text{C}}^{ - 1}}{\text{)}}\)

\((F = Eq = ){\text{ }}4.3 \times {10^5} \times 2 \times 1.6 \times {10^{ - 19}}\); (allow ECF from (b)(i))

\(1.4 \times {10^{ - 13}}{\text{ (N)}}\);

Award [2] for a bald correct answer.

\(\Delta {E_{\text{P}}} = q\Delta V\)\(\,\,\,\)or\(\,\,\,\)\(3.2 \times {10^{ - 19}} \times 5.2 \times {10^3}\);

\(1.7 \times {10^{ - 15}}{\text{ (J)}}\);

negative/loss;

\(\left( {_{\;88}^{226}{\text{Ra}} \to _{\;86}^{222}{\text{Rn}} + _2^4{\text{He}} + _0^0\gamma } \right)\)

\(_{\;86}^{222}{\text{Rn}}\)\(\,\,\,\)or\(\,\,\,\)\(_2^4{\text{He}}\);

numbers balance top and bottom on right-hand side;

\(\lambda  = \frac{{\ln 2}}{{1600}} = 4.33 \times {10^{ - 4}}{\text{ (y}}{{\text{r}}^{ - 1}}{\text{)}}\);

\(0.05 = {{\text{e}}^{ - \lambda t}}\);

6900 (years);

Award [3] for a bald correct answer.

Award [2 max] for 2.18 \( \times \) 10¹¹ (s).

Award [1 max] to a candidate who identifies time as about 4.3 half-lives but cannot get further or gives an approximate reasoned answer.

However award [3] if number n of half-lives is calculated from 0.05 = 2^–n (= 4.32 usually from use of log₂ working) and time shown.

Field patterns were often negligently drawn. Lines did not meet both plates, edge effects were ignored, and the (vital) equality of spacing between drawn lines was not considered. Candidates continue to show their inadequacy in responding to questions that demand a careful and accurate diagram.

This sequence of calculations was often undertaken well with appropriate figures carried through from part to part.

This sequence of calculations was often undertaken well with appropriate figures carried through from part to part.

This sequence of calculations was often undertaken well with appropriate figures carried through from part to part. The only common error was the omission of a consideration of the gain or loss of the energy change in part (iii).

As one of the easiest questions on the paper this was predictably well done.

In the past candidates have found calculation involving exponential change difficult. On this occasion, however examiners saw a large number of correct and well explained solutions from candidates.

State Newton’s universal law of gravitation.

A satellite of mass m orbits a planet of mass M. Derive the following relationship between the period of the satellite T and the radius of its orbit R (Kepler’s third law).

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

A polar orbiting satellite has an orbit which passes above both of the Earth’s poles. One polar orbiting satellite used for Earth observation has an orbital period of 6.00 × 10³s.
                                                                   Mass of Earth                = 5.97 × 10²⁴ kg
                                                                   Average radius of Earth = 6.37 × 10⁶ m

(i) Using the relationship in (b), show that the average height above the surface of the Earth for this satellite is about 800 km.

(ii) The satellite moves from an orbit of radius 1200 km above the Earth to one of radius 2500 km. The mass of the satellite is 45 kg.

Calculate the change in the gravitational potential energy of the satellite.

(iii) Explain whether the gravitational potential energy has increased, decreased or stayed the same when the orbit changes, as in (c)(ii).

the (attractive) force between two (point) masses is directly proportional to the product of the masses;
and inversely proportional to the square of the distance (between their centres of mass); 
Use of equation is acceptable:
Award [2] if all five quantities defined. Award [1] if four quantities defined.

\(G\frac{{Mm}}{{{R^2}}} = \frac{{m{v^2}}}{R}\) so \({v^2} = \frac{{Gm}}{R}\);
\(v = \frac{{2\pi R}}{T}\);
\({v^2} = \frac{{4{\pi ^2}{R^2}}}{{{T^2}}} = \frac{{Gm}}{R}\);

or

\(G\frac{{Mm}}{{{R^2}}} = m{\omega ^2}R\);
\({\omega ^2} = \frac{{4{\pi ^2}}}{{{T^2}}}\);
\(\frac{{4{\pi ^2}}}{{{T^2}}} = \frac{{GM}}{{{R^3}}}\);
Award [3] to a clear response with a missing step.

(i) \({R^3} = \frac{{6.67 \times {{10}^{ - 11}} \times 5.97 \times {{10}^{24}} \times {{6000}^2}}}{{4 \times {\pi ^2}}}\);
R=7.13x10⁶(m);
h=(7.13x10⁶-6.37x10⁶)=760(km);
Award [3] for an answer of 740 with π taken as 3.14.

