2024 OCR A-Level Physics B (Advancing Physics) - H557 Practice Paper with Answers

(a) The equation is of the form \(a = -\omega^2 x\). Since acceleration \(a\) is directly proportional to displacement \(x\) and is in the opposite direction (as indicated by the negative sign), this represents simple harmonic motion.
By comparison, \(\omega^2 = 2\pi^2\), which gives \(\omega = \sqrt{2}\pi \approx 4.44\text{ rad s}^{-1}\).
Since \(\omega = 2\pi f\), the frequency is \(f = \frac{\omega}{2\pi} = \frac{\sqrt{2}\pi}{2\pi} = \frac{\sqrt{2}}{2} \approx 0.707\text{ Hz}\) (or \(0.71\text{ Hz}\) to 2 s.f.).

(b) Using the formula for the angular frequency of a mass-spring system: \(\omega = \sqrt{\frac{k}{m}}\).
Substituting the given values:
\(\omega = \sqrt{\frac{40\text{ N m}^{-1}}{0.25\text{ kg}}} = \sqrt{160} \approx 12.65\text{ rad s}^{-1}\).
This rounds to \(13\text{ rad s}^{-1}\) (to 2 s.f.), as required.

(c)(i) The initial acceleration is given by:
\(a_0 = -\frac{k}{m} x_0 = -\frac{40}{0.25} \times 0.050 = -160 \times 0.050 = -8.0\text{ m s}^{-2}\).

(c)(ii) The next velocity is:
\(v_1 = v_0 + a_0 \Delta t = 0 + (-8.0\text{ m s}^{-2} \times 0.020\text{ s}) = -0.16\text{ m s}^{-1}\).

(c)(iii) The next displacement is:
\(x_1 = x_0 + v_1 \Delta t = 0.050\text{ m} + (-0.16\text{ m s}^{-1} \times 0.020\text{ s}) = 0.050 - 0.0032 = 0.0468\text{ m}\) (or \(0.047\text{ m}\) to 2 s.f.).

Part (a) [3 marks]:
- [1 mark] Explains that acceleration is directly proportional to displacement and directed towards the equilibrium position (or opposite to displacement).
- [1 mark] Identifies \(\omega^2 = 2\pi^2\) to find \(\omega = \sqrt{2}\pi \approx 4.44\text{ rad s}^{-1}\).
- [1 mark] Correctly calculates \(f = 0.71\text{ Hz}\) (accept \(0.707\text{ Hz}\)).

Part (b) [2 marks]:
- [1 mark] Recalls and uses \(\omega = \sqrt{\frac{k}{m}}\).
- [1 mark] Shows calculation leading to \(\approx 12.6\text{ rad s}^{-1}\) or \(12.7\text{ rad s}^{-1}\) and rounds to \(13\text{ rad s}^{-1}\).

Part (c) [5 marks]:
- [1 mark] (i) Calculates \(a_0 = -8.0\text{ m s}^{-2}\).
- [1 mark] (ii) Recognizes formula \(v_1 = v_0 + a_0 \Delta t\).
- [1 mark] Calculates \(v_1 = -0.16\text{ m s}^{-1}\).
- [1 mark] (iii) Recognizes formula \(x_1 = x_0 + v_1 \Delta t\).
- [1 mark] Calculates \(x_1 = 0.0468\text{ m}\) (or \(0.047\text{ m}\)).

(a) Using the potential divider equation:
\(V_{out} = V_{in} \times \frac{R_{th}}{R + R_{th}}\)
\(6.0 = 9.0 \times \frac{2.4\text{ k}\Omega}{R + 2.4\text{ k}\Omega}\)
Divide both sides by 9.0:
\(\frac{2}{3} = \frac{2.4}{R + 2.4}\)
Cross-multiply:
\(2(R + 2.4) = 3(2.4)\)
\(2R + 4.8 = 7.2\)
\(2R = 2.4 \implies R = 1.2\text{ k}\Omega\).

