IB Physics HL D.1 Gravitational fields Questions

SLSL HLHL

Practise D.1 by using gravitational-field and potential models to analyse planetary orbits, satellite energy changes, equipotential diagrams and escape conditions.

172 questions in this syllabus scope, with question parts counted together.

Exam points & syllabus

Exam points

Derive T² = 4π²r³/(GM) by equating gravitational and centripetal force or acceleration, then substituting v = 2πr/T or ω = 2π/T.
For a common central mass, T²/r³ is constant; compare radius using period to the power 2/3 and period using radius to the power 3/2.
Find the central mass from M = 4π²r³/(GT²), converting days or years to seconds before calculation.
For different central bodies compare M ∝ r³/T²; the same T²/r³ constant applies only when the central mass is the same.
At fixed central mass and orbital radius, period and circular speed are independent of the orbiting body’s mass.
For elliptical orbits, locate the central body at a focus and associate maximum orbital speed with the nearest point; keep this qualitative.
Different free-orbit periods can make an inner satellite move ahead; a linked system may need propulsion to maintain relative position.
State universal gravitation as attraction between point masses proportional to their mass product and inversely proportional to centre separation squared.
Identify the source of gravitational attraction and link it to the centripetal acceleration; gravity supplies the required force.
In force ratios retain both masses; at equal density mass follows volume, while separation enters inversely squared.
A spherical body of uniform density can be represented externally by its mass at its centre; a satellite’s spatial size must also justify a point model.
Test a point-mass potential model with at least two constant VR products or inferred masses, using centre distance Rplanet + height.
Define gravitational field strength as force per unit mass on a small test mass at the point; free-fall acceleration has the same numerical value.
Use source mass and centre distance in g = GM/r²; compare surface fields with M/R² and use the body’s own radius.
For a zero-field point on the line between two masses, equate the opposing GM/r² fields using the two actual centre distances.
Use F = mg and Newton’s third law to compare forces on interacting bodies; equal opposite forces need not mean equal accelerations.
Estimate average gravity from |Δv|/Δt only after checking that the body is coasting under gravity with negligible drag.
To justify a negligible gravitational influence, calculate it at the actual separation and compare it explicitly with the dominant field or force.
Distinguish electric-to-gravitational field ratios from force ratios; cancel a common r² only after writing each field correctly.
Draw each individual gravitational vector straight towards its own source; equal masses at equal distances require equal arrow lengths.
Interpret decreasing radial field strength through increasing field-line separation farther from the surface.
Use signed two-body potential energy Ep = −GMm/r and zero at infinite separation; a less negative value is an increase.
Derive circular kinetic energy GMm/(2r), then add Ep to obtain total energy ET = −GMm/(2r).
For circular energy graphs, EK is positive, Ep negative and twice EK in magnitude, and ET is negative with ET = Ep/2.
Calculate finite ΔEp as mΔV or GMm(1/r₁ − 1/r₂), converting surface heights to centre radii.
For a transfer between circular orbits, required net work is ΔET = ΔEp/2; include the kinetic-energy decrease or increase.
For launch from rest on a non-rotating surface into orbit, include both the potential-energy increase and final circular kinetic energy.
To escape from an existing circular orbit, supply the magnitude of its negative total energy, retaining the kinetic energy already present.
Define gravitational potential as external work per unit mass in bringing a small test mass from infinity to the specified point.
Explain negative potential with zero at infinity and attractive gravity: positive work is needed to take a mass out to infinity.
Apply V = −GM/r with the correct centre distance and units J kg⁻¹ or m² s⁻²; compare potential as inverse distance, not inverse square.
For a single external point-mass field, derive V = −gr from the potential and field formulas; do not cancel finite differences in g = −ΔV/Δr.
Add gravitational potentials as signed scalars with each source’s own distance; zero resultant field does not imply zero potential.
A potential maximum between two bodies has zero gradient: use equal opposing fields there to infer position or mass.
For a curved potential graph use the local tangent for field at a point; use |ΔV|/Δr for a stated interval average.
The signed gravitational field is the negative potential gradient and points towards lower potential; acceleration is independent of the test mass.
Read potential values, rather than graph area or gradient, for potential energy or gravitational work between endpoints.
Separate work by gravity, Wg = −mΔV, from external work at unchanged kinetic energy, Wext = mΔV.
Draw isolated spherical equipotentials as concentric spheres or circles in section, with radii obtained from inverse potential ratios.
For equal potential increments, increasing outward equipotential gaps indicate a weakening field; field lines cross the surfaces normally.
Define escape speed as the minimum launch speed to reach infinity without further propulsion; leaving an atmosphere or orbit is insufficient.
Use energy conservation at escape threshold to derive vesc = √(2GM/r) = √(2gr) = √(−2V), with all local quantities referring to the same point.
Compare escape speeds through √(M/R), or R√ρ for uniform planets, keeping diameter and radius ratios consistent.
Square a launch-speed multiplier in kinetic energy; below escape speed conserve inverse-radius energy to the zero-speed turning point.
Justify escape failure by negative total energy or a launch speed below escape speed; reverse the same energy relation for infall from infinity.
Use complete supplied model relations with current escape energy; in a uniform-density sphere substitute M = (4/3)πρr³.
Use a supplied molecular-speed comparison to discuss likely atmospheric retention qualitatively, without claiming every molecule has the rms speed.
For circular speed v = √(GM/r), use central mass and centre radius; add height, convert units and retain precision for show-that calculations.
Compare orbital and escape speeds at their stated radii; at height R above a radius-R planet, orbital speed is half the surface escape speed.
For an orbit just above a uniform spherical body, combine density and the circular period relation to obtain T = √[3π/(Gρ)].
Small drag lowers total orbital energy and radius; the resulting smaller near-circular orbit has greater speed and kinetic energy.
With a supplied ρv² drag model, increasing density and speed can strengthen orbital decay and lead qualitatively to re-entry or impact.