(ii) clear use of \(\Delta V = \frac{{\Delta E}}{m}\) and \(V =  - \frac{{Gm}}{r}\) or \(\Delta E = GMm\left( {\frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right)\);
one value of potential energy calculated (2.37×10⁹ or 2.02×10⁹ );
3.5×10⁸ (J);
Award [3] for a bald correct answer.
Award [2] for 7.7x10⁹. Award [1] for 7.7x10¹².
Award [0] for answers using mgΔh.

(iii) increased;
further from Earth / closer to infinity / smaller negative value;
Award [0] for a bald correct answer.

(a) Many candidates stated that the Newton’s law force is proportional to the masses of the objects in question, rather than the product of the masses.

This part was generally well done with most candidates coming to the correct outcome; too often steps were missed out in the derivations and this cost candidates mark. It is essential that they realise that a derivation must include every step. The presentation of this part left much to be desired in quite a large minority with mathematically incorrect statements being given.

(i) Most candidates were able to substitute values into the equation and rearrange it to find a value for R. Many then fail to subtract the radius of the Earth.

(ii) Very few candidates completed this part correctly. Confusion between potential and potential energy was common as were adding the height in kilometres to the radius of the Earth in metres. A sizeable minority of candidates attempted to use the mgh equation.

(iii) This part was quite well answered with most candidates realising that the increase in height meant an increase in potential energy. Several argued that the magnitude decreased but being a negative quantity this meant an increase.

The table gives the gravitational potential V for various distances r from the surface of Earth. The radius of Earth is \(6.4 \times {10^3}{\text{ km}}\).

Show that the data are consistent with Earth acting as a point mass with its mass concentrated at its centre.

(i)     the gravitational potential energy.

(ii)     kinetic energy

(iii)     total energy.

recognize that \(V \times \) a distance is constant / \(\frac{{{V_1}}}{{{V_2}}} = \frac{{{R_2}}}{{{R_1}}}{\text{ (}} = 1.625)\);

use of \({R_E} + r\);

evaluates at least two data points \(4.0 \times {10^{14}}\); } (allow \(V \times R\) to be expressed in any consistent unit)

Award [2 max] if answer fails to use radius of Earth but infers wrong conclusion from correct evaluation of two or more data points.

Award [3 max] if answer evaluates mass of Earth and shows that it is the same for two or more data points.

(i)     change in gravitational potential \( = ( + )28{\text{ }}({\text{MJ}}\,{\text{k}}{{\text{g}}^{ - 1}})\);

change in gravitational potential energy \( = (28 \times {10^6} \times 780 = ){\text{ }}2.2 \times {10^{10}}{\text{ (J)}}\);

Award [2] for a bald correct answer.

(ii)     (change in) \(\frac{1}{2}m{v^2} = \) (change in) \(\frac{{GMm}}{2}{\text{ }}\left( { =  - {\text{change in }}\frac{{mV}}{2}} \right)\);

\(( - )1.1 \times {10^{10}}{\text{ (J)}}\);

Award [2] for a bald correct answer.

(iii)     change in kinetic energy \( =  - 1.1 \times {10^{10}}{\text{ (J)}}\); (recognition that KE is lost)

change in gravitational potential energy \( =  + 2.2 \times {10^{10}}{\text{ (J)}}\); } recognition that gpe is gained)

\(( + )1.1 \times {10^{10}}{\text{ (J)}}\); (direction of change required for this mark)

Award [1 max] for a bald correct answer with sign.

Award [0] for a bald correct answer with no sign.

or

change in total energy = change in potential energy + change in kinetic energy;

adds (e)(i) to (e)(ii); (allow ECF in value and sign)

\(( + )1.1 \times {10^{10}}{\text{ (J)}}\); (do not allow negative answer)

Accept correct answer only without ECF for this approach.

This was a simple test to establish whether candidates could check the consistency of data. Most could not. The obvious solution (to evaluate Vr twice having first incorporated the radius of the Earth) was rare. Many candidates preferred to calculate the mass of the earth twice and check that the values obtained were similar. In principle this could have scored full marks but three marks were only rarely seen by examiners.

(i)     It is quite clear that candidates do not understand the relationship between gravitational potential and gravitational potential energy. A consideration of units could have led them in the correct direction. Many could only tackle this by carrying out a full solution from the gravitational potential equation printed in the data booklet.

(ii)     Similarly candidates were unable to use the known change in gravitational potential energy to establish the change in kinetic energy. Most felt that they had to calculate the kinetic energies from first principles and usually obtained the wrong answer.

