Astrophysics and Rocket Science Lectures PDF

All Questions posed in Astrophystech + Rocket Science Lectures 1. What is Blackbody Radiation? Distribution of energy between a finite number of particles in a fixed volume is defined by Maxwell Boltzmann distribution. Heated objects radiate photons. Peak emission wavelength depends on temperature....

All Questions posed in Astrophystech + Rocket Science Lectures 1. What is Blackbody Radiation? Distribution of energy between a finite number of particles in a fixed volume is defined by Maxwell Boltzmann distribution. Heated objects radiate photons. Peak emission wavelength depends on temperature. Intensity of emission depends on the number of photons emitted per second. As the temperature increases, mean photon energy shifts to lower wavelengths. 2. What are the differences between Thermal Bremsstrahlung and Synchrotron Radiation? In TB radiation charged particles undergo acceleration as a result of interaction with other charged particles. Rate of acceleration determines the frequency produced. Higher acceleration leads to higher frequencies of emission. E = hf. i.e. high photon energy. In SR charged particles are accelerated in magnetic and electric fields in circles or straight lines. 3. What is electron degeneracy pressure? Normal gas ( in which electrons orbit nuclei) exert higher pressure when heated causing expansion, but this is not the case for degenerate matter. In degenerate matter, the atoms have been stripped of their electrons. The pressure on the matter forces the electrons together to such an extent that quantum mechanical effects come into play. In normal matter, electrons can move between orbital levels, but in degenerate matter the electron energy levels are determined by the Pauli Exclusion Principle. This is that no two electrons are allowed to exist in the same energy level unless they have opposite spins. Thus degenerate matter exerts a pressure resisting gravitational collapse, as all the lower energy levels are filled with electrons and therefore very high energy is needed to promote additional electrons to free orbital levels. 4. What are the main differences between a type II and a type Ia supernova? A type II SN emits around 1046J over a week or so whereas a type Ia releases 1044J. A type II SN is created when a main sequence star of 8 solar masses or more, fuses iron in the core at which point power generation stops and the star collapses. As temperature increases along with pressure, a point is reached at which the protons and electrons combine to create neutrons. For a type II SN, a neutron star is formed if the core is less than 3 solar masses. Otherwise a black hole results. The light curve (intensity of light received as a function of time) depends on the size of the initial star. A shockwave passes out from the core into the outer radii of the star’s original matter, creating hydrogen emission lines within the spectra observed. This is also where nucleosynthesis leads to the formation of heavier elements. A type Ia SN is formed in a binary system consisting of a red giant and a white dwarf. No fusion processes exist in a white dwarf and it is supported by only electron degeneracy pressure which sets a maximum of 1.4 solar masses (the Chandrasekhar limit). If matter is drawn from the companion star, the white dwarf’s mass will increase above 1.4 solar masses which is sufficient to initiate runaway fusion. Because the conditions for a type Ia are so much more clearly defined, the light curve is extremely predictable with no hydrogen emission lines in the spectra. 