Covered syllabus points

D.1.1—Kepler’s three laws of orbital motion D.1.2—Universal gravitation D.1.3—Point-mass approximation D.1.4—Gravitational field strength D.1.5—Gravitational field lines D.1.6 (HL)—Gravitational potential energy D.1.7 (HL)—Two-body potential energy D.1.8 (HL)—Gravitational potential D.1.9 (HL)—Potential gradient D.1.10 (HL)—Work in gravitational fields D.1.11 (HL)—Gravitational equipotentials D.1.12 (HL)—Equipotentials and field lines D.1.13 (HL)—Escape speed D.1.14 (HL)—Orbital speed D.1.15 (HL)—Atmospheric drag on orbits

[Maximum number: 6]

This question is in two parts. Part 1 is about gravitational force fields. Part 2 is about properties of a gas.

(a)

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}}{G M}

[ 3 ]

$G \frac{M m}{R^{2}}=\frac{m v^{2}}{R}$ so $v^{2}=\frac{G m}{R}$ ;
$v=\frac{2 \pi R}{T} ;$ $\left(v^{2}=\right) \frac{4 \pi^{2} R^{2}}{T^{2}}=\frac{G m}{R} ;$
or
$G \frac{M m}{R^{2}}=m \omega^{2} R ;$ $\omega^{2}=\frac{4 \pi^{2}}{T^{2}} ;$ $\frac{4 \pi^{2}}{T^{2}}=\frac{G M}{R^{3}} ;$

Marking guidance:

Award [3] to a clear response with a missing step.

(b)

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 $6.00 \times 10^{3} \mathrm{~s}$ .

\begin{array}{ll} \text { Mass of Earth } & =5.97 \times 10^{24} \mathrm{~kg} \\ \text { Average radius of Earth } & =6.37 \times 10^{6} \mathrm{~m} \end{array}

[ 3 ]

(i)

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

[ 3 ]

$\quad R^{3}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 6000^{2}}{4 \times \pi^{2}}$ ;

\begin{aligned} & R=7.13 \times 10^{6}(\mathrm{~m}) \\ & h=\left(7.13 \times 10^{6}-6.37 \times 10^{6}\right)=760(\mathrm{~km}) \end{aligned}

Award [3] for an answer of 740 with $\pi$ taken as 3.14.

[Maximum number: 3]

There is a proposal to place a satellite in orbit around planet Mars.

The satellite is to have an orbital time T equal to the length of a day on Mars. It can be shown that

T^{2}=k R^{3}

where R is the orbital radius of the satellite and k is a constant.