(iii)     Most candidates did not continue from (i) and (ii) to this part question; it was either left blank or an attempt was made at a statement of conservation of energy. This was another case where knowledge of a principle does not necessarily mean that a candidate can operate it effectively. Even the most able candidates seemed to struggle with this standard piece of bookwork.

(i)     Ignoring air resistance, calculate the horizontal distance travelled by the clay block before it strikes the ground.

(ii)     The diagram in (c) shows the path of the clay block neglecting air resistance. On the diagram, draw the approximate shape of the path that the clay block will take assuming that air resistance acts on the clay block.

Define gravitational potential energy of a mass at a point.

(i)     calculate the change in gravitational potential energy of the rocket at a distance 4R from the centre of Mars.

(ii)     show that the magnitude of the gravitational field strength at a distance 4R from the centre of Mars is \({\text{0.23 N}}\,{\text{k}}{{\text{g}}^{ - 1}}\).

(i)     use of kinematic equation to yield time;

\(t = \sqrt {\frac{{2s}}{g}} {\text{ (}} = 0.42{\text{ s)}}\);

\(s = {\text{horizontal speed\( \times \)time}}\);

\( = 1.8{\text{ m}}\);

Accept g = 10 m\(\,\)s\(^{ - 2}\) equivalent answers 1.79 from 9.8, 1.77 from 10.

(ii)     initial drawn velocity horizontal; (judge by eye)

reasonable shape; (i.e. quasi-parabolic)

horizontal distance moved always decreasing when compared to given path / range less than original;

work done in moving mass from infinity to a point;

(i)     read offs \( - 12.6\) and \( - 3.2\);

gain in \({\text{gpe }}1.2 \times {10^4} \times [12.6 - 3.2]\) or gain in g potential \(\left[ {12.6 \times {{10}^6} - 3.2 \times {{10}^6}} \right]\);

\( = 1.13 \pm 0.05 \times {10^5}{\text{ MJ}}\) or \(1.13 \pm 0.05 \times {10^{11}}{\text{ J}}\);

(ii)     use of gradient of graph to determine \(g\);

values substituted from drawn gradient \(\left( {{\text{typically }}\frac{{6.7 \times {{10}^6}}}{{7 \times 3.3 \times {{10}^6}}}} \right)\);

\( = 0.23{\text{ N}}\,{\text{k}}{{\text{g}}^{ - 1}}\) (allow answers in the range of 0.20 to 0.26 N\(\,\)kg–1)

Award [0] for solutions from \(\frac{V}{r}\).

(i)     Candidates were required to determine the time taken to fall to the floor and then use this time to evaluate the distance travelled horizontally. Many managed this with more or less success.

(ii)     Many candidates produced poor attempts at the sketch. Initial trajectories were not horizontal and the general shapes of the curves were usually not quasi-parabolic.

Some candidates gave a definition of gravitational potential, i.e. they related the energy to that of a unit mass.

Throughout this part candidates were instructed to use the graph, those who used other non-graphical methods were penalised.

(i)     There were many good evaluations with complete and well presented solutions.

(ii)     Use of the Data Booklet equation for gravitational field strength without reference to the graph was common.

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.

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 = –\(\frac{{\Delta V}}{{\Delta r}}\) 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]

Show that the energy of photons from the UV lamp is about 10 eV.

Calculate, in J, the maximum kinetic energy of the emitted electrons.

Suggest, with reference to conservation of energy, how the variable voltage source can be used to stop all emitted electrons from reaching the collecting plate.

The variable voltage can be adjusted so that no electrons reach the collecting plate. Write down the minimum value of the voltage for which no electrons reach the collecting plate.

On the diagram, draw and label the equipotential lines at –0.4 V and –0.8 V.

An electron is emitted from the photoelectric surface with kinetic energy 2.1 eV. Calculate the speed of the electron at the collecting plate.

E₁ = –13.6 «eV» E₂ = – \(\frac{{13.6}}{4}\) = –3.4 «eV»

energy of photon is difference E₂ – E₁ = 10.2 «≈ 10 eV»

Must see at least 10.2 eV.

[2 marks]

10 – 5.1 = 4.9 «eV»

4.9 × 1.6 × 10^–19 = 7.8 × 10^–19 «J»

Allow 5.1 if 10.2 is used to give 8.2×10⁻¹⁹ «J».

EPE produced by battery

exceeds maximum KE of electrons / electrons don’t have enough KE

For first mark, accept explanation in terms of electric potential energy difference of electrons between surface and plate.

