3.9 Astrophysics (A-level only) — coverage pack

Answer

• $40.0$

Method: Convert the eyepiece focal length: $f_e=30.0\,\text{mm}=0.0300\,\text{m}$. Hence $M=f_o/f_e=1.20/0.0300=40.0$.

Answer

• $1.2\times10^{-2}\,\text{rad}$ and $1.525\,\text{m}$

Method: The angular magnification is $M=1.500/0.0250=60.0$. The angle at the eye is therefore $60.0\times2.0\times10^{-4}=1.2\times10^{-2}\,\text{rad}$. In normal adjustment the separation is $f_o+f_e=1.500+0.0250=1.525\,\text{m}$.

Answer

• $30.0\,\text{mm}$ and $0.990\,\text{m}$

Method: The required magnification is at least $M=(4.80\times10^{-3})/(1.50\times10^{-4})=32.0$. Since $M=f_o/f_e$, the greatest acceptable focal length is $f_e=0.960/32.0=0.0300\,\text{m}=30.0\,\text{mm}$. At this limiting eyepiece focal length, the normal-adjustment length is $f_o+f_e=0.960+0.0300=0.990\,\text{m}$; smaller acceptable eyepiece focal lengths give shorter telescopes. A longer-focal-length eyepiece would give less than the required angular magnification.

Answer

• Chromatic aberration.

Method: A mirror reflects different visible wavelengths without refracting them by different amounts, so it does not form wavelength-dependent focal lengths. The avoided aberration is chromatic aberration.

Answer

• The parabolic concave primary reflects the rays towards its focus; before they meet, the convex secondary reflects them back through the central hole in the primary and towards the eyepiece.

Method: Award the chain in order: the concave parabolic primary starts to converge the parallel rays; the convex secondary intercepts them before the primary focus; the secondary reflects them back through the hole in the primary so that they continue to the eyepiece.

Answer

• The reflector has no chromatic aberration, and a parabolic primary avoids the on-axis spherical aberration of a spherical surface. Its mirror can be supported from behind, so a large aperture can be made thinner and with less distortion than a lens supported only at its edge. A refractor can also absorb light in its thick objective, although the reflector's secondary obstruction can reduce the collected intensity and affect the image.

Method: Build the comparison from linked points: reflection does not disperse colours, so there is no chromatic aberration; a parabolic primary focuses on-axis parallel rays together, unlike a spherical surface; a mirror is supported across its back, whereas a lens is supported around its edge; this makes large reflector apertures mechanically practical. A balanced discussion may add the reflector disadvantage that the secondary obstructs part of the incoming beam.

Answer

• Earth's atmosphere absorbs astronomical X-rays.

Method: X-rays from space are strongly absorbed by the atmosphere, so a detector must be placed above most or all of the atmosphere.

Answer

• $7.64\times10^{4}\,\text{m}$

Method: Equal resolution requires $\lambda_{\rm optical}/D_{\rm optical}=\lambda_{\rm radio}/D_{\rm radio}$. Therefore $D_{\rm radio}=0.210\times0.200/(550\times10^{-9})=7.64\times10^4\,\text{m}$. This very large value follows from the much longer radio wavelength.

Answer

• A radio dish can work on the ground, preferably away from human-made radio interference. An infrared telescope should be high and dry to reduce absorption by atmospheric water vapour. Ultraviolet and X-ray instruments should be above the atmosphere, normally on satellites, because those radiations are strongly absorbed. Radio wavelengths are much longer than visible wavelengths, so a much larger $D$ is needed for comparable resolution through $\theta\approx\lambda/D$; the large diameter also gives greater collecting power because this is proportional to $D^2$.

Method: Link each location to atmospheric transmission: radio reaches the ground, so choose a radio-quiet ground site; infrared is absorbed by water vapour, so use a high, dry site; ultraviolet is largely absorbed, so place its telescope above the atmosphere; X-rays are also absorbed, so use a satellite. Then compare apertures: $\theta\approx\lambda/D$ means the much longer radio wavelength requires a much larger $D$ for similar angular resolution, and collecting power proportional to $D^2$ also helps detect weak radio signals.

Answer

• $4.0$

Method: Collecting power is proportional to $D^2$, so the factor is $(0.60/0.30)^2=2^2=4.0$.

Answer

• $2.00\times10^{-7}\,\text{rad}$; a larger diameter gives a smaller minimum angle.