5. Why do we get runaway fusion in a type Ia supernova even though the centre of the star comprises carbon and oxygen? In the centre of a white dwarf all fusion has stopped and the matter has become degenerate. Degenerate matter does not obey gas laws. Increased temperature in the centre of a main sequence star causes expansion of its outer layers due to increased pressure and this results in cooling and slowing down of fusion processes. This is a negative feedback loop. In the centre of the white dwarf, the increased temperature just leads to faster fusion without any expansion and cooling. 6. Explain the origin of an X ray burster. A neutron star in close proximity to another star may be able to pull hydrogen onto its surface. The high temperature allows hydrogen to fuse on the surface of the neutron star. Normally this increase in temperature would then cause the pressure to increase, the hydrogen to expand and the reaction to stop. Gas Laws however don’t apply on a neutron star and the increased temperature just causes fusion to accelerate. This leads to runaway thermonuclear fusion on the surface of the star causing a burst of X rays via blackbody radiation. This lasts around 1 minute and releases around 1031 J of energy. 7. What are the key differences between the signals observed from an X ray pulsar and an X ray burster? X ray pulsar X ray burster Origin of radiation Synchrotron radiation Blackbody radiation Emission mechanism EM radiation from charged Thermonuclear explosion of particles moving in a magnetic material captured from a field of the neutron star companion star on the surface of the neutron star Polarised Yes No Duration Very brief (milliseconds) Up to a minute Regularity Extremely regular due to time Irregular (houses to months period of star’s rotation between bursts due to rate of accretion) 8. Explain the origin of long and short gamma ray bursters. Long GRBs are formed when stars of >30 solar masses collapse as a hypernova. A black hole forms in the centre before much of the star has collapsed inward from the outer radii. Conservation of angular momentum leads to the creation of a fast spinning accretion disc. Most infalling material is absorbed by the black hole but some is ejected in back to back jets along the axis of rotation. These jets blow the star apart. Particles annihilate with their antiparticles to create high energy gammas which are detected on Earth if a beam is directed towards us. They last a few minutes until the accretion disc passes into the black hole. Short GRBs are created via the coalescence of 2 neutron stars in a binary orbit. Losing energy via gravitational radiation they eventually rip each other apart forming an accretion disc around the centre of mass which can be either a large neutron star or more likely a black hole. With less material, these short GRBs are very short lived (few seconds). 9. What is the difference between a quasar and a long GRB? A long GRB emits gamma rays for several minutes and occurs as a result of the disintegration of a supermassive star going hypernova, the gamma rays produced from the accretion disc formed out of the original material of the star. A quasar emits a full spectrum of radiation over a huge time period as a result of the accretion of stars, etc. within an active galaxy onto a central supermassive black hole. 10. Imagine a ray of green light of wavelength λ = 530 nm incident on a detector with a work function of 1.1eV. What is the kinetic energy given to a photoelectron ejected from this target? The frequency of green light is found using c =f. where c is the velocity of light, so 3  10 8 f  9  5.66  1014 Hz. Einstein said this ray was made up of photons each of energy 530  10 E photon  h f  6.626  10 34  5.66  1014  3.54  10 19 J. 3.54  10 19 So our green photons each have energy  2.3 eV (electron volts). 