[2 marks]

4.9 «V»

Allow 5.1 if 10.2 is used in (b)(i).

Ignore sign on answer.

[1 mark]

two equally spaced vertical lines (judge by eye) at approximately 1/3 and 2/3

labelled correctly

[2 marks]

kinetic energy at collecting plate = 0.9 «eV»

speed = «\(\sqrt {\frac{{2 \times 0.9 \times 1.6 \times {{10}^{ - 19}}}}{{9.11 \times {{10}^{ - 31}}}}} \)» = 5.6 × 10⁵ «ms^–1»

Allow ECF from MP1

[2 marks]

Outline, with reference to the energy of the rocket, why the speed of the rocket is changing between P and Q.

Estimate the average gravitational field strength of the planet between P and Q.

(i) An object is a distance r from the centre of a planet. Show that the minimum speed required to escape the gravitational field is equal to
\[\sqrt {2g'r} \]
where g′ is the gravitational field strength at distance r from the centre of a planet.

(ii) Discuss, using a calculation, whether the rocket at P can completely escape the gravitational field of the planet without further use of the engine.

A space station is in orbit at a distance r from the centre of the planet in (e)(i). A satellite is launched from the space station so as just to escape from the gravitational field of the planet. The launch takes place in the same direction as the velocity of the space station. Outline why the launch velocity relative to the space station can be less than your answer to (e)(i).

gravitational potential energy is being gained;
this is at the expense of kinetic energy (and speed falls);

\(\left( {{\rm{acceleration}} = \frac{{\left( {v - u} \right)}}{t} = \frac{{4.25 \times {{10}^3} - 4.38 \times {{10}^3}}}{{60}} = } \right)\left(  -  \right)2.17\left( {{\rm{m}}{{\rm{s}}^{ - 2}}} \right)\);

(i) \(\left( {{\rm{gravitational force}}} \right){\rm{ = }}mg'{\rm{ = }}\frac{{{\rm{GMm}}}}{{{r^2}}}\);
\(\left( {{\rm{kinetic energy}}} \right){\rm{ = }}\frac{{{\rm{m}}{{\rm{v}}^{\rm{2}}}}}{2} = \frac{{GMm}}{r}\);
\({v^2} = 2\frac{{GM}}{{{r^2}}}r\left( { = 2g'r} \right)\);

Satellite X orbits 6600 km from the centre of the Earth.

Mass of the Earth = 6.0 x 10²⁴ kg

Show that the orbital speed of satellite X is about 8 km s^–1.

the orbital times for X and Y are different.

satellite Y requires a propulsion system.

The cable between the satellites cuts the magnetic field lines of the Earth at right angles.

Explain why satellite X becomes positively charged.

Satellite X must release ions into the space between the satellites. Explain why the current in the cable will become zero unless there is a method for transferring charge from X to Y.

The magnetic field strength of the Earth is 31 μT at the orbital radius of the satellites. The cable is 15 km in length. Calculate the emf induced in the cable.

Estimate the value of k in the following expression.

T = \(2\pi \sqrt {\frac{m}{k}} \)

Give an appropriate unit for your answer. Ignore the mass of the cable and any oscillation of satellite X.

Describe the energy changes in the satellite Y-cable system during one cycle of the oscillation.

«\(v = \sqrt {\frac{{G{M_E}}}{r}} \)» = \(\sqrt {\frac{{6.67 \times {{10}^{ - 11}} \times 6.0 \times {{10}^{24}}}}{{6600 \times {{10}^3}}}} \)

7800 «m s^–1»

Full substitution required

Must see 2+ significant figures.

Y has smaller orbit/orbital speed is greater so time period is less

Allow answer from appropriate equation

Allow converse argument for X

to stop Y from getting ahead

to remain stationary with respect to X

otherwise will add tension to cable/damage satellite/pull X out of its orbit

cable is a conductor and contains electrons

electrons/charges experience a force when moving in a magnetic field

use of a suitable hand rule to show that satellite Y becomes negative «so X becomes positive»

Alternative 2

cable is a conductor

so current will flow by induction flow when it moves through a B field

use of a suitable hand rule to show current to right so «X becomes positive»

Marks should be awarded from either one alternative or the other.