Mars has a mass of $6.4 \times 10^{23} \mathrm{~kg}$ . Show that, for Mars, k is about $9 \times 10^{-13} \mathrm{~s}^{2} \mathrm{~m}^{-3}$ .

$\frac{mv^2}{r}=\frac{GMm}{r^2}$ , hence $v=\sqrt{\frac{GM}{R}}$ . Also $v=\frac{2\pi R}{T}$ .

OR

$m\omega^2r=\frac{GMm}{r^2}$ , 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 $k=\frac{4\pi^2}{GM}$ , $k=9.2\times10^{-13}\ /\ 9.3\times10^{-13}$

[Maximum number: 2]

There is a proposal to place a satellite in orbit around planet Mars.

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

[Maximum number: 3]

This question is in two parts

In a simple model of the hydrogen atom, the electron is regarded as being in a circular orbit about the proton. The magnitude of the electric field strength at the electron due to the proton is $E_{\mathrm{p}}$ . The magnitude of the gravitational field strength at the electron due to the proton is $g_{\mathrm{p}}$ .

Determine the order of magnitude of the ratio shown below.

\frac{E_{\mathrm{p}}}{g_{\mathrm{p}}}

$E_{\mathrm{p}}=\frac{e}{4 \pi \varepsilon_{0} r^{2}}$ and $g_{\mathrm{p}}=\frac{G m_{\mathrm{p}}}{r^{2}}$ ; (both needed)
$\frac{e}{4 \pi \varepsilon_{0} G m_{\mathrm{p}}}\left(=\frac{9 \times 10^{9} \times 1.6 \times 10^{-19}}{6.7 \times 10^{-11} \times 1.7 \times 10^{-27}}\right) ;$ $\approx 10^{28}$ ;

Marking guidance:

Award [2 max] if response calculates ratio of force as this is an ECF from the first marking point ( $10^{39}$ ).
Award [3] for solution that correctly evaluates field strengths separately and then divides.

[Maximum number: 1]

The diagram shows the gravitational field lines of planet X.

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

[Maximum number: 2]

This question is in two parts. Part 1 is about the use of renewable energy sources. Part 2 is about the gravitational potential of the Earth.

A satellite of mass 780 kg is put into circular orbit $4.0 \times 10^{3} \mathrm{~km}$ above the surface of the Earth. It is then moved to a higher orbit $3.2 \times 10^{4} \mathrm{~km}$ above the surface.

Calculate, for the satellite as it is moved from the lower orbit to the higher orbit, the change in

the gravitational potential energy.

change in gravitational potential $=(+) 28\left(\mathrm{MJ} \mathrm{kg}^{-1}\right)$ ;
change in gravitational potential energy $=\left(28 \times 10^{6} \times 780=\right) 2.2 \times 10^{10}(\mathrm{~J})$ ;

Marking guidance:

Award [2] for a bald correct answer.

Exam points

Use Ep = −GMm/r with zero at infinite separation, keeping the negative sign and measuring r between the centres of the bodies.
Obtain GM from a known surface potential using GM = −VsurfaceR, then use the required orbital radius and mass for potential energy.
For circular orbits, derive EK = GMm/(2r) from gravitational centripetal motion before adding Ep to obtain ET = −GMm/(2r).
Keep EK positive, Ep negative and ET negative, with EK = −Ep/2 and ET = Ep/2 in a circular orbit.
Sketch orbital energy against centre distance as inverse-radius curves approaching zero; total energy approaches zero from below.
Raising a circular orbit increases potential and total energy while decreasing kinetic energy; a less negative energy is an increase.
Calculate ΔEp = GMm(1/r₁ − 1/r₂) using centre radii, adding the planet’s radius when the stated distances are heights above its surface.
Work needed to transfer between circular orbits equals the change in total energy, which is half the potential-energy change.
For a small height change with approximately constant g, use ΔEp ≈ mgΔr, then include the kinetic-energy change for an orbital transfer.
At twice the centre distance, gravitational weight is one-quarter but potential energy is one-half of its original negative value.
Use the requested significant figures and show the energy steps in a derivation; a correct unrounded or unexplained value may lose marks.

[Maximum number: 2]

A satellite of mass m is in a circular orbit of radius r around Earth. Earth has mass M.