1.6  10 19 11. We collide a gamma ray photon (λ = 3×10-14 m) with an electron. What is the momentum of the photon before the collision? What is the energy lost by the photon if following the collision its direction changes by 60 degrees? h Calculate the momentum of the photon using p .  We always use SI units so mass is in kg, distance is in m, time is in s, energy is in joules. So h 6.626  10 34 Js p   2.21  10 20 kgms 1.  3  10 14 m  h  f  i  (1  cos  ) so  f  i  2.45  10 12 (1  cos 60 )  1.23  10 12. me c ch So E photon before  hf   6.63  10 12 joules. before ch So E photon after  hf   1.58  10 13 joules. after Energy lost is 6.47  10 12 joules. 12. Explain why a detector configured to use the photoelectric effect would provide a more accurate measurement of total energy of a photon compared to one configured for Compton scattering. If measurement of the photon’s total energy is important, then the photoelectric effect provides a means of achieving this easily, as energies of photons are of order 1 MeV and work functions of a target are of order eV. Therefore all photon energy is transferred to target electrons. Compton scattering deposits a variable amount of energy depending on the scattering angle. 13. Deduce the minimum energy of a photon in eV in order for it to create an electron – positron pair by pair production. We need to create these two leptons, each with mass 9.1x10-31 kg. E = 2mc2 = 1.64x10-13 J. 1.64×10−13 This equates to = 1.022 × 106 eV or 1022 keV. 1.6×10−19 14. What thickness of 1 g/cm3 body tissue is needed to reduce the flux of 20 keV X-rays by 50% ? I( x )     exp    x  I0     exp 11 x  1 2 x  0.69 cm 15. A lead wall of thickness 1 cm is placed between a doctor of thickness 20 cm and a flux of 10 MeV gammas. What fraction of the initial gamma beam line flux interacts with the doctor’s body? (density of lead is 11.4 g/cm3). We want to first find the fraction of intensity which pass through the lead and then the fraction of this flux which interact with the doctor.    x   exp 0.05 ( 11.4 )( 1 )  0.57 I( x ) Fraction which passes through lead  exp    lead I0     lead  Fraction of flux incident on doctor which passes through her     tissue x  exp 0.022( 1 )( 20 )  0.64. I( x )  exp   I0     tissue  Therefore if 64% of the flux incident on the doctor passes through him, 36% must have interacted within him by absorption or scattering off the beam line. This represents 36% of the flux incident on the doctor but as only 57% make it through the lead, the fraction of original gammas interacting with the doctor is 0.36  57  20.5%. 16. Describe the process whereby a photon interacts with a scintillator coupled to a photomultiplier tube to produce a signal. An incident photon interacts within the scintillator by photoelectric effect, Compton scattering or pair production producing a high energy electron. This electron then passes through the scintillator generating a path of ionisation. After a very short time (1 microsecond) this ionisation recombines generating visible photons. This light is guided to a photomultiplier tube where it interacts with the photocathode releasing electrons. The photocathode is a negatively charged electrode coated with caesium iodide which has a very low work function. It therefore subsequently emits electrons via the photoelectric effect. The electrons produced are then guided by a strong electric field within the PMT to the 1 st dynode stage. On colliding with this, more electrons are generated by collisional ionisation. These are attracted to the 2nd dynode stage and so on until eventually over 1,000,000 electrons are collected at the anode of the PMT and amplified to produce a signal. 