(b) At \(60^\circ\text{C}\), the thermistor resistance is \(R_{th} = 450\ \Omega = 0.45\text{ k}\Omega\).
The fixed resistor is \(R = 1.2\text{ k}\Omega = 1200\ \Omega\).
The new output voltage is:
\(V_{out} = V_{in} \times \frac{R_{th}}{R + R_{th}} = 9.0\text{ V} \times \frac{450\ \Omega}{1200\ \Omega + 450\ \Omega}\)
\(V_{out} = 9.0 \times \frac{450}{1650} = 9.0 \times 0.2727 = 2.45\text{ V}\) (or \(2.5\text{ V}\) to 2 s.f.).

(c) In a semi-conductor such as an NTC thermistor, there is a relatively small band gap between the valence and conduction bands. As temperature increases:
1. Charge carriers (electrons) gain thermal energy.
2. More electrons are excited across the energy gap into the conduction band, significantly increasing the charge carrier density \(n\).
3. Although lattice ion vibrations increase and cause more scattering, the exponential increase in the number density \(n\) of free charge carriers dominates, greatly increasing the electrical conductivity and thus reducing the resistance.

(d) Modification: Measure the output voltage \(V_{out}\) across the fixed resistor instead of the thermistor (or swap the physical positions of the fixed resistor and the thermistor in the potential divider).
Explanation: Since the thermistor's resistance decreases as temperature increases, the fixed resistor will take a larger share of the total \(9.0\text{ V}\) potential difference, meaning the voltage across the fixed resistor increases as temperature rises.

Part (a) [2 marks]:
- [1 mark] Recalls potential divider formula: \(V_{out} = V_{in} \times \frac{R_{th}}{R + R_{th}}\).
- [1 mark] Shows complete algebraic steps leading to \(R = 1.2\text{ k}\Omega\).

Part (b) [3 marks]:
- [1 mark] Expresses both resistances in consistent units (e.g. \(450\ \Omega\) and \(1200\ \Omega\)).
- [1 mark] Correctly substitutes values into the potential divider equation: \(9.0 \times \frac{450}{1650}\).
- [1 mark] Calculates \(2.5\text{ V}\) (or \(2.45\text{ V}\)).

Part (c) [3 marks]:
- [1 mark] Mentions that semi-conductors have a band gap and that thermal energy excites electrons from the valence to the conduction band.
- [1 mark] Explains that the number density of charge carriers \(n\) increases (exponentially) with temperature.
- [1 mark] Clarifies that the increase in \(n\) far outweighs the increase in lattice scattering, resulting in a net decrease in resistance.

Part (d) [2 marks]:
- [1 mark] State modification: measure output voltage across the fixed resistor (or swap positions of the two components).
- [1 mark] Explanation: as temperature rises, thermistor resistance decreases, so a larger fraction of the supply voltage is dropped across the fixed resistor.

(a) Using Newton's second law for horizontal motion:
\(F_{net} = m a = 0.80\text{ kg} \times 1.5\text{ m s}^{-2} = 1.2\text{ N}\).
The net force is the difference between the forward thrust and the drag force:
\(F_{net} = F_{thrust} - F_{drag}\)
\(1.2\text{ N} = 4.5\text{ N} - F_{drag} \implies F_{drag} = 4.5\text{ N} - 1.2\text{ N} = 3.3\text{ N}\).

(b)(i)
- According to Newton's Third Law, when the drone's rotors exert a downward force on the air, the air exerts an equal and opposite upward force (thrust) on the rotors.
- According to Newton's First Law (or Second Law with \(a = 0\)), for the drone to remain stationary (hover), the net vertical force must be zero.
- Therefore, the upward force exerted by the air must exactly balance the downward force of gravity (weight) acting on the drone.

(b)(ii) The vertical thrust must equal the weight of the drone:
\(W = m g = 0.80\text{ kg} \times 9.81\text{ m s}^{-2} = 7.85\text{ N}\) (or \(7.8\text{ N}\) to 2 s.f.).