Show that the total energy of the satellite is given by $E_{\mathrm{T}}=-\frac{G M m}{2 r}$ .

Identifies orbital velocity as $v=\sqrt{\frac{G M}{r}}$ $E_{\kappa}=\frac{1}{2} m v^{2}=\frac{G M m}{2 r} \text { so } E_{\mathrm{T}}=\frac{G M m}{2 r}-\frac{G M m}{r} \text { Hence result }$

Exam points

Define gravitational potential as external work per unit mass in bringing a small test mass from infinity to the stated point.
Explain negative potential using zero at infinity and attractive gravity: positive work is required to move a mass out to infinity.
Use V = −GM/r with the source mass and centre distance; potential has units J kg⁻¹ or m² s⁻².
Add the planet’s radius to surface height before using inverse-distance potential scaling.
Distinguish inverse-distance potential from inverse-square field strength: at four times the centre radius, V becomes V/4 while g becomes g/16.
For an isolated spherical body, derive V = −gr from V = −GM/r and g = GM/r²; do not cancel finite differences in a gradient formula.
Sketch potential against height with a finite negative surface intercept, increasing towards zero from below as height grows.
Add gravitational potentials as signed scalars; opposing fields can cancel while their negative potentials still add.
Use the distance to each source separately in potential sums, and distinguish a body’s own radius from the orbital separation.
At a potential maximum between two masses, the gradient is zero, so equate their opposing inverse-square fields to find position or mass.
Equal surface potentials imply equal M/R, whereas for equal-density planets the magnitude of surface potential grows with R².
When using a graph’s potential as a body’s own surface potential, check that other sources are negligible or explicitly accounted for.
Obtain a body’s potential energy by multiplying its mass by the potential value at that point, rather than by a graph gradient or area.
For show-that calculations, display substitution or a sufficiently precise value before rounding to the stated estimate.

[Maximum number: 3]

Define gravitational potential at a point in a gravitational field.

Exam points

Relate signed gravitational field to the negative potential gradient, g = −dV/dr; the field points towards decreasing potential.
For the field at a specified point on a curved potential graph, use a tangent there and take its gradient magnitude when magnitude is requested.
Estimate an average field over a finite interval with |ΔV|/Δr, converting MJ to J and km or Mm to metres.
Use the separation between equipotential surfaces, rather than either centre radius, as Δr in an average gradient.
Determine gravitational acceleration from the potential gradient directly; the test mass does not change the acceleration.
Read the signed potential labels before choosing field direction: motion initially accelerates towards the more negative potential.

[Maximum number: 2]

A uniform, spherical planet P has mass M and radius R.

The surface of P is an equipotential surface $E_{0}$ . The dotted circle labeled $E_{1}$ is an equipotential line at a distance $\frac{3 R}{2}$ from the centre of P . The potential difference between $E_{1}$ and $E_{0}$ is $2.2 \times 10^{6} \mathrm{Jkg}^{-1}$ .

The radius R of P is $8.5 \times 10^{6} \mathrm{~m}$ .

Estimate the magnitude of the gravitational field strength between $E_{1}$ and $E_{0}$ .

$\begin{aligned} g=\pi \frac{\Delta V_{\mathrm{g}}}{\Delta r}=>\frac{2.2 \times 10^{6}}{\frac{8.5 \times 10^{6}}{2}} g=0.52 \ll \mathrm{~N} \mathrm{~kg}^{-1} \gg \end{aligned}$

Ignore negative sign

[2]

Exam points

Calculate gravitational potential-energy change as ΔEp = m(Vfinal − Vinitial), keeping the signs of both potentials.
Distinguish work by gravity, Wg = −mΔV, from external work against gravity at unchanged kinetic energy, Wext = mΔV.
Convert MJ kg⁻¹ to J kg⁻¹ before multiplying by mass, then convert J to GJ only if required by the answer scale.
Use potential values at the two endpoints of a potential graph, rather than its shaded area or gradient, to calculate work or energy change.
Moving along one equipotential requires no net gravitational work; inward motion gives positive work by gravity and outward motion gives negative work.
For greatest external work, identify the largest increase in potential using the actual equipotential levels and symmetry of the diagram.
When inferring mass from work, use the correct agent sign or consistent magnitudes: m = |Wg|/|ΔV|.
For transfers in a specified source field, use its mass and the two source-centred radii; omit other bodies only when instructed.
For a transfer between circular orbits, distinguish total energy change from potential-energy change and include the kinetic-energy change when required.