17. How does a microchannel plate detector produce an image of sources? A single X ray photon interacting in a channel, produces a charge pulse of about 1000 electrons. This originates from the rear of the plate. The individual holes confine the electrons, preserving the spatial pattern of the image. The electrons exit the holes and under the influence of a high voltage positive potential, are attracted to a small conducting pad directly beneath the hole of a similar diameter to the hole. This produces an image when the charge is read out. 18. What is the main difference between an MPPC and a semiconductor diode detector? In a semiconductor diode, the reverse voltage is low enough that ionisation is prevented from recombining and is collected by the readout electronics within breakdown of the diode. In an MPPC the voltage is such (-70V) that any ionisation will lead to avalanche breakdown of one of the photodiodes, yielding a “fired” signal. Since the MPPC consists of 100s or 1000s of photodiodes connected together in an array, the number of photons striking the detector is given by the number of fired pixels rather than the current from a single photon. 19. Explain what is meant by a grazing incidence telescope. A GIT features concentric mirrors nested inside each other arranged at suitably shallow angles that energetic X rays are reflected towards a focal plane. These mirrors slowly focus the photons via multiple reflections onto a focal plane. The mirror surface is ceramic coated with a thin iridium metal layer. 20. What is a coded aperture mask? The CAM consists of areas that are transparent or opaque to the desired photon energy range, distributed in a predetermined pattern which casts a shadow on a detector plate. Each point source in the sky creates a unique projection. Since the pattern of the aperture mask is known, the shadow cast by each source on the detector can be used to reconstruct the nature of the original sky. After a certain illumination period, the accumulated detector image may be deconvolved using reconstruction software to reveal the original image of the sky. 21. How are the CCD detectors on the JWST configured to detect infrared radiation? The greatest challenge for the NIRCam is that infrared photon energy is lower than that of visible light requiring the use of semiconductor materials with narrower energy bandgaps. The 6MP CCD is manufactured from mercury-cadmium-telluride. To improve signal to noise ratio – and unlike optical CCDs – the charge collected at each pixel is amplified in situ and read out directly. 22. How does a radio telescope work? Dish focuses the incident radio frequency wave onto the low noise blockdown converter (LNB) at the focal point. Within the LNB a waveguide directs the radio wave past protruding small metal antennae which pick up the signal. The signal is weak so must be amplified. An ultra low noise preamp is used at the first amplification stage to limit noise. High frequency radio waves are difficult to manipulate electronically so the signal must be down converted to lower frequencies. This is achieved by mixing the radio signal with a local oscillator signal of similar frequency to produce beats. The beat frequency output is then amplified and directed through a low pass filter to remove high frequencies. The ac signal is then rectified into a dc signal and recorded. 