Method: Convert $500\,\text{nm}=5.00\times10^{-7}\,\text{m}$. Then $\theta\approx\lambda/D=(5.00\times10^{-7})/2.50=2.00\times10^{-7}\,\text{rad}$. Since $\theta$ is inversely proportional to $D$, increasing $D$ improves resolution by reducing the minimum resolvable angle.

Answer

• Only the $10.0\,\text{m}$ telescope resolves the pair; it has $6.25$ times the collecting power; a CCD has higher quantum efficiency and can integrate and store the exposure.

Method: For the $4.00\,\text{m}$ telescope, $\theta=600\times10^{-9}/4.00=1.50\times10^{-7}\,\text{rad}$, which is greater than the separation, so it cannot resolve the pair. For the $10.0\,\text{m}$ telescope, $\theta=600\times10^{-9}/10.0=6.0\times10^{-8}\,\text{rad}$, which is smaller than the separation, so it can resolve the pair. The collecting-power ratio is $(10.0/4.00)^2=6.25$. A CCD converts a larger fraction of incident photons into a signal and can accumulate a long exposure and store it, unlike a momentary visual observation.

Answer

• The star with apparent magnitude $+1.0$.

Method: On the magnitude scale, the smaller apparent magnitude denotes the brighter object, so the $+1.0$ star appears brighter.

Answer

• $15.8$

Method: The magnitude difference is $5.00-2.00=3.00$. P has the lower magnitude and is therefore brighter, so $I_P/I_Q=2.51^3=15.8$.

Answer

• $5.26$; yes, because this is brighter than the magnitude-$6$ limit.

Method: Let the magnitude increase be $\Delta m$. Since the intensity has fallen by a factor of $20.0$, $2.51^{\Delta m}=20.0$. Thus $\Delta m=\log(20.0)/\log(2.51)=3.26$. The new apparent magnitude is $2.00+3.26=5.26$. This is below the approximate limiting magnitude $6$, so the star should still be visible under suitable dark conditions.

Answer

• $M=4.3$

Method: Absolute magnitude is defined as the apparent magnitude at $10\,\text{pc}$, so at this distance $M=m=4.3$.

Answer

• $100\,\text{pc}$

Method: $m-M=8.4-3.4=5.0$. Therefore $5.0=5\log(d/10)$, so $\log(d/10)=1.0$. Hence $d/10=10$ and $d=100\,\text{pc}$.

Answer

• $3.98\times10^3\,\text{pc}$ and $1.30\times10^4\,\text{ly}$

Method: $m-M=12.0-(-1.0)=13.0$. Hence $13.0=5\log(d/10)$ and $\log(d/10)=2.60$. Therefore $d=10\times10^{2.60}=3.98\times10^3\,\text{pc}$. In light years this is $3.98\times10^3\times3.26=1.30\times10^4\,\text{ly}$.

Answer

• $5.0\times10^3\,\text{K}$

Method: $\lambda_{\max}=580\,\text{nm}=5.80\times10^{-7}\,\text{m}$. Thus $T=(2.9\times10^{-3})/(5.80\times10^{-7})=5.0\times10^3\,\text{K}$.

Answer

• $1.64$

Method: For spherical stars, $P\propto R^2T^4$. Therefore $P_B/P_A=(2.00)^2(0.800)^4=4.00\times0.4096=1.6384\approx1.64$.

Answer

• $6.0\times10^3\,\text{K}$ and $3.2\times10^9\,\text{m}$

Method: Wien's law gives the guard value $T=(2.9\times10^{-3})/(480\times10^{-9})=6.04\times10^3\,\text{K}$. The distance is $d=50.0\times3.086\times10^{16}=1.543\times10^{18}\,\text{m}$. From the inverse-square law, $P=4\pi d^2I=4\pi(1.543\times10^{18})^2(3.20\times10^{-10})=9.57\times10^{27}\,\text{W}$. Stefan's law gives $P=4\pi R^2\sigma T^4$, so the guard value is $R=\sqrt{P/(4\pi\sigma T^4)}=3.18\times10^9\,\text{m}$. To the two significant figures supported by Wien's constant, the final values are $6.0\times10^3\,\text{K}$ and $3.2\times10^9\,\text{m}$.

Answer

• The class B star.

Method: The sequence OBAFGKM runs from highest to lowest temperature. Of B, G and M, B occurs first and is therefore hottest.

