5.6 - Astrophysics A2 and Cosmology
- Syllabus
- 2021
- Topic
- 5.6
- Level
- A2
A gravitational field is a region in which a mass experiences a force. The field describes the influence of the source mass throughout the surrounding space, whether or not a test mass is currently placed there.
At any point, the field direction is the direction of the force on a small positive test mass. Around an isolated spherical mass, this is radially inward because gravity is attractive.
| Role | Meaning |
|---|---|
| source mass | creates the gravitational field |
| test mass | samples the field by experiencing a force |
| field line | shows local force direction; closer spacing represents a stronger field |
If several masses are present, their gravitational forces add as vectors. Equivalently, add their field contributions vectorially at the point of interest.
A gravitational field is not the same as the force itself. The field can exist at a point before a test mass is placed there; the force appears when a mass occupies that point.
g=mFg
Gravitational field strength g at a point is the gravitational force Fg per unit mass m placed at that point. It is a vector and points in the direction of the force.
| Quantity | Unit | Note |
|---|---|---|
| Fg | N | gravitational force |
| m | kg | test mass |
| g | Nkg−1 | numerically equivalent to ms−2 |
If a 2.5kg object experiences a gravitational force of 20N, then g=20/2.5=8.0Nkg−1. A different small test mass at the same point has the same g but a different force Fg=mg.
Do not treat g as a property of the test mass. The source masses and position determine g; the test mass only sets the force it experiences.
F=Gr2m1m2
Here m1 and m2 are the interacting masses, r is the centre-to-centre separation, and G=6.67imes10−11Nm2kg−2. Each mass experiences an equal-magnitude force directed towards the other.
Choose SI units, find the centre separation, square it, then substitute. For a uniform spherical body, its external gravitational effect can be modelled as if all its mass were concentrated at its centre.
| Change | New force |
|---|---|
| one mass doubles | 2F |
| both masses double | 4F |
| separation doubles | F/4 |
| separation triples | F/9 |
The r in the law is not the gap between two surfaces. It is the distance between the mass centres; also, the two forces form a Newton's-third-law pair rather than cancelling on either one object.
Place a test mass m a distance r from a point or spherical source mass M. Newton's law gives Fg=GMm/r2; dividing by the test mass in g=Fg/m removes m.
g=r2GM
The magnitude falls as 1/r2, while the vector direction is radially inward. Outside a spherical planet or star, measure r from its centre, so at altitude h use r=R+h.
At a distance 2R from the centre of a planet, g=GM/(2R)2=gsurface/4. This comparison avoids recalculating G and M.
The formula is valid for a point mass and outside a spherically symmetric mass. It cannot be applied through an extended object's interior as though all interior points were outside a point mass.
Vgrav=−rGM
Gravitational potential Vgrav is potential energy per unit mass, measured in Jkg−1. The conventional zero is at infinity. Because gravity is attractive, a finite point in the field has negative potential.
ΔEp=mΔV=m(Vfinal−Vinitial)
As r increases, V rises towards zero: it becomes less negative. The gradient is steep close to the source and shallow far away. Field strength and potential are linked by g=−dV/dr in the radial direction.
A value such as −20MJkg−1 is lower than −5MJkg−1. Moving away requires an increase in potential energy even though the numerical value approaches zero.
| Feature | Gravitational field | Electric field |
|---|---|---|
| source property | mass M | charge Q |
| force on test object | F=mg | F=qE |
| point-source dependence | g=GM/r2 | E=kQ/r2 in magnitude |
| interaction | always attractive between masses | attractive or repulsive |
| field direction | force on a test mass | force on a positive test charge |
| potential | zero at infinity; negative around an isolated mass | sign depends on source charge |
Both are long-range force fields, obey superposition, and have inverse-square radial fields for point sources. Their field lines show force direction, while their potentials are scalar quantities.
Gravity is normally much weaker than electrostatic interaction at particle scale, but astronomical bodies contain enormous mass and are often nearly electrically neutral, so gravity dominates their large-scale motion.
Do not copy the sign behaviour of electric fields into gravity. There is no negative mass in this model, so gravitational interaction does not become repulsive.
For a mass m in a circular orbit of radius r around a much larger central mass M, the inward gravitational force supplies the required centripetal force.
r2GMm=rmv2=mω2r
v=rGM,T=2πGMr3,T2=GM4π2r3
The orbiting mass cancels from the equations. For the same central mass, a larger circular orbit has a lower speed but a longer period. The velocity is tangential while acceleration and force point inward.