23. (i) Assuming that no signal delay is applied, what is the minimum baseline separation to ensure correct operation given the angle theta is 60 degrees? (ii) Both dishes are moved to look at an object at theta is 30 degrees. If the baseline doesn’t change, what time delay is now needed on dish 2? (i) (ii) 24. The pp reaction below releases a total of 0.43 MeV of energy. How much energy does the electron neutrino receive? p  p d  e   e There are 3 particles after the reaction. As always, the distribution of momentum and energy cannot violate conservation laws. It is possible to set up a model in which the deuteron and positron have effectively zero velocity and the neutrino has very high velocity in the opposite direction. If this were the case, the neutrino would possess all the energy. But it would also be possible to imagine a model in which the neutrino was stationary and the deuteron and positron had non-zero but different velocities, each in the opposite direction to the other such that linear momentum was conserved. In this case the neutrino would have none of the energy. Therefore the neutrino could possess any energy from 0 to 0.43 MeV. 25. Calculate the ratio of kinetic energies of two reaction products with masses m and M travelling at u and v respectively following a reaction assuming M>>m. 𝑢 𝑀 Conservation of linear momentum gives 𝑚𝑢 = 𝑀𝑣 so = 𝑣 𝑚 𝑘𝑒 𝑜𝑓 𝑠𝑚𝑎𝑙𝑙 𝑚𝑎𝑠𝑠 𝑚 0.5 𝑚𝑢2 𝑚 𝑢 2 𝑚 𝑀 2 𝑀 Ratio of ke = = = ( )( ) = ( )( ) = 𝑘𝑒 𝑜𝑓 𝑙𝑎𝑟𝑔𝑒 𝑚𝑎𝑠𝑠 𝑀 0.5𝑀𝑣 2 𝑀 𝑣 𝑀 𝑚 𝑚 The smaller particle gets most of the kinetic energy. 26. Explain what is meant by an active veto. An active veto comprises a scintillation material positioned around a target such that an incident photon of energy significant enough to pass into the target material will also deposit energy within the veto usually via Compton scattering. This allows the event to be recorded by the target to be vetoed based on the presence of a simultaneous event in the veto of a suitable amplitude. 27. What was the solar neutrino problem? Based on the solar luminosity and energy released in the various PP cycles believed to occur in the Sun, a clear prediction for the number of neutrinos expected at the Earth could be made. Early neutrino detectors in the 1960s consistently recorded fewer solar neutrinos than expected. Neutrinos were believed at that time to be massless meaning that their flavours were fixed when they were produced and therefore all the neutrinos originating from the Sun were believed to be electron neutrinos. All the neutrino detectors at that time were therefore set up to be only sensitive to electron neutrinos and therefore all measurements of the solar neutrino flux were only between a third and a half of the expected result. 28. What were the key discoveries made by Goddard which advanced Rocket Science? He was the first to use turbopumps to pressurise liquid propellant. He was the first to appreciate that although stability necessitated the use of fins at the rear of the craft, fins by themselves were not enough (particularly at low velocities) to control the orientation of the craft. He pioneered the use of gyroscopes connected through linkages to moving surfaces on the fins and also to deflectors located in the exhaust stream of the engine. The latter proved much more effective. 