[Maximum number: 2]

The graph shows how the gravitational potential V of Earth varies with distance R from the centre of Earth in the range $R=2.0 \times 10^{8} \mathrm{~m}$ to $R=5.0 \times 10^{8} \mathrm{~m}$ .

The Moon is at a distance of $4.0 \times 10^{8} \mathrm{~m}$ from the centre of Earth. At some time in the past it was at a distance of $2.7 \times 10^{8} \mathrm{~m}$ from the centre of Earth.

Use the graph opposite to determine

by how much the potential energy of the Moon has changed as a result of moving from $R=2.7 \times 10^{8} \mathrm{~m}$ to $R=4.0 \times 10^{8} \mathrm{~m}$ . The mass of the Moon is $7.4 \times 10^{22} \mathrm{~kg}$ .

change in $V=0.45 \times 10^{6}-0.50 \times 10^{6} \mathrm{~J} \mathrm{~kg}^{-1}$ ;
change in $\mathrm{PE}=\left(0.5 \times 10^{6} \times 7.4 \times 10^{22}=\right) 3.3-3.7 \times 10^{28} \mathrm{~J}$ ;

Exam points

Represent equipotential surfaces around an isolated spherical mass by concentric spheres, drawn as concentric circles in cross-section.
Locate a requested equipotential using V = −GM/r; potential ratios give inverse radius ratios rather than linear spacing between labelled values.
For equal potential increments moving outward, increase the radial gaps as the gravitational field becomes weaker.
If successive equal potential steps take radius from R to 3R/2, the following equipotential is at 3R from the same centre.

[Maximum number: 2]

The diagram shows the gravitational field lines of planet X.

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

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

Exam points

Field lines are perpendicular to equipotential surfaces in both gravitational and electric fields.
A mass released from rest initially accelerates normal to the equipotential towards lower, more negative gravitational potential.
For equal potential steps, closer equipotential spacing means stronger field and greater free-fall acceleration along the direction of motion.

[Maximum number: 1]

The diagram shows 5 gravitational equipotential lines. The gravitational potential on each line is indicated. A point mass m is placed on the middle line and is then released. Values given in $\mathrm{MJ} \mathrm{kg}^{-1}$ .

Which is correct about the direction of motion and the acceleration of the point mass?

Direction

Acceleration

to the right

decreasing

to the right

increasing

to the left

decreasing

to the left

increasing

Exam points

Define escape speed as the minimum launch speed needed to reach infinity without further propulsion; leaving an atmosphere or orbit is insufficient.
Use energy conservation at the escape threshold: ½mv² − GMm/r = 0, giving vesc = √(2GM/r) independent of the escaping mass.
Use vesc = √(2gr) with g and r referring to the same point; r is measured from the source centre.
Relate escape speed to potential using vesc = √(−2V) for signed negative potential, or √(2|V|) for its magnitude.
Compare escape speeds using the square root of M/r; changing diameter by a given factor changes radius by that same factor.
For uniform planets, combine M ∝ ρR³ with escape speed to obtain vesc ∝ R√ρ before comparing density and radius changes.
When launch speed is a multiple of escape speed, square that factor in the initial kinetic energy.
For a launch below escape speed, set final kinetic energy to zero at the turning point and conserve inverse-radius potential energy.
Subtract the planet’s radius from the turning-point centre distance to obtain maximum height above the surface.
Justify failure to escape by negative total energy or launch speed below escape speed; a bare conclusion does not establish the argument.
A body falling from negligible speed at infinity reaches the local escape speed in the reverse, loss-free process.
For several gravitating sources, add their negative potentials before applying the escape-energy condition in a consistent reference frame.
Use the supplied molecular speed and escape speed to judge likely atmospheric loss qualitatively; an rms speed is not every molecule’s speed.
In a supplied uniform-density model, substitute M = (4/3)πρr³ and the given speed–radius relation into the current escape condition.

[Maximum number: 2]

A uniform, spherical planet P has mass M and radius R.