Centripetal force is not an additional force alongside gravity: it is the name for the resultant inward force. An orbiting body feels weightless because it is in continuous free fall, not because gravity is absent.
A black body is an ideal object that absorbs all incident electromagnetic radiation and is also a perfect emitter. Its emission spectrum depends only on its absolute temperature.
| When temperature increases | What the radiation curve does |
|---|---|
| peak position | moves to a shorter wavelength |
| peak height | increases |
| total area under the curve | increases, so emitted power per unit area rises |
| spectrum | remains continuous over a broad range of wavelengths |
To compare two curves, check that the axes and scales match. Locate each λmax for temperature information and compare areas for total emitted power per unit area.
A star can be approximated as a black-body radiator. Its colour does not mean it emits one wavelength: it emits a distribution whose peak and total area change with temperature.
The wavelength at the peak is the most intense part of a continuous distribution, not the only wavelength emitted. A peak outside visible red does not by itself mean that no red light is emitted.
L=σAT4,σ=5.67×10−8Wm−2K−4
Luminosity L is the total power radiated, A is emitting surface area, and T is absolute temperature. For a spherical star, A=4πR2.
L=4πR2σT4
At fixed radius, doubling T multiplies L by 24=16. At fixed temperature, doubling radius multiplies surface area and luminosity by 22=4. Ratio methods often remove σ and 4π.
Temperature must be in kelvin, and A is surface area rather than cross-sectional area πR2. The fourth-power dependence makes early rounding especially costly.
λmaxT=2.898×10−3mK
Read or calculate the peak wavelength, convert it to metres, then rearrange to T=(2.898imes10−3)/λmax. The relation is inverse: hotter black bodies peak at shorter wavelengths.
If λmax=5.00imes10−7m, then T=2.898imes10−3/(5.00imes10−7)=5.80imes103K.
| Check | Expected |
|---|---|
| wavelength unit | metres |
| temperature unit | kelvin |
| hotter source | smaller λmax |
| answer order | stellar temperatures are typically thousands of kelvin |
Do not substitute nanometres directly into the constant stated in mK. Wien's law identifies the spectral peak; it does not give luminosity unless combined with other information.
If a source radiates uniformly in all directions, its luminosity L passes through a spherical surface of area 4πd2 at distance d. Intensity is power received per unit area.
I=4πd2L
| Quantity | Unit |
|---|---|
| luminosity L | W |
| distance d | m |
| intensity I | Wm−2 |
At twice the distance, the same luminosity is spread over four times the area, so intensity is one quarter. Rearranging gives L=4πd2I or d=L/(4πI).
Intensity is not luminosity: two stars can have different intrinsic luminosities yet produce the same intensity at Earth because their distances differ. The equation assumes isotropic spreading and no unmodelled absorption.
Observe a nearby star against very distant background stars, then repeat six months later from the opposite side of Earth's orbit. The nearby star appears to shift relative to the fixed background.
Half the total angular shift is the parallax angle p. With the known radius of Earth's orbit as the baseline, right-triangle geometry gives the distance.
d=tanp1AU≈p1AU(p in radians),d(pc)=p(arcsec)1
Measure the angular displacement carefully, halve it to obtain p, select one consistent unit relation, and calculate d. More distant stars have smaller parallax angles.
The six-month baseline is the diameter of Earth's orbit, but the triangle used with parallax angle p has a one-AU side. Do not use Pythagoras: the distance comes from angular trigonometry.
A standard candle is an astronomical object whose luminosity L is known. Find such an object in a distant cluster or galaxy and measure the intensity I received at Earth.
I=4πd2L⟹d=4πIL
| Step | Evidence used |
|---|---|
| identify and calibrate the standard candle | establishes known luminosity L |
| measure received intensity | supplies I |
| apply inverse-square spreading | calculates distance d |
Nearby distance methods can calibrate the luminosity of a standard-candle class; that calibrated class can then extend the astronomical distance scale to much greater distances.
A bright-looking object is not automatically a standard candle. Its intrinsic luminosity must be known, and uncorrected absorption or a wrong classification would make the inferred distance unreliable.