29. What is the difference between actual exhaust velocity and effective exhaust velocity? The actual exhaust velocity is the exit velocity of the hot gases as they leave the rocket. The effective exhaust velocity should be used for all calculations involving rocket flight as it takes into account the additional thrust which results from the pressure difference between the hot gases and the local atmospheric pressure. 𝐹 𝑣𝑒𝑓𝑓 = 𝑡𝑜𝑡𝑎𝑙 for which the total thrust 𝐹𝑡𝑜𝑡𝑎𝑙 = 𝑚̇𝑝 𝑣𝑒 + (𝑃𝑒 − 𝑃𝑎 )𝐴𝑒 where 𝑣𝑒 is effective velocity. 𝑚̇𝑝 30. A spacecraft’s engine ejects mass at a rate of 50 kg/s with an exhaust velocity of 2, 800 ms−1. The pressure at the nozzle exit is 50 kPa and the exit area is 4 m2. What is the effective exhaust velocity of the engine in the vacuum of space? 𝐹𝑡𝑜𝑡𝑎𝑙 = 𝑚̇𝑝 𝑣𝑒 + (𝑃𝑒 − 𝑃𝑎 )𝐴𝑒 = 50(2800) + (50000 − 0)4 = 340,000 N. 𝐹𝑡𝑜𝑡𝑎𝑙 340000 𝑣𝑒𝑓𝑓 = = = 6800 m/s. 𝑚̇𝑝 50 31. A spacecraft’s engine ejects mass at a rate of 50 kg/s with a burn time of 35 seconds. What is the total impulse if the effective exhaust velocity is 6800 m/s? 𝐼 = 𝑀𝑝 𝑣𝑒𝑓𝑓 = (50)35(6800) = 11900000 Ns 32. A spacecraft’s engine ejects mass at a rate of 50 kg/s with a burn time of 35 seconds and its total impulse is 11900000 Ns. What is the motor’s specific impulse? 𝐼 𝑀𝑝 𝑣𝑒𝑓𝑓 𝑣𝑒𝑓𝑓 11900000 6800 𝐼𝑠𝑝 = = = = = = 693.2 𝑠 𝑀𝑝 𝑔 𝑀𝑝 𝑔 𝑔 (50)(35)(9.81) 9.81 33. Falcon 9 achieves a velocity of 500 m/s at an altitude of 17 km in a local atmospheric density of 0.15 kg/m3. If the drag coefficient is 1.2 and the effective frontal area is 11 m 2, what is the power required to maintain this velocity at this altitude? 1 𝐷 = 𝜌𝑣 2 𝐴𝐷 𝐶𝐷 = 0.5 (0.15)5002 (11)(1.2) = 247,500 N 2 At constant velocity forces balance. So thrust = drag. 𝑃𝑜𝑤𝑒𝑟 = 𝐹𝑜𝑟𝑐𝑒 × 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦. Equating the forces gives 𝑃𝑜𝑤𝑒𝑟 = 247,500 × 500 = 123750000 W. 34. The Space Shuttle has an effective wing area of 250 m2 and an effective frontal area of 312 m2. If the 𝐶 ratio 𝐶 𝐿 = 3.4 find the required glide slope angle. 𝐷 1 𝐿 = 𝜌𝑣 2 𝐴𝐿 𝐶𝐿 2  1 2 𝐷= 𝜌𝑣 𝐴𝐷 𝐶𝐷 2 𝐿 𝐴𝐿 𝐶𝐿 250 = = (3.4) = 2.72 𝐷 𝐴𝐷 𝐶𝐷 312 𝐿 1 = = 2.72 𝐷 tan 𝑎 So tan 𝑎 = 0.367 and a = 20.2 degrees. 35. A spacecraft leaves the launch pad accelerating vertically at constant rate of 10 m/s 2. If the air density ℎ is given by the expression 𝜌 = 𝜌0 𝑒𝑥𝑝 (− ℎ ) where ℎ0 = 8 km find the time at which max Q occurs. 0 1 1 ℎ 𝐷 = 2 𝜌𝑣 2 𝐴𝐷 𝐶𝐷 = 2 𝑣 2 𝐴𝐷 𝐶𝐷 𝜌0 𝑒𝑥𝑝 (− ℎ ) 0 1 1 𝑎𝑡 2 𝑣 = 0 + 𝑎𝑡 and ℎ = 0 + 2 𝑎𝑡 2 Sub in gives: 𝐷 = 2 𝑎2 𝑡 2 𝐴𝐷 𝐶𝐷 𝜌0 𝑒𝑥𝑝 (− 2ℎ ) 0 Collect all the constants… 1 2 2 𝑎𝑡 2 𝐷 = 𝑎 𝐴𝐷 𝐶𝐷 𝜌0 𝑡 𝑒𝑥𝑝 (− ) 2 2ℎ0 Call all the constants K. 2 𝑎𝑡 2 𝐷 = 𝐾𝑡 𝑒𝑥𝑝 (− ) 2ℎ0 𝑑𝐷 Max D will be when = 0 i.e. a turning point. 𝑑𝑡 𝑑𝐷 𝑎𝑡 2 2 𝑎𝑡 𝑎𝑡 2 = (2𝐾𝑡)𝑒𝑥𝑝 (− ) + 𝐾𝑡 [− 𝑒𝑥𝑝 (− )] 𝑑𝑡 2ℎ0 ℎ0 2ℎ0 Set this equal to zero. 𝑎𝑡 3 0 = 2𝑡 − ℎ0 2ℎ0 2(8000) So 𝑡 = √ =√ = 40 seconds. 𝑎 10 36. Why following launch must the location of the centre of pressure be aft of the centre of mass in order to ensure stability? The total mass of the spacecraft is said to act at the centre of mass. The total lift and drag forces act at the centre of pressure. Stability is lost when the spacecraft is acted on by an external force which changes the orientation of the spacecraft relative to the oncoming airstream. When this occurs a restoring torque must act and this is provided by the lift and drag forces generated as a result of the increased angle of attack with the oncoming airstream. In order for this torque to act to restore equilibrium the centre of pressure must be aft of the centre of mass. The greater the distance between these centres, the greater the stability. 