Answer

• $7.1\times10^3\,\text{K}$; class F

Method: $T=(2.9\times10^{-3})/(410\times10^{-9})=7.07\times10^3\,\text{K}$. The class F range is approximately $6000$-$7500\,\text{K}$, so the most likely class is F.

Answer

• Balmer absorption requires bound hydrogen atoms already in the $n=2$ state. In a cool M star too few collisions excite hydrogen to $n=2$. In a very hot O star much of the hydrogen is ionised, so fewer bound atoms can absorb Balmer photons. An A-star temperature gives a large population of bound hydrogen atoms in $n=2$, producing the strongest Balmer lines; the difference is an excitation and ionisation effect, not evidence of different hydrogen abundance.

Method: Start with the transition condition: Balmer absorption begins from $n=2$. For an M star, thermal excitation is insufficient, so the $n=2$ population is small. For an O star, the high temperature ionises much of the hydrogen, leaving fewer bound atoms. At class A temperatures there is a favourable balance: hydrogen remains bound and many atoms occupy $n=2$. Therefore line strength depends on temperature-dependent excitation and ionisation rather than simply on hydrogen abundance.

Answer

• The main sequence; spectral class G.

Method: The Sun is an ordinary hydrogen-burning main-sequence star and its temperature of about $5800\,\text{K}$ places it in spectral class G.

Answer

• The white-dwarf region: the star is hot but intrinsically faint.

Method: $12000\,\text{K}$ places the star on the hot, left side of the diagram. $M=+12$ is intrinsically faint and therefore low on the vertical axis. Hot but faint stars occupy the lower-left white-dwarf region.

Answer

• A forming protostar contracts towards the main sequence. It then spends most of its lifetime near the Sun's class-G main-sequence position. When core hydrogen is depleted, it expands and moves to the red-giant region: its surface becomes cooler but its luminosity rises, so its absolute magnitude decreases. After its outer layers are lost, the hot exposed remnant moves to the white-dwarf region; it is hot but faint and then cools and fades.

Method: Give the ordered marking chain: formation and contraction towards the main sequence; a long stable main-sequence stage; movement to the cool but luminous red-giant region, which is rightwards and upwards on the usual axes; loss of the outer envelope leaving a hot compact remnant; arrival in the hot but faint lower-left white-dwarf region, followed by cooling and fading.

Answer

• $7.71\times10^3\,\text{m}$

Method: $R_s=2GM/c^2=2(6.67\times10^{-11})(5.20\times10^{30})/(3.00\times10^8)^2=7.71\times10^3\,\text{m}$.

Answer

• $3.98\times10^8\,\text{pc}$

Method: $m-M=18.5-(-19.5)=38.0$. Hence $38.0=5\log(d/10)$, so $\log(d/10)=7.60$. Therefore $d=10\times10^{7.60}=3.98\times10^8\,\text{pc}$. The standard-candle assumption supplies the absolute magnitude.

Answer

• $8.9\times10^3\,\text{m}$ and $3.3\times10^{10}\,\text{years}$

Method: $R_s=2GM/c^2=2(6.67\times10^{-11})(6.0\times10^{30})/(3.00\times10^8)^2=8.89\times10^3\,\text{m}$. The equivalent solar emission time is $t=E/P=(4.0\times10^{44})/(3.8\times10^{26})=1.05\times10^{18}\,\text{s}$. Converting gives $t=(1.05\times10^{18})/(3.156\times10^7)=3.34\times10^{10}\,\text{years}$, or $3.3\times10^{10}\,\text{years}$ to two significant figures.

Answer

• $8.0\times10^{-4}$

Method: $\Delta\lambda=500.40-500.00=0.40\,\text{nm}$. Therefore $z=\Delta\lambda/\lambda=0.40/500.00=8.0\times10^{-4}$.

Answer

• $2.11\times10^5\,\text{m s}^{-1}$, receding

Method: $\Delta f=1.41900-1.42000=-0.00100\,\text{GHz}$. Using $\Delta f/f=-v/c$, $v=-c\Delta f/f=-(3.00\times10^8)(-0.00100/1.42000)=2.11\times10^5\,\text{m s}^{-1}$. The received frequency is lower, so the cloud is receding. Also $v/c=7.04\times10^{-4}\ll1$, so the approximation is valid.