(c)(i) Using the equation of motion \(v^2 = u^2 + 2as\) with \(u = 0\), \(a = g = 9.81\text{ m s}^{-2}\), and \(s = 35\text{ m}\):
\(v^2 = 0 + 2 \times 9.81\text{ m s}^{-2} \times 35\text{ m} = 686.7\text{ m}^2\text{ s}^{-2}\).
\(v = \sqrt{686.7} = 26.2\text{ m s}^{-1}\) (or \(26\text{ m s}^{-1}\) to 2 s.f.).

(c)(ii) Using work-energy principles:
Initial gravitational potential energy lost: \(\Delta E_p = m g h = 0.80\text{ kg} \times 9.81\text{ m s}^{-2} \times 35\text{ m} = 274.7\text{ J}\).
Final kinetic energy gained: \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.80\text{ kg} \times (18\text{ m s}^{-1})^2 = 0.40 \times 324 = 129.6\text{ J}\).
Work done against the vertical resistive force (drag): \(W_{drag} = \Delta E_p - E_k = 274.7\text{ J} - 129.6\text{ J} = 145.1\text{ J}\).
Since \(W_{drag} = F_{drag} \times s\):
\(F_{drag} = \frac{145.1\text{ J}}{35\text{ m}} = 4.145\text{ N}\) (or \(4.1\text{ N}\) to 2 s.f.).

Alternatively, using kinematic acceleration:
\(v^2 = u^2 + 2as \implies 18^2 = 2 \times a \times 35 \implies a = 4.63\text{ m s}^{-2}\).
Using Newton's Second Law: \(m g - F_{drag} = m a\)
\(F_{drag} = m(g - a) = 0.80 \times (9.81 - 4.63) = 0.80 \times 5.18 = 4.14\text{ N}\) (or \(4.1\text{ N}\) to 2 s.f.).

Part (a) [2 marks]:
- [1 mark] Calculates the net force: \(F_{net} = 0.80 \times 1.5 = 1.2\text{ N}\).
- [1 mark] Correctly finds drag force: \(4.5 - 1.2 = 3.3\text{ N}\).

Part (b) [4 marks]:
- [1 mark] (i) Mentions Newton's 3rd Law: action-reaction pair of forces (rotors push down on air, air pushes up on rotors).
- [1 mark] Mentions Newton's 1st/2nd Law: net force is zero for static equilibrium (hovering).
- [1 mark] Explains that the upward lift force balances the downward force of gravity (weight).
- [1 mark] (ii) Calculates total vertical thrust: \(0.80 \times 9.81 = 7.8\text{ N}\) or \(7.85\text{ N}\).

Part (c) [4 marks]:
- [1 mark] (i) Uses \(v^2 = 2gs\) to find \(v = \sqrt{2 \times 9.81 \times 35}\).
- [1 mark] Calculates speed: \(26\text{ m s}^{-1}\) (accept \(26.2\text{ m s}^{-1}\)).
- [1 mark] (ii) Uses work-energy theorem (or calculates actual downward acceleration \(a = 4.63\text{ m s}^{-2}\)).
- [1 mark] Calculates the average resistive force: \(4.1\text{ N}\) (accept range \(4.1 - 4.15\text{ N}\)).

(a) Use \(E = \frac{hc}{\lambda}\):
\(E = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{380 \times 10^{-9}\text{ m}} = 5.234 \times 10^{-19}\text{ J}\).
To convert to eV: \(\frac{5.234 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} = 3.27\text{ eV}\).

(b) Using the photoelectric equation:
\(E_{k,\text{max}} = E - \Phi = 3.27\text{ eV} - 2.14\text{ eV} = 1.13\text{ eV}\).
In Joules: \(1.13 \times 1.60 \times 10^{-19}\text{ J} = 1.81 \times 10^{-19}\text{ J}\).
Using \(E_{k,\text{max}} = \frac{1}{2} m_e v_{\text{max}}^2\):
\(v_{\text{max}} = \sqrt{\frac{2 \times 1.81 \times 10^{-19}\text{ J}}{9.11 \times 10^{-31}\text{ kg}}} = 6.30 \times 10^5\text{ m s}^{-1}\).