A simple Hertzsprung-Russell diagram plots stellar luminosity vertically, increasing upward, against surface temperature horizontally. Temperature conventionally decreases from left to right.
| Region | Temperature and luminosity |
|---|---|
| main sequence | diagonal band from hot, luminous upper left to cool, dim lower right |
| red giants | cool but luminous, upper right |
| white dwarfs | hot but dim, lower left |
Luminosity depends on both surface temperature and radius through L=4πR2σT4. A cool star can therefore be highly luminous if it has a very large radius; a hot star can be dim if it is very small.
Label both axes and their directions first, then place the three regions. If luminosity is expressed relative to the Sun, it is commonly shown on a logarithmic scale.
The horizontal axis runs in the opposite direction to an ordinary number line when it is temperature. Do not place red giants at the lower right merely because they are cool: their large radii make them luminous.
A main-sequence star is stable while hydrogen fusion in its core supplies energy and pressure balances gravitational contraction. Its mass determines how rapidly it evolves and what later stages are possible.
| Lower-mass path | Higher-mass path |
|---|---|
| core hydrogen becomes depleted | core hydrogen becomes depleted |
| core contracts and heats; outer layers expand and cool into a red giant | core contracts; outer layers expand into a red supergiant |
| helium fusion occurs; eventually the outer layers are lost | successive fusion stages build heavier nuclei |
| hot, small remnant is a white dwarf | core collapse and supernova leave a neutron star or black hole |
On the H-R diagram, a main-sequence star moves toward the giant region as its surface cools and its radius and luminosity change. A white dwarf lies in the hot-but-dim region because its surface is hot but its area is small.
The sequence is driven by fuel depletion: reduced core fusion permits contraction, contraction raises core temperature, and new fusion stages or collapse change the star's radius, temperature and luminosity.
Stars do not all follow one identical route. Initial mass is decisive: a Sun-like star does not undergo the same final core-collapse sequence as a much more massive star.
When a wave source moves relative to an observer, successive wavefronts are emitted from different positions. Motion towards the observer compresses their spacing; motion away spreads them out.
| Source motion relative to observer | Observed wavelength | Observed frequency | Light description |
|---|---|---|---|
| towards | shorter | higher | blueshift |
| away | longer | lower | redshift |
Astronomers compare known spectral lines with their observed wavelengths. A periodic change between redshift and blueshift can reveal a star moving back and forth because of an orbiting companion.
The effect depends on radial relative motion—the component along the observer's line of sight. Sideways motion alone does not produce the same first-order wavelength shift.
A Doppler shift changes the received frequency and wavelength; it does not mean the source has emitted a different chemical fingerprint. For light, no material medium is required.
z=λΔλ≈−fΔf≈cv
Use Δλ=λobserved−λrest. A receding source has a longer observed wavelength, so z>0. Its frequency falls, so Δf=fobserved−frest<0 and the minus sign makes the same positive z. The speed approximation applies when v≪c.
v=H0d
For galaxies at cosmological distances, recession speed v is proportional to distance d. A graph of v against d has gradient H0; the widespread redshift-distance relation is evidence that the universe is expanding.
Keep the reference quantities consistent: λ and f are rest values in these fractional changes. Do not use vpprox zc without recognising its low-speed approximation, and do not confuse a red spectral colour with measured displacement of known lines.
tH≈H01
If expansion is approximated as steady, distance divided by recession speed gives d/v=1/H0, a characteristic age. A larger measured H0 gives a smaller age estimate. Convert H0 to s−1 before taking its reciprocal in seconds.
Galaxy motions and gravitational effects can indicate more mass than is directly visible. This proposed dark matter increases the total gravitational influence and therefore affects models of how expansion may change.
| Uncertain evidence | Consequence |
|---|---|
| different determinations of H0 | different inferred expansion ages |
| amount and distribution of dark matter | different total density and gravitational slowing |
| assumptions about how expansion changed over time | 1/H0 is a model estimate, not an exact stopwatch reading |
The ultimate fate depends on the competition represented in the model between cosmic expansion and the gravitational effect of the universe's matter. Because H0 and unseen mass are inferred from observations with assumptions and uncertainties, conclusions about age and fate have been controversial.
Dark matter is not simply ordinary matter that is faint, and a flat galaxy rotation curve is evidence for additional gravitational mass rather than a direct photograph of it. Keep conclusions conditional on the measured H0, the possible dark matter and the model used.