Answer

• $1.37\times10^5\,\text{m s}^{-1}$, $1.51\times10^{10}\,\text{m}$, and $3.02\times10^{10}\,\text{m}$

Method: The maximum shift magnitude is $\lvert\Delta\lambda\rvert=0.300\,\text{nm}$. Hence $v=c\lvert\Delta\lambda\rvert/\lambda=(3.00\times10^8)(0.300/656.300)=1.37\times10^5\,\text{m s}^{-1}$; $v/c=4.57\times10^{-4}$ validates the approximation. The period is $T=8.00\times86400=6.912\times10^5\,\text{s}$. For circular motion, $v=2\pi r/T$, so $r=vT/(2\pi)=1.51\times10^{10}\,\text{m}$. Equal masses orbit at equal radii on opposite sides, so their separation is $2r=3.02\times10^{10}\,\text{m}$.

Answer

• $8.4\times10^3\,\text{km s}^{-1}$

Method: $v=Hd=70\times120=8400\,\text{km s}^{-1}=8.4\times10^3\,\text{km s}^{-1}$.

Answer

• $1.44\times10^{10}\,\text{years}$

Method: Convert $H$: $H=(68.0\times10^3)/(3.086\times10^{22})=2.20\times10^{-18}\,\text{s}^{-1}$. Then $t\approx1/H=4.54\times10^{17}\,\text{s}$. In years, $t=(4.54\times10^{17})/(3.156\times10^7)=1.44\times10^{10}\,\text{years}$. This assumes a constant expansion rate.

Answer

• $64.4\,\text{Mpc}$; $v/c=0.0150$, so $v\ll c$ is a reasonable approximation.

Method: $\Delta\lambda=493.40-486.10=7.30\,\text{nm}$, so $z=7.30/486.10=0.0150$. The recession speed is $v\approx zc=0.0150(3.00\times10^5)=4.51\times10^3\,\text{km s}^{-1}$. Hubble's law gives $d=v/H=(4.51\times10^3)/70.0=64.4\,\text{Mpc}$. Since $v/c=0.0150\ll1$, use of the non-relativistic approximation is reasonable.

Answer

• Accretion of matter onto an active supermassive black hole.

Method: A quasar is powered by gravitational energy released as matter accretes onto a supermassive black hole in an active galactic nucleus.

Answer

• $2.4\times10^{37}\,\text{W}$

Method: $d=900\times3.086\times10^{22}=2.777\times10^{25}\,\text{m}$. For isotropic emission, $P=4\pi d^2F=4\pi(2.777\times10^{25})^2(2.5\times10^{-15})=2.42\times10^{37}\,\text{W}$.

Answer

• $85.7\,\text{Mpc}$, $1.41\times10^{37}\,\text{W}$, and $3.67\times10^{10}L_{\odot}$

Method: $v\approx zc=0.0200(3.00\times10^5)=6.00\times10^3\,\text{km s}^{-1}$; the ratio $v/c=0.0200$ is small enough for this estimate. Hubble's law gives $d=v/H=6000/70.0=85.7\,\text{Mpc}=2.645\times10^{24}\,\text{m}$. Then $P=4\pi d^2F=4\pi(2.645\times10^{24})^2(1.60\times10^{-13})=1.41\times10^{37}\,\text{W}$. Using the unrounded power, $P/L_{\odot}=3.67\times10^{10}$.

Answer

• The transit method.

Method: A periodic dip occurs when a planet repeatedly passes across the stellar disc, so the observation indicates the transit method.

Answer

• $6.21\times10^7\,\text{m}$

Method: The fractional depth is $0.810\%=0.00810$. Using $\Delta I/I=(R_p/R_*)^2$, $R_p=R_*\sqrt{0.00810}=(6.90\times10^8)(0.0900)=6.21\times10^7\,\text{m}$.

Answer

• $3.2\times10^7\,\text{m}$ and $7.0\times10^3\,\text{m s}^{-1}$; matching periodic transit and Doppler signals are independent evidence for the same orbiting companion.

Method: The transit depth is $0.160\%=0.00160$, so $R_p=R_*\sqrt{0.00160}=(8.0\times10^8)(0.0400)=3.2\times10^7\,\text{m}$. The maximum radial speed is $v\approx c\Delta\lambda/\lambda=(3.00\times10^8)(0.014/600.0)=7.0\times10^3\,\text{m s}^{-1}$, with $v/c=2.33\times10^{-5}$. Repeated transits show an orbiting body crossing the stellar disc, while the alternating Doppler shift independently shows the host star moving towards and away from Earth. If both variations share the $5.2$-day period, one orbiting companion explains both signals and reduces the chance that brightness variability alone caused the dips.