(c) In the phasor model, every possible path from the source to a point on the screen is associated with a phasor (a rotating arrow). The phase angle of the phasor depends on the path length / travel time. At the screen, the phasors for all paths (e.g., through Slit 1 and Slit 2) are added tip-to-tail. The resultant phasor's length squared is proportional to the probability of the photon arriving at that point. This directly maps to classical superposition where wave amplitudes (phasor lengths) and phase differences determine the resultant wave amplitude, whose intensity (amplitude squared) corresponds to light brightness.

(d) Initially, individual localized impacts (dots) will appear randomly on the screen, showing the particle-like nature of photons. Over a long time, the cumulative distribution of these dots will form a clear double-slit interference pattern (alternating bright and dark bands), demonstrating that the probability of arrival is governed by the wave-like phasor model.

(a)
- M1: Recall and substitute into \(E = hf\) or \(E = \frac{hc}{\lambda}\)
- A1: Correct photon energy in Joules: \(5.23 \times 10^{-19}\text{ J}\)
- A1: Correct conversion to eV: \(3.27\text{ eV}\) (accept \(3.26\text{ eV}\) to \(3.28\text{ eV}\))

(b)
- M1: Use of \(E_{k,\text{max}} = hf - \Phi\) to get \(1.13\text{ eV}\)
- A1: Correct conversion of \(E_{k,\text{max}}\) to Joules: \(1.81 \times 10^{-19}\text{ J}\)
- M1: Equating kinetic energy to \(\frac{1}{2} m v^2\) and rearranging for \(v\)
- A1: Correct maximum speed: \(6.30 \times 10^5\text{ m s}^{-1}\) (accept \(6.3 \times 10^5\text{ m s}^{-1}\))

(c) Level of Response:
- **Level 3 (5-6 marks)**: Detailed and coherent explanation covering: (1) phasor definition/rotation linked to path length, (2) tip-to-tail phasor addition, (3) probability proportional to (resultant amplitude)\(^2\), and (4) link to wave superposition (constructive/destructive interference matching phase alignment).
- **Level 2 (3-4 marks)**: Clear explanation but lacks either the quantitative link to probability (amplitude squared) or the explicit connection to classical wave superposition.
- **Level 1 (1-2 marks)**: Basic mention of phasors or path difference without explaining how they are combined or how they relate to probability.

(d)
- B1: Mentions that individual, distinct dots/clicks are recorded one by one (particle-like behaviour).
- B1: Explains that the accumulated dots form the classical double-slit interference pattern (wave-like probability distribution).

(a) Using the approximation \(p = \frac{E}{c}\):
\(p = \frac{350 \times 10^6 \times 1.60 \times 10^{-19}\text{ J}}{3.00 \times 10^8\text{ m s}^{-1}} = 1.867 \times 10^{-19}\text{ kg m s}^{-1}\).
Using de Broglie's relation \(\lambda = \frac{h}{p}\):
\(\lambda = \frac{6.63 \times 10^{-34}\text{ J s}}{1.867 \times 10^{-19}\text{ kg m s}^{-1}} = 3.55 \times 10^{-15}\text{ m} \approx 3.5 \times 10^{-15}\text{ m}\).

(b) Electrons are lepton point-particles (no internal structure) and do not experience the strong nuclear force; they interact purely through the electromagnetic force. This allows clean mathematical modeling of the charge distribution. In contrast, alpha-particles are hadrons (consisting of 4 nucleons with their own size) and experience the strong force at close proximity, which complicates the interpretation of the scattering data.