Do not allow discussion of positive charges moving towards X

electrons would build up at satellite Y/positive charge at X

preventing further charge flow

by electrostatic repulsion

unless a complete circuit exists

«ε = Blv =» 31 x 10^–6 x 7990 x 15000

3600 «V»

Allow 3700 «V» from v = 8000 m s^–1.

use of k = «\(\frac{{4{\pi ^2}m}}{{{T^2}}} = \)» \(\frac{{4 \times {\pi ^2} \times 350}}{{{{5.2}^2}}}\)

510

N m^–1 or kg s^–2

Allow MP1 and MP2 for a bald correct answer

Allow 500

Allow N/m etc.

E_p in the cable/system transfers to E_k of Y

and back again twice in each cycle

Exclusive use of gravitational potential energy negates MP1

The kinetic energy \({E_{\text{K}}}\) given to the shuttle at its launch is given by the expression

\[{E_{\text{K}}} = \frac{{7GMm}}{{8{R_{\text{E}}}}}\]

where \(G\) is the gravitational constant, \(M\) is mass of the Earth and \({R_{\text{E}}}\) is the radius of the Earth. Deduce that the shuttle cannot escape the gravitational field of the Earth.

Show that the total energy of the shuttle in its orbit is given by \( - \frac{{GMm}}{{2R}}\). Air resistance is negligible.

Using the expression for \({E_{\text{K}}}\) in (a) and your answer to (b)(i), determine \(R\) in terms of \({R_{\text{E}}}\).

In practice, the total energy of the shuttle decreases as it collides with air molecules in the upper atmosphere. Outline what happens to the speed of the shuttle when this occurs.

KE needs to be \( \ge \) (magnitude of) GPE at surface \(\left( { - \frac{{GMm}}{{{R_E}}}} \right)\);

But KE is \(\frac{{7GMm}}{{8{R_{\text{E}}}}} < \frac{{GMm}}{{{R_{\text{E}}}}}\) / OWTTE;

or

shows that total energy at launch \( =  - \frac{{GMm}}{{8{R_{\text{E}}}}}\); (appropriate working required)

this is \( < 0\), so escape impossible;

or

states that escape velocity needed is \(\sqrt {\frac{{2GM}}{{{R_{\text{E}}}}}} \);

shows launch velocity is only \(\sqrt {\frac{{7GM}}{{4{R_{\text{E}}}}}} \); (appropriate working required)

\({E_{{\text{tot}}}} = {\text{PE}} + {\text{KE}}\);

shows that kinetic energy \( = (\frac{1}{2}m{v^2} = ){\text{ }}\frac{{GMm}}{{2R}}\); (appropriate working required)

adds PE \(\left( { - \frac{{GMm}}{R}} \right)\) and KE to get given answer; (appropriate working required)

\( - \frac{{GMm}}{{{R_{\text{E}}}}} + \frac{{7GMm}}{{8{R_{\text{E}}}}} =  - \frac{{GMm}}{{2R}}\); (equating total energy at launch and in orbit)

\(\frac{1}{{8{R_{\text{E}}}}} = \frac{1}{{2R}}\);

\(R = 4{R_{\text{E}}}\);

Award [0] for an answer such as \(R = \frac{{4{R_{\text{E}}}}}{7}\).

total energy decreases/becomes a greater negative value, so R decreases;

as \(R\) decreases kinetic energy increases;

speed increases;

Allow third marking point even if reasoning is incorrect.

Candidates were asked to show that the shuttle could not escape the Earth’s field. There are many ways of approaching this, but in general answers were good, with only the weakest candidates failing to know where to start.

The determination of total energy of a mass in Earth orbit is a standard classroom derivation. Most were able to reproduce it, but not everyone explained how the formula for orbital KE was derived.

Determining the radius of the orbit proved difficult for most candidates with many obtaining negative values or orbits with a radius less than the radius of the Earth. This was due to carelessness with the symbols used for the two different radii.

Few candidates equated a decrease in total energy with an increasingly negative value. The consequent fact that the radius of the orbit decreases and velocity increases was counter-intuitive for most candidates. Most incorrectly opted for a decrease in KE due to resistive forces.

Outline what is meant by escape speed.

A probe is launched vertically upwards from the surface of a planet with a speed

\[v = \frac{3}{4}{v_{{\rm{esc}}}}\]

where v_esc is the escape speed from the planet. The planet has no atmosphere.

Determine, in terms of the radius of the planet R, the maximum height from the surface of the planet reached by the probe.

The total energy of a probe in orbit around a planet of mass M is \(E =  - \frac{{GMm}}{{2r}}\) where m is the mass of the probe and r is the orbit radius. A probe in low orbit experiences a small frictional force. Suggest the effect of this force on the speed of the probe.