37. Immediately following launch, a spacecraft’s engine produces a thrust of 200,000 N. If the distance between the centre of mass of the spacecraft and the exit of the rocket nozzle is 10 m, what torque would be produced if the nozzle were to vector its thrust to an angle of 5 degrees from the longitudinal axis of the spacecraft? 5 Torque = FR = 200000 (10sin5)=174311 Nm. 38. A spacecraft’s engine has a specific impulse of 363 seconds and a total thrust of 2 MN. The fully fuelled spacecraft has a total mass of 35,000 kg. You may assume the spacecraft is not in the gravitational field of any planet or its atmosphere, and that its initial velocity is zero. (a) Find the mass ratio required to reach a ∆𝑣 of 7700 m/s. (b) Find the required burn time. ∆𝑣 7700 (a) ∆𝑣 = 𝑣𝑏𝑢𝑟𝑛 − 𝑣0 = 𝐼𝑠𝑝 𝑔 𝑙𝑛[𝑀𝑅] so 𝑀𝑅 = 𝑒𝑥𝑝 𝐼 = 𝑒𝑥𝑝 363×9.81 = 8.69. 𝑠𝑝 𝑔 𝑀𝑝 (b) To find the burn time it is quickest to use 𝑚̇𝑝 = 𝑡. But we need to find the total mass of propellant and the 𝑏𝑢𝑟𝑛 rate of burning it of course. We can find the former from the mass ratio, and we can find the latter from the expression for the total thrust of the rocket. 𝐹 2000000 𝐹 = 𝑚̇𝑝 𝑣𝑒𝑓𝑓 = 𝑚̇𝑝 𝐼𝑠𝑝 𝑔 so 𝑚̇𝑝 = = = 562 kg/s. 𝐼𝑠𝑝 𝑔 3561 1 1 1 1 = 𝑀𝑅 so MR = 1 − 𝑃𝐹 and 𝑃𝐹 = 1 − 𝑀𝑅 = 1 − 8.69 = 0.885 1−𝑃𝐹 𝑀𝑝 𝑃𝐹 = 𝑚 so 𝑀𝑝 = 𝑃𝐹 × 𝑚𝑣𝑜 = 0.885 × 35000 = 30973 kg 𝑣𝑜 𝑀𝑝 𝑀𝑝 30973 𝑚̇𝑝 = 𝑡 so 𝑡𝑏𝑢𝑟𝑛 = 𝑚̇ = 562 = 55.1 seconds. 𝑏𝑢𝑟𝑛 𝑝 39. What are the escape velocities from Earth and Mars given that their respective radii are 6380 km and 3390 km and their masses are 5.97x1024 kg and 6.39x1023 kg. The gravitational constant is 6.67x10-11 Nkg-2m2. 2𝐺𝑀 2×6.67×10−11 ×5.97×1024 Earth 𝑉𝑒𝑠𝑐𝑎𝑝𝑒 = √ =√ = 11.17 km/s 𝑅𝑝 6380000 2𝐺𝑀 2×6.67×10−11 ×6.39×1023 Mars 𝑉𝑒𝑠𝑐𝑎𝑝𝑒 = √ =√ = 5.01 km/s 𝑅𝑝 3390000 40. What velocity in km/s is required to maintain the International Space Station in its orbit 418 km above the Earth’s surface? Radius of Earth is 6380 km. 𝐺𝑀 6.67×10−11 ×5.97×1024 𝑉𝑐 = √ =√ = 7.653 km/s 𝑅 6380000+418000 41. Deduce the radius at the apoapsis (greatest distance between Earth and satellite) and the periapsis (least distance between Earth and satellite) of a satellite that orbits the Earth in an elliptical orbit with eccentricity e = 0.5 given that the semi major axis a is 12,000 km. 𝑎(1−𝑒 2 ) 𝑅 = 1+𝑒 cos 𝜃 At periapsis theta is 0 and at apoapsis theta is 180. 𝑎(1−𝑒 2 ) 12000000(1−0.52 ) Periapsis: 𝑅𝑝 = 1+𝑒 cos 𝜃 = = 6000000 m or 6,000 km. 1+0.5 cos 0 𝑎(1−𝑒 2 ) 12000000(1−0.52 ) Apoapsis: 𝑅𝑎 = 1+𝑒 cos 𝜃 = = 18000000 m or 18,000 km. 1+0.5 cos 180 42. Deduce the velocity at apoapsis and periapsis of the satellite in Q41 that orbits the Earth in an elliptical orbit with eccentricity e = 0.5 given that the gravitational constant is 6.67x10-11 Nkg-2m2 and the mass of the Earth is 5.97x1024 kg, and the semi major axis a is 12,000 km. 2 1 𝑉0 = √𝐺𝑀 ( − ) 𝑅 𝑎 2 1 2 1 Periapsis: 𝑉p = √𝐺𝑀 (𝑅 − 𝑎) = √6.67 × 10−11 × 5.97 × 1024 (6000000 − 12000000) = 9977 m/s 𝑝 2 1 2 1 Apoapsis: 𝑉a = √𝐺𝑀 (𝑅 − 𝑎) = √6.67 × 10−11 × 5.97 × 1024 (18000000 − 12000000) = 3326 m/s 𝑎 43. Calculate the radius of the geostationary orbit from first principles given the gravitational constant is 6.67x10-11 Nkg-2m2 and the radius of the Earth is 6380 km, the mass of the Earth is 5.97x1024 kg, and the length of the sidereal day is 23.93 hours.  𝐺𝑀𝑚 𝐺𝑀 𝐺𝑀 3 𝐺𝑀 3 6.67×10−11 ×5.97×1024 𝐹= = 𝑚𝜔2 𝑅 and thus 𝑅 3 = = 2𝜋 2 𝑅 = √ 2𝜋 2 = √ 2 = 42150000 m 𝑅2 𝜔2 ( ) ( ) ( 2𝜋 ) 𝑇 𝑇 23.93×60×60 Altitude above Earth’s surface is 42,150,000 – 6,380,000 = 35,770,000 m or 35,770 km. 44. This question asks you to occupy the role of mission control on a mission back to the Moon using a Hohmann transfer orbit. You will need to calculate the energy associated with multiple orbits, and also determine the ∆𝑣 associated with the transfer from one to another. Equations you will need….  