(c) Comparison:
- **Interactions**: Electrons scatter electromagnetically (diffraction due to wave nature). Alpha particles scatter via electrostatic repulsion (Coulomb scattering as a charged particle), but at high energies can experience the strong nuclear force.
- **Extracting dimensions**: Electron diffraction uses the angle of the first minimum (\(\sin \theta \approx 1.22 \lambda / d\)) to find the nuclear diameter directly. Rutherford scattering uses the distance of closest approach (equating initial KE to electrostatic PE) to find an upper limit for the nuclear radius.
- **Limitations**: Electron diffraction requires extremely high energies (relativistic) to match the nuclear scale wavelength. Alpha backscattering only provides an *upper limit* of the radius (as the alpha particle may not touch the nucleus) and is distorted if the alpha particle enters the range of the strong force.

(d) Rearranging \(\sin \theta \approx 1.22 \frac{\lambda}{d}\) for \(d\):
\(d = \frac{1.22 \lambda}{\sin \theta} = \frac{1.22 \times 3.55 \times 10^{-15}\text{ m}}{\sin(38.5^\circ)} = \frac{4.331 \times 10^{-15}\text{ m}}{0.6225} = 6.96 \times 10^{-15}\text{ m} \approx 7.0 \times 10^{-15}\text{ m}\) (or \(7.0\text{ fm}\)).

(a)
- M1: Use of \(p = E/c\) with correct energy conversion: \(p = 1.87 \times 10^{-19}\text{ kg m s}^{-1}\)
- M1: Use of \(\lambda = h/p\)
- A1: Correct calculation showing \(\lambda = 3.55 \times 10^{-15}\text{ m}\) (which rounds to \(3.5 \times 10^{-15}\text{ m}\))

(b)
- B1: Electrons are fundamental/point-like (whereas alpha particles have structure/finite size).
- B1: Electrons only interact via the electromagnetic force (no strong nuclear force interaction).
- B1: The strong interaction complicates the scattering analysis of alpha particles when they get very close.
- B1: Electron scattering directly probes the nuclear charge distribution.

(c) Level of Response:
- **Level 3 (5-6 marks)**: Detailed comparison addressing all three prompts (forces, extraction of size/charge, and limitations) with accurate physics for both methods.
- **Level 2 (3-4 marks)**: Explains both methods but misses one of the prompts or contains minor physical inaccuracies (e.g. confusing the forces or the equations used).
- **Level 1 (1-2 marks)**: Basic descriptive points about one or both methods without structured comparison or mathematical links.

(d)
- M1: Rearrangement of \(d = \frac{1.22 \lambda}{\sin \theta}\) and substitution of values
- A1: Correct evaluation to \(7.0 \times 10^{-15}\text{ m}\) (accept range \(6.9 \times 10^{-15}\text{ m}\) to \(7.0 \times 10^{-15}\text{ m}\))

(a) The equation calculates the change in nuclei \(\Delta N\) in a small interval \(\Delta t\), which is subtracted from the current \(N\) to find the new \(N\) for the next step. If \(\Delta t\) is too large, \(N\) changes significantly during the step, meaning the approximation that the rate of decay (\(\frac{\Delta N}{\Delta t}\)) remains constant during the step becomes highly inaccurate.

(b)
(i) At \(t = 0\), \(N_0 = 1000\).
\(\Delta N_1 = -\lambda N_0 \Delta t = -0.015 \times 1000 \times 2.0 = -30\).
At \(t = 2.0\text{ s}\), \(N(2) = 1000 - 30 = 970\).
(ii) At \(t = 2.0\text{ s}\), \(N(2) = 970\).
\(\Delta N_2 = -\lambda N(2) \Delta t = -0.015 \times 970 \times 2.0 = -29.1\).
At \(t = 4.0\text{ s}\), \(N(4) = 970 - 29.1 = 940.9\) (or \(941\)).

(c) Analytical value at \(t = 4.0\text{ s}\):
\(N(4) = 1000 e^{-0.015 \times 4.0} = 1000 e^{-0.06} = 941.764\).
Percentage error:
\(\text{Error} = \frac{|940.9 - 941.764|}{941.764} \times 100\% = \frac{0.864}{941.764} \times 100\% = 0.0917\% \approx 0.092\%\) (accept \(0.09\%\)).
Explanation of systematic error: The model uses the number of nuclei at the *start* of each interval to compute the decay rate for the whole interval. Because \(N\) is actually decreasing continuously, the rate of decay used is too high, leading to more simulated decays than occur in reality, thus underestimating the remaining nuclei \(N\).