«kinetic energy at take off» =\(\frac{9}{{16}} \times \frac{{GMm}}{R}\)

kinetic energy at take off + «gravitational» potential energy = «gravitational» potential energy at maximum height

OR
\(\frac{9}{{16}} \times \frac{{GMm}}{R} - \frac{{GMm}}{R} =  - \frac{{GMm}}{R}\)


solves for r and subtracts R from answer =\(\frac{{9R}}{7}\)


Award [0] for work that assumes constant g.

Explain what is meant by escape speed.

Titania is a moon that orbits the planet Uranus. The mass of Titania is 3.5×10²¹kg. The radius of Titania is 800 km.

(i) Use the data to calculate the gravitational potential at the surface of Titania.

(ii) Use your answer to (b)(i) to determine the escape speed for Titania.

An astronaut visiting Titania throws an object away from him with an initial horizontal velocity of 1.8 m s^–1. The object is 1.5 m above the moon’s surface when it is thrown. The gravitational field strength at the surface of Titania is 0.37 N kg^–1.

Calculate the distance from the astronaut at which the object first strikes the surface.

(minimum) speed of object to escape gravitational field of a planet/travel to infinity;
at surface of planet;
without (further) energy input;

(i) \( - \frac{{6.67 \times {{10}^{ - 11}} \times 3.5 \times {{10}^{21}}}}{{8.0 \times {{10}^5}}}\);
−2.9×10⁵Jkg^–1; (allow Nmkg⁻¹)
Award [1 max] if negative sign omitted.

(ii) \(\frac{1}{2}m{v^2} = mV\);
speed=\( = \sqrt {2 \times 2.9 \times {{10}^5}} \); (allow ECF from (b)(i))
7.6 ×10²ms^-1;
Ignore sign.
Award [3] for a bald correct answer.

time to hit surface \( = \sqrt {\frac{{2.0 \times 1.5}}{{0.37}}} \left( { = 2.85{\rm{s}}} \right)\);
distance to impact = 2.85×1.8;
5.1m;

Explain what is meant by the gravitational potential at the surface of a planet.

An unpowered projectile is fired vertically upwards into deep space from the surface of planet Venus. Assume that the gravitational effects of the Sun and the other planets are negligible.

The following data are available.

(i) Determine the initial kinetic energy of the projectile.

(ii) Describe the subsequent motion of the projectile until it is effectively beyond the gravitational field of Venus.

the «gravitational» work done «by an external agent» per/on unit mass/kg

Allow definition in terms of reverse process of moving mass to infinity eg “work done on external agent by…”.
Allow “energy” as equivalent to “work done”

in moving a «small» mass from infinity to the «surface of» planet / to a point

N.B.: on SL paper Q5(a)(i) and (ii) is about “gravitational field”.

i
escape speed
Care with ECF from MP1.

v = «\(\sqrt {\left( {\frac{{2\,{\text{GM}}}}{R}} \right)}  = \)»

\(\sqrt {\left( {\frac{{2 \times 6.67 \times {{10}^{ - 11}} \times 4.87 \times {{10}^{24}}}}{{6.05 \times {{10}^6}}}} \right)} \) or 1.04×10⁴«m s^–1»

or «1.1 × 1.04 × 10⁴ m s^-1»= 1.14 × 10⁴ «m s^–1»

KE = «0.5 × 3500 × (1.1 × 1.04 × 10⁴ m s^–1)² =» 2.27×10¹¹ «J»

Award [1 max] for omission of 1.1 – leads to 1.88×10¹¹ m s^-1.
Award [2] for a bald correct answer.

ii
Velocity/speed decreases / projectile slows down «at decreasing rate»

«magnitude of» deceleration decreases «at decreasing rate»
Mention of deceleration scores MP1 automatically.

velocity becomes constant/non-zero
OR
deceleration tends to zero

Accept “negative acceleration” for “deceleration”.

Must see “velocity” not “speed” for MP3.

State why the work done by the gravitational force during one full revolution of the probe is zero.

Deduce for the probe in orbit that its

(i) speed is \(v = \sqrt {\frac{{GM}}{r}} \).

(ii) total energy is \(E =  - \frac{{GMm}}{{2r}}\).

It is now required to place the probe in another circular orbit further away from the planet. To do this, the probe’s engines will be fired for a very short time.

State and explain whether the work done on the probe by the engines is positive, negative or zero.

because the force is always at right angles to the velocity / motion/orbit is an equipotential surface; 
Do not accept answers based on the displacement being zero for a full revolution.