A projectile of mass m is launched with kinetic energy $\frac{2 G M m}{3 R}$ from the surface of P.

Determine whether the projectile will escape the gravitational field of P.

ALTERNATE 1
$\begin{aligned} & E_{\mathrm{T}}=\lessdot \frac{2 G M m}{3 R}-\frac{G M m}{R}=»-\frac{1}{3} \frac{G M m}{R} & E_{\mathrm{T}}<0 \text { hence it will not escape } \end{aligned}$

ALTERNATE 2
$\frac{1}{2} m v^{2}=\frac{2 G M m}{3 R} \Rightarrow v=\sqrt{\frac{4 G M}{3 R}}$ $v<v_{\text {esc }}$ hence it will not escape

Argument necessary to get second MP. Do not accept a bald statement that it will not escape.

Allow ECF from MP1.

[2]

Exam points

Derive circular orbital speed by equating gravitational and centripetal force, cancelling the orbiting mass and obtaining v = √(GM/r).
Use the central mass and orbital centre radius in v = √(GM/r); convert km to m and show enough precision for a show-that result.
Add surface height to the body’s radius before finding speed or orbital kinetic energy.
At a common orbital radius, speed scales with √M; at a common central mass it scales with 1/√r and is independent of satellite mass.
For fixed satellite and central masses, circular kinetic energy varies as 1/r; read scaled orbit radii from the centre rather than from the surface.
A higher circular orbit has lower speed and kinetic energy but greater, less negative gravitational potential energy.
Lowering a circular orbit requires negative net engine work while final orbital speed and kinetic energy increase.
Combine v = 2πr/T with the gravitational orbital-speed relation to derive or use T² = 4π²r³/(GM).
Compare orbital and escape speeds at their stated radii; at height R above a planet of radius R, orbital speed is half the surface escape speed.
Use ET = −GMm/(2r) to infer a circular orbital radius from total energy, distinguishing radius from height above the surface.
Read average height from the actual time graph, then add the body radius before estimating orbital speed.
For a circular orbit just above a uniform spherical body, substitute M = (4/3)πρR³ to obtain T = √[3π/(Gρ)].

[Maximum number: 2]

The graph shows the variation with time of the height of the International Space Station (ISS) during a 30-day period.

The following data are available:

\begin{aligned} \text { Mass of Earth } & =5.98 \times 10^{24} \mathrm{~kg} \\ \text { Mass of the ISS } & =4.20 \times 10^{5} \mathrm{~kg} \\ \text { Radius of the Earth } & =6.38 \times 10^{6} \mathrm{~m} \end{aligned}

For the 30-day period indicated in the graph the loss of total energy of the ISS is $4.5 \times 10^{9} \mathrm{~J}$ .

Show, using the average height during the 30-day period, that the speed of the ISS is about $8 \times 10^{3} \mathrm{~ms}^{-1}$ .

Average height 417.5 km
$v=\sqrt{\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{(6.38+0.4175) \times 10^{6}}} \text { OR } 7.7 \times 10^{3} \mathrm{~ms}^{-1}$

Accept $417.5 \pm 0.3$ for MP1

Exam points

Explain small atmospheric drag through loss of total orbital mechanical energy, rather than assuming that orbital kinetic energy must decrease.
Use E = −GMm/(2r): when total orbital energy becomes more negative, the radius and height of a slowly decaying near-circular orbit decrease.
At the smaller near-circular radius, v = √(GM/r) is larger, so orbital speed and kinetic energy increase despite total energy loss.
With a supplied drag model proportional to ρv², greater orbital speed and denser atmosphere can increase drag as altitude falls.
Connect energy loss, falling orbit, increasing speed and stronger drag to eventual re-entry, heating or impact when asked for the satellite’s likely fate.

[Maximum number: 4]

A satellite of mass m is in a circular orbit of radius r around Earth. Earth has mass M.

The satellite experiences a drag force due to the atmosphere of Earth. With reference to the results in (a) and (b)(i), state and explain the likely fate of this satellite.

The total energy is reduced OR the satellite gets closer to the surface
«From $v=\sqrt{\frac{G M}{r}}$ » the satellite moves faster
«From $\rho S v^{2}$ » the opposing force increases « because v and $\rho$ increase »
The satellite burns / crashes / falls to Earth

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