*Question*

a.

(i) Define* gravitational field strength.*

(ii) State the SI unit for gravitational field strength.[2]

A planet orbits the Sun in a circular orbit with orbital period *T* and orbital radius *R*. The mass of the Sun is *M*.

(i) Show that \(T = \sqrt {\frac{{4{\pi ^2}{R^3}}}{{GM}}} \).

(ii) The Earth’s orbit around the Sun is almost circular with radius 1.5×10^{11} m. Estimate the mass of the Sun.[4]

**Answer/Explanation**

### Markscheme

a.

(i) «gravitational» force per unit mass on a «small **or** test» mass

(ii) N kg^{–1 }

*Award mark if N kg ^{-1} is seen, treating any further work as neutral.Do not accept bald m s^{–2}*

i

clear evidence that *v* in \({v^2} = \frac{{4{\pi ^2}{R^2}}}{{{T^2}}}\) is equated to orbital speed \(\sqrt {\frac{{GM}}{R}} \)**OR**

clear evidence that centripetal force is equated to gravitational force**OR**

clear evidence that *a* in \(a = \frac{{{v^2}}}{R}\) etc is equated to *g* in \(g = \frac{{GM}}{{{R^2}}}\) with consistent use of symbols*Minimum is a statement that \(\sqrt {\frac{{GM}}{R}} \) is the orbital speed which is then used in \(v = \frac{{2\pi R}}{T}\)**Minimum is F _{c} = F_{g} ignore any signs.*

*Minimum is g = a.*

substitutes and re-arranges to obtain result*Allow any legitimate method not identified here.**Do not allow spurious methods involving equations of shm etc*

\( \ll T = \sqrt {\frac{{4{\pi ^2}R}}{{\left( {\frac{{GM}}{{{R^2}}}} \right)}}} = \sqrt {\frac{{4{\pi ^2}{R^3}}}{{GM}}} \gg \)

ii

«*T *= 365 × 24 × 60 × 60 = 3.15 × 10^{7 }s»

\(M = \, \ll \frac{{4{\pi ^2}{R^3}}}{{G{T^2}}} = \gg \,\, = \frac{{4 \times {{3.14}^2} \times {{\left( {1.5 \times {{10}^{11}}} \right)}^3}}}{{6.67 \times {{10}^{ – 11}} \times {{\left( {3.15 \times {{10}^7}} \right)}^2}}}\)

2×10^{30}«kg»

*Allow use of 3.16 x 10 ^{7} s for year length (quoted elsewhere in paper).*

*Condone error in power of ten in MP1.*

*Award*

**[1 max]**if incorrect time used (24 h is sometimes seen, leading to 2.66 x 10^{35}kg).*Units are not required, but if not given assume kg and mark POT accordingly if power wrong.*

*Award*

**[2]**for a bald correct answer.*No sf penalty here.*

*Question*

The two arrows in the diagram show the gravitational field strength vectors at the position of a planet due to each of two stars of equal mass *M*.

Each star has mass *M*=2.0×10^{30}kg. The planet is at a distance of 6.0×10^{11}m from each star.

a.

Show that the gravitational field strength at the position of the planet due to **one** of the stars is *g*=3.7×10^{–4}Nkg^{–1}.[1]

Calculate the magnitude of the resultant gravitational field strength at the position of the planet.[2]

**Answer/Explanation**

### Markscheme

a.

\(g = \frac{{GM}}{{{r^2}}} = \frac{{6.67 \times {{10}^{ – 11}} \times 2.0 \times {{10}^{30}}}}{{{{\left( {6.0 \times {{10}^{11}}} \right)}^2}}}\)

**OR**

3.71×10^{-4}Nkg^{−1}

«*g*_{net} = 2cos34» 2*g* **OR***g*cos34 **OR***g*sin56 * OR* vector addition diagram shown

«*g*_{net }=«2×3.7×10^{−4} ×cos34^{o} =**»** 6.1×10^{−4} Nkg^{−1}

*Question*

This question is in **two** parts. **Part 1** is about kinematics and gravitation. **Part 2** is about radioactivity.

**Part 1** Kinematics and gravitation

A ball is released near the surface of the Moon at time *t*=0. The point of release is on a straight line between the centre of Earth and the centre of the Moon. The graph below shows the variation with time *t* of the displacement s of the ball from the point of release.