(d) Experimental Details:
- **Safety**: Keep the source (protactinium generator) in its sealed container; handle only with tongs; wear safety glasses and lab coat; keep at a safe distance when not taking measurements.
- **Background Correction**: Measure the background count rate for a few minutes before bringing the source near. Subtract this background rate from all subsequent experimental readings.
- **Data Collection**: Place the GM tube close to the protactinium generator. Record the counts in short intervals (e.g., every \(10\text{ s}\)) for about \(5\text{ minutes}\).
- **Analysis**: Convert count data to corrected activity (or corrected count rate). Plot a graph of \(\ln(\text{Activity})\) against time. The gradient will equal \(-\lambda\), from which the half-life can be calculated as \(T_{1/2} = \frac{\ln 2}{\lambda}\) and compared to the computational model.

(a)
- B1: Explains that \(\Delta N\) is calculated and subtracted iteratively to get the next value of \(N\).
- B1: Identifies that the decay rate decreases continuously as \(N\) decreases.
- B1: Explains that \(\Delta t \ll T_{1/2}\) ensures \(N\) (and hence decay rate) does not change significantly during a single step.

(b)
- M1: Correctly calculates \(\Delta N_1 = -30\) and \(N(2) = 970\).
- M1: Uses the new value \(970\) to calculate \(\Delta N_2 = -29.1\).
- A1: Correctly calculates \(N(4) = 940.9\) (or \(941\)).

(c)
- M1: Correctly calculates the analytical value: \(941.8\) (or \(941.76\)).
- A1: Correctly calculates the percentage error: \(0.092\%\) (or \(0.09\%\)).
- B1: Explains that the starting value of \(N\) is used to estimate decay rate over the interval.
- B1: Identifies that because actual \(N\) decreases continuously, this start-of-step rate is an overestimate, causing an underestimate of the remaining \(N\).

(d)
- B1: Mentions safety precaution (e.g., using tongs, keeping distance, sealed source).
- B1: Mentions measuring and subtracting background radiation.
- B1: Mentions appropriate data gathering method (recording counts in small, regular intervals over several half-lives).
- B1: Identifies plotting \(\ln(\text{Activity})\) vs \(t\) yields a straight line with gradient \(-\lambda\).
- B1: Mentions checking for agreement with the half-life derived from the numerical model.

(a) For a circular orbit, the gravitational force provides the centripetal force:

\(\frac{GMm}{r^2} = \frac{mv^2}{r}\)

Simplify this to find an expression for orbital speed \(v\):

\(v^2 = \frac{GM}{r}\)

Since the orbital speed is also the distance around the circle divided by the period:

\(v = \frac{2\pi r}{T}\)

Substitute this back into the equation:

\(\left(\frac{2\pi r}{T}\right)^2 = \frac{GM}{r}\)

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

Rearranging for \(T^2\):

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

Taking the square root gives the desired expression:

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

(b) Rearrange the period formula to solve for the radius \(r\):

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

Substitute the given values:

\(r^3 = \frac{(6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}) \times (3.20 \times 10^{13}\text{ kg}) \times (4.32 \times 10^4\text{ s})^2}{4\pi^2}\)

\(r^3 = \frac{2134.4 \times 1.866 \times 10^9}{39.478}\)

\(r^3 \approx 1.009 \times 10^{11}\text{ m}^3\)

Taking the cube root of both sides:

\(r = (1.009 \times 10^{11})^{1/3} \approx 466\text{ m}\).

(a) [Total: 2.25 marks]
- 1 mark: Equating gravitational force to centripetal force (\(GMm/r^2 = mv^2/r\)).
- 1 mark: Correct substitution of \(v = 2\pi r / T\) into the simplified force relation.
- 0.25 marks: Clear algebraic steps leading to the final given equation.