(i) equating gravitational force \(\frac{{GMm}}{{{r^2}}}\);
to centripetal force \(\frac{{m{v^2}}}{r}\) to get result;

(ii) kinetic energy is \(\frac{{GMm}}{{2r}}\);
addition to potential energy −\(\frac{{GMm}}{{r}}\) to get result;

the total energy (at the new orbit) will be greater than before/is less negative;
hence probe engines must be fired to produce force in the direction of motion / positive work must be done (on the probe); 
Award [1] for mention of only potential energy increasing.

Define electric potential at a point in an electric field.

A positive point charge is moving towards a small, charged metal sphere along a radial path.

At the position shown in the diagram, the point charge has a speed of 3.50 m s^–1 and is at a distance of 2.78 m from the centre of the metal sphere. The charge on the sphere is +4.00 μC.

(i) State the direction of the velocity of the point charge with respect to an equipotential surface due to the metal sphere.

(ii) Show that the electric potential V due to the charged sphere at a distance of 2.78 m from its centre is 1.29 ×10⁴ V.

(iii) The electric potential at the surface of the sphere is 7.20 ×10⁴ V. The point charge has a charge of +0.0320 μC and its mass is 1.20 ×10^{– 4} kg. Determine if the point charge will collide with the metal sphere.

the work done per unit charge;
when a small/test/point positive charge; (charge sign is essential)
is moved from infinity to the point;

(i) perpendicular / at right angles / at 90° / normal;

(ii) \(V = \frac{{8.99 \times {{10}^9} \times 4.00 \times {{10}^{ - 6}}}}{{2.78}}\) or \(1.2935 \times {10^4}{\rm{V}}\); (use of \(\frac{1}{{4\pi {\varepsilon _0}}}\) gives 1.29378×10⁴)
(≈1.29×10⁴V)

There were three marks for this question and most scored 1 or 2. The full definition of electric potential at a point was simply not well enough remembered. Many forgot that it is (i) the work done per unit charge on a (ii) positive (test) charge (iii) as the charge is moved from infinity to the point in question.

(i) Most failed to read the question and did not realize that the question required the direction with respect to an equipotential surface. Most gave the direction relative to the diagram.
(ii) This was well done by a great many, however examiners did expect to see a full substitution, it was not acceptable to leave the permittivity of free space as a symbol in this “show that” question.
(iii) Even where candidates could negotiate a path through this testing question, solutions were seldom clear.

Define gravitational potential at a point in a gravitational field.

The graph shows how the gravitational potential V of Earth varies with distance R from the centre of Earth in the range R=2.0×10⁸ m to R=5.0×10⁸ m.

The Moon is at a distance of 4.0×10⁸ m from the centre of Earth. At some time in the past it was at a distance of 2.7×10⁸ m from the centre of Earth.

Use the graph opposite to determine

(i) the present day magnitude of the acceleration of the Moon.

(ii) by how much the potential energy of the Moon has changed as a result of moving from R=2.7×10⁸ m to R=4.0×10⁸ m. The mass of the Moon is 7.4×10²² kg.

State why the change of potential energy in (f)(ii) is an increase.

work done per unit mass;
in bringing (test) mass from infinity to point;
reference to small/point (test) mass;

(i) tangent construction attempted at R=4.0×10⁸ m;
triangle/pair of coordinates used in calculation;
attempt to calculate gradient;
2.5×10^–3 ms⁻²; (accept answers in the range of 2.2 to 2.7)
Award [1 max] for \(\frac{V}{R}\) to give (–)2.1×10^–3.

(ii) change in V=0.45×10⁶-0.50×10⁶ Jkg⁻¹;
change in PE = (0.5×10⁶×7.4×10²²=)3.3-3.7×10²⁸ J;

work is done against the gravitational field of Earth / Moon is now closer to infinity/further from Earth / \(\frac{{ - GMm}}{R}\) means that as R increases potential increases/becomes less negative;

State the Stefan-Boltzmann law for a black body.

Deduce that the solar power incident per unit area at distance d from the Sun is given by

\(\frac{{\sigma {R^2}{T^4}}}{{{d^2}}}\).

Calculate, using the data given, the solar power incident per unit area at distance d from the Sun.

State two reasons why the solar power incident per unit area at a point on the surface of the Earth is likely to be different from your answer in (c).

The average power absorbed per unit area at the Earth’s surface is 240Wm^–2. By treating the Earth’s surface as a black body, show that the average surface temperature of the Earth is approximately 250K.

Explain why the actual surface temperature of the Earth is greater than the value in (e).

(i) Identify the force that causes the centripetal acceleration of the spaceship.

(ii) Explain why astronauts inside the spaceship would feel “weightless”, even though there is a force acting on them.