**Part 2** Radioactivity

Two isotopes of calcium are calcium-40 \(\left( {\frac{{40}}{{20}}{\rm{Ca}}} \right)\) and calcium-47 \(\left( {\frac{{47}}{{20}}{\rm{Ca}}} \right)\). Calcium-40 is stable and calcium-47 is radioactive with a half-life of 4.5 days.

a.

State the significance of the negative values of *s*.[1]

Use the graph to

(i) estimate the velocity of the ball at *t* \( = \) 0.80 s.

(ii) calculate a value for the acceleration of free fall close to the surface of the Moon.[6]

The following data are available.

Mass of the ball = 0.20 kg

Mean radius of the Moon = 1.74 \( \times \) 10^{6} m

Mean orbital radius of the Moon about the centre of Earth = 3.84 \( \times \) 10^{8} m

Mass of Earth = 5.97 \( \times \) 10^{24} kg

Show that Earth has no significant effect on the acceleration of the ball.[4]

Calculate the speed of an identical ball when it falls 3.0 m from rest close to the surface of Earth. Ignore air resistance.[1]

Sketch, on the graph, the variation with time* t* of the displacement* s* from the point of release of the ball when the ball is dropped close to the surface of Earth. (For this sketch take the direction towards the Earth as being negative.)[3]

Explain, in terms of the number of nucleons and the forces between them, why calcium-40 is stable and calcium-47 is radioactive.[3]

Calculate the percentage of a sample of calcium-47 that decays in 27 days.[3]

The nuclear equation for the decay of calcium-47 into scandium-47 \(\left( {{}_{21}^{47}{\rm{Sc}}} \right)\) is given by

\[{}_{20}^{47}{\rm{Ca}} \to {}_{21}^{47}{\rm{Sc + }}{}_{ – 1}^0{\rm{e + X}}\]

(i) Identify X.

(ii) The following data are available.

Mass of calcium-47 nucleus = 46.95455 u

Mass of scandium-47 nucleus = 46.95241 u

Using the data, determine the maximum kinetic energy, in MeV, of the products in the decay of calcium-47.

(iii) State why the kinetic energy will be less than your value in (h)(ii).[4]

**Answer/Explanation**

### Markscheme

a.

upwards (or away from the Moon) is taken as positive / downwards (or towards the Moon) is taken as negative / towards the Earth is positive;

(i) tangent drawn to curve at 0.80s;

correct calculation of gradient of tangent drawn;

−1.3 ±0.1m s–1 or 1.3 ±0.1m s–1 downwards;

or

correct coordinates used from the graph; substitution into a correct equation;

−1.3 ±0.1m s–1 or 1.3 ±0.1m s–1 downwards;

(ii) any correct method used;

correct reading from graph;

1.6 to 1.7 m s–2;

values for masses, distance and correct G substituted into Newton’s law;

see subtraction (ie r value = 3.84 × 108 − 1.74 × 106 =3.82 × 108 m);

F=5.4 to 5.5 × 10–4 N / a=2.7 × 10–3 m s–2;

comment that it’s insignificant compared with (0.2 × 1.63 =) 0.32 to 0.33 N / 1.63 m s–2;

7.7 m s\(^{ – 1}\);

curve permanently below Moon curve;

smooth parabola; (judge by eye)

line passing through s = −3.00 m, t = 0.78 s or s = −3.50 m, t = 0.84 s (±1mm);

Ca-40 has 20 protons and 20 neutrons, Ca-47 has 20 protons and 27 neutrons / Ca-47 has 7 additional neutrons;

mention of strong/nuclear and coulomb/electrostatic/electromagnetic forces;

excess neutrons/too high a neutron-to-proton ration leads to the coulomb/electrostatic’ electromagnetic force being greater than the strong/nuclear force (so the nucleus is unstable);

Award [1 max] for an answer stating that Ca-47 has more neutrons so is bigger and less stable.

six half-lives occurred;

\(\left( {{{\left( {\frac{1}{2}} \right)}^6} = } \right)1.6\% \) remaining;

98.4 / 98% decayed;

(i)(electron) anti-neutrino / \(\overline v \) ;

(ii) 46.95455 u − (46.95241 u + 0.00055 u) = 0.00159 u;

1.48 MeV;

(iii) does not account for energy of (anti) neutrino/gamma ray photons;