(b) [Total: 4.0 marks]
- 1 mark: Correct rearrangement of the formula for \(r\) or \(r^3\).
- 1 mark: Correct substitution of all numerical constants and values, including the time in seconds.
- 1 mark: Finding intermediate value for \(r^3\) (approx. \(1.0 \times 10^{11}\)).
- 1 mark: Correct final answer to 3 significant figures: \(466\text{ m}\) (accept range \(460\text{--}470\text{ m}\)).

(a) Gravitational potential energy is defined to be zero at infinity (where the gravitational field strength is zero). Because the gravitational force is attractive, work must be done on the telescope to move it from any point in the gravitational field to infinity. Since energy is added to reach zero, the potential energy at any point within the field must be negative.

(b) The formula for gravitational potential energy is \(E_p = -\frac{GMm}{r}\).

The change in gravitational potential energy \(\Delta E_p\) is:

\(\Delta E_p = E_{p,\text{final}} - E_{p,\text{initial}}\)

\(\Delta E_p = -\frac{GM_E m}{r_f} - \left(-\frac{GM_E m}{r_i}\right) = GM_E m \left(\frac{1}{r_i} - \frac{1}{r_f}\right)\)

Substitute the given values into the equation:

\(GM_E m = (6.67 \times 10^{-11}) \times (5.97 \times 10^{24}) \times 1200 = 4.778 \times 10^{17}\text{ J m}\)

\(\Delta E_p = 4.778 \times 10^{17} \times \left(\frac{1}{6.37 \times 10^6} - \frac{1}{4.22 \times 10^7}\right)\)

Calculate the reciprocals:

\(\frac{1}{6.37 \times 10^6} \approx 1.570 \times 10^{-7}\text{ m}^{-1}\)

\(\frac{1}{4.22 \times 10^7} \approx 2.370 \times 10^{-8}\text{ m}^{-1}\)

\(\Delta E_p = 4.778 \times 10^{17} \times (1.570 \times 10^{-7} - 0.237 \times 10^{-7})\)

\(\Delta E_p = 4.778 \times 10^{17} \times (1.333 \times 10^{-7}) \approx 6.37 \times 10^{10}\text{ J}\).

(a) [Total: 2.25 marks]
- 1 mark: Stating that gravitational potential energy is defined as zero at infinity.
- 1 mark: Stating that the gravitational force is attractive, meaning work must be done on the mass to move it away to infinity.
- 0.25 marks: Linking this to why the energy inside the field is less than zero (negative).

(b) [Total: 4.0 marks]
- 1 mark: Selecting and stating the correct potential energy formula \(E_p = -GMm/r\).
- 1 mark: Calculating the constant factor \(GM_E m\) correctly (\(4.78 \times 10^{17}\)) or writing the complete algebraic difference equation.
- 1 mark: Correct calculation of the initial potential energy (\(-7.50 \times 10^{10}\text{ J}\)) or the final potential energy (\(-1.13 \times 10^{10}\text{ J}\)).
- 1 mark: Stating the correct final positive change in energy of \(6.37 \times 10^{10}\text{ J}\) (accept \(6.3\text{--}6.4 \times 10^{10}\text{ J}\)).

(a) The Doppler formula for electromagnetic radiation at speeds much lower than the speed of light is:

\(\frac{\Delta\lambda}{\lambda_0} = \frac{v}{c}\)

Rearranging to find the orbital speed \(v\):

\(v = c \cdot \frac{\Delta\lambda}{\lambda_0}\)

Substitute the given values:

\(v = (3.00 \times 10^8\text{ m s}^{-1}) \times \frac{0.145\text{ nm}}{656.3\text{ nm}}\)

\(v = 3.00 \times 10^8 \times 2.2093 \times 10^{-4} \approx 6.63 \times 10^4\text{ m s}^{-1}\).