Deduce that the speed of the spaceship is \(v = \sqrt {\frac{{GM}}{r}} \).

The table gives equations for the forms of energy of the orbiting spaceship.

The spaceship passes through a cloud of gas, so that a small frictional force acts on the spaceship.

(i) State and explain the effect that this force has on the total energy of the spaceship.

(ii) Outline the effect that this force has on the speed of the spaceship.

power/energy per second emitted is proportional to surface area;
and proportional to fourth power of absolute temperature / temperature in K;
Accept equation with symbols defined.

solar power given by \(4\pi {R^2}\sigma {T^4}\);
spreads out over sphere of surface area \(4\pi {d^2}\);
Hence equation given.

\(\left( {\frac{{\sigma {R^2}{T^4}}}{{{d^2}}} = } \right)\frac{{5.7 \times {{10}^{ - 8}} \times {{\left[ {7.0 \times {{10}^8}} \right]}^2} \times {{\left[ {5.8 \times {{10}^3}} \right]}^4}}}{{{{\left[ {1.5 \times {{10}^{11}}} \right]}^2}}}\);

=1.4×10³(Wm^-2);

Award [2] for a bald correct answer.

some energy reflected;
some energy absorbed/scattered by atmosphere;
depends on latitude;
depends on time of day;
depends on time of year;
depends on weather (eg cloud cover) at location;
power output of Sun varies;
Earth-Sun distance varies;

power radiated=power absorbed;
\(T = {}^4\sqrt {\frac{{240}}{{5.7 \times {{10}^{ - 8}}}}} \left( { = 250{\rm{K}}} \right)\);

Accept answers given as 260 (K).

radiation from Sun is re-emitted from Earth at longer wavelengths;
greenhouse gases in the atmosphere absorb some of this energy;
and radiate some of it back to the surface of the Earth;

(i) gravitational force / gravitational attraction / weight; (do not accept gravity)

(ii) astronauts and spaceship have the same acceleration;
acceleration is towards (centre of) planet;
so no reaction force between astronauts and spaceship;

or

astronauts and spaceships are both falling towards the (centre of the) planet;
at the same rate;
so no reaction force between astronauts and spaceship;

gravitational force equated with centripetal force / \(\frac{{GmM}}{{{r^2}}} = \frac{{m{v^2}}}{r}\);

\( \Rightarrow {v^2} = \frac{{GM}}{r} \Rightarrow \left( {v = \sqrt {\frac{{GM}}{r}} } \right)\);

(i) thermal energy is lost;
total energy decreases;

(ii) since E decreases, r also decreases;
as r decreases v increases / E_k increases so v increases;

To show the given value there is the requirement for an explanation of why the incident power absorbed by the Earth’s surface is equal to the power radiated by the Earth, few candidates were successful in this aspect. Although most could substitute into the Stefan-Boltzmann equation they needed to either show that the fourth root was used or to find the temperature to more significant figures than the value given.

Outline how this diagram shows that the gravitational field strength of planet X decreases with distance from the surface.

The diagram shows part of the surface of planet X. The gravitational potential at the surface of planet X is –3V and the gravitational potential at point Y is –V.

Sketch on the grid the equipotential surface corresponding to a gravitational potential of –2V.

A meteorite, very far from planet X begins to fall to the surface with a negligibly small initial speed. The mass of planet X is 3.1 × 10²¹ kg and its radius is 1.2 × 10⁶ m. The planet has no atmosphere. Calculate the speed at which the meteorite will hit the surface.

At the instant of impact the meteorite which is made of ice has a temperature of 0 °C. Assume that all the kinetic energy at impact gets transferred into internal energy in the meteorite. Calculate the percentage of the meteorite’s mass that melts. The specific latent heat of fusion of ice is 3.3 × 10⁵ J kg^–1.

the field lines/arrows are further apart at greater distances from the surface

circle centred on Planet X
three units from Planet X centre

loss in gravitational potential = \(\frac{{6.67 \times {{10}^{ - 11}} \times 3.1 \times {{10}^{21}}}}{{1.2 \times {{10}^6}}}\)

«= 1.72 × 10⁵ JKg⁻¹»

equate to \(\frac{1}{2}\)v²

v = 590 «m s⁻¹»

Allow ECF from MP1.

available energy to melt one kg 1.72 × 10⁵ «J»

fraction that melts is \(\frac{{1.72 \times {{10}^5}}}{{3.3 \times {{10}^5}}}\) = 0.52 OR 52%

Allow ECF from MP1.

Allow 53% from use of 590 ms^-1.