(b) For a circular orbit of radius \(R\) and orbital period \(T\), the speed is given by:

\(v = \frac{2\pi R}{T}\)

Rearranging to solve for the orbit radius \(R\):

\(R = \frac{v T}{2\pi}\)

Substitute the speed from part (a) and the period in seconds:

\(R = \frac{(6.628 \times 10^4\text{ m s}^{-1}) \times (7.34 \times 10^5\text{ s})}{2\pi}\)

\(R = \frac{4.865 \times 10^{10}}{6.283} \approx 7.74 \times 10^9\text{ m}\).

(a) [Total: 3.0 marks]
- 1 mark: Stating the Doppler shift equation \(\Delta\lambda / \lambda_0 = v/c\).
- 1 mark: Correctly substituting values with consistent units (no conversion needed as units of nm cancel out).
- 1 mark: Calculating the correct speed of \(6.63 \times 10^4\text{ m s}^{-1}\) (accept \(6.6 \times 10^4\text{ m s}^{-1}\)).

(b) [Total: 3.25 marks]
- 1 mark: Stating the relation between circular speed, period, and radius (\(v = 2\pi R / T\)).
- 1 mark: Rearranging the equation to make \(R\) the subject.
- 1.25 marks: Calculating the final radius of \(7.74 \times 10^9\text{ m}\) (accept \(7.7\text{--}7.8 \times 10^9\text{ m}\)).

This is a Level of Response question where candidates are expected to integrate gravitational physics with satellite applications.

**Physics of Gravitational Fields & Orbits:**
1. For any circular orbit, the gravitational force acts as the centripetal force:

\(\frac{GMm}{r^2} = \frac{mv^2}{r} \implies v = \sqrt{\frac{GM}{r}}\)

This shows that orbital speed \(v\) is inversely proportional to the square root of the orbital radius \(r\).

2. The orbital period is related by Kepler's Third Law: \(T^2 \propto r^3\). Therefore, larger orbits have much longer orbital periods.

**Low Earth Orbit (LEO) Satellites:**
- **Characteristics:** Located at low altitudes (typically \(200\text{ km}\) to \(2000\text{ km}\) above the surface). They have high orbital speeds (around \(7.8\text{ km s}^{-1}\)) and short orbital periods (around 90 to 120 minutes).
- **Scientific Advantages:** Being close to the Earth's surface allows high-resolution imaging and detailed sensing of environmental parameters (such as soil moisture, ice sheet thickness, and vegetation index). If placed in polar orbits, they eventually sweep over the entire globe, providing global coverage over several days.

**Geostationary Earth Orbit (GEO) Satellites:**
- **Characteristics:** Positioned at a very specific altitude (about \(36,000\text{ km}\) above the equator, orbital radius \(r \approx 4.2 \times 10^7\text{ m}\)) where the orbital period is exactly equal to the Earth's rotational period (24 hours). This means their orbital speed is lower (around \(3.1\text{ km s}^{-1}\)).
- **Scientific Advantages:** They remain fixed relative to a point on the Earth's surface, allowing for continuous, uninterrupted monitoring of weather systems, atmospheric patterns, and natural disasters (like wildfires) over a wide geographic region.

For this Level of Response question, marks are awarded across three levels based on the quality of explanation and inclusion of physical principles:

- **Level 3 (5.0 - 6.25 marks):** Coherent, detailed comparison of both orbits. Explains the physics showing \(v \propto 1/\sqrt{r}\) and \(T^2 \propto r^3\). Correctly identifies approximate altitudes/radii for both LEO and GEO. Thoroughly links these physical properties to their scientific observational advantages (e.g., LEO for high spatial resolution and global coverage over time; GEO for continuous temporal coverage of a specific hemisphere).

- **Level 2 (2.5 - 4.9 marks):** Logical comparison of LEO and GEO orbits. Mentions the relationship between altitude and speed qualitatively (higher orbit means lower speed). Identifies at least one scientific advantage of each orbit type with some scientific detail.

- **Level 1 (0.5 - 2.4 marks):** Basic description of LEO and GEO orbits. May include disjointed facts about their altitudes or speeds but lacks mathematical justification or logical link to their scientific roles.