The velocity of a spacecraft of mass 𝑚 in a stable circular orbit around Earth of mass 𝑀𝐸 at a 𝐺𝑀𝐸 distance 𝑅 between their centres of mass, is given by 𝑉𝑐 = √. 𝑅  The total energy 𝐸 associated with a circular orbit comprising the kinetic and potential energy terms for a spacecraft of mass 𝑚 orbiting around Earth of mass 𝑀𝐸 at a distance 𝑅 𝐺𝑀𝐸 𝑚 between their centres of mass is given by 𝐸 = −. 2𝑅  The gravitational potential energy for a spacecraft of mass 𝑚 orbiting around Earth of mass 𝐺𝑀𝐸 𝑚 𝑀𝐸 at a distance 𝑅 between their centres of mass is given by 𝑈 = − 𝑅 𝐺𝑀𝐸 𝑚  The total energy of an elliptical orbit is given by the expression 𝐸 = − where 𝑎 is 2𝑎 the semi-major axis of the ellipse. B2 B1  The mass of the Earth is 𝑀𝐸 = 6 × 1024 kg and the mass of Space X Starship is 𝑚 = 1320,000 kg and the radius of Earth is 𝑅𝐸 = 6400 km and 𝑅 = 𝑅𝐸 + ℎ where ℎ is the altitude and 𝑅 is a distance from the centre of the Earth. (a) Imagine a Space X Starship in a stable low Earth circular orbit of 500 km altitude as indicated by the blue circle in the figure. Determine its orbital velocity and the total energy associated with this orbit. (b) The spacecraft needs to undergo a Hohmann transfer orbit in order to move into a larger circular orbit of altitude 384,400 km as defined by the red circle in the figure. The radius of this orbit is equal to the distance between the Earth and the Moon. Determine the orbital velocity required to follow this larger circular orbit marked in red, and the total energy associated with this orbit. (c) In order for Starship to move from one circular orbit to another, it must transfer temporarily into an elliptical orbit indicated in green in the figure. Deduce a value for this total energy of the transfer orbit. (d) We can now calculate the velocity injection required at B1 in the diagram. HINT: Imagine that the burn has just taken place at B1 and instantaneously the spacecraft has achieved the higher velocity required to move into the elliptical orbit. At this moment, it is still located at the radius equal to its low Earth orbit, but it now has the energy required by the elliptical orbit. Use this knowledge to determine its new kinetic energy and hence its new velocity and therefore ∆𝑣1. (e) Apply what you have understood from part (d) at B2 in order to deduce the ∆𝑣2 required at this point in order for the spacecraft to assume the larger orbit indicated by the red circle in the figure. 𝑎3 (f) Kepler’s 3rd law expresses the time period for an elliptical orbit as 𝑇 = 2𝜋√𝐺𝑀 where 𝑎 is the 𝐸 semi-major axis. How long will it take Starship to complete the transit orbit from the low Earth orbit to an orbit equivalent to that of the Moon? (g) How much energy will be required to move from the original low Earth orbit (blue) to the lunar orbit (red)? BEFORE  45. Compare the ∆𝑣 , the burn time and the total force produced by (a) a chemical liquid propellant rocket engine and (b) an electromagnetic Hall effect rocket engine in deep space. The amount of propellant is identical (1,200,000 kg) as is the total mass of the spacecraft (1,320,000 kg). The specific impulse is 350 and 4000 respectively. The rate of propellant ejection is 𝑚̇𝑝 = 1500 kg per second for the chemical engine and 𝑚̇𝑝 = 0.1 kg per second for the EM engine. (a) Taking the liquid propellant rocket engine first: 𝑀𝑝 = 𝑚𝑣𝑜 − 𝑚𝑣𝑏𝑢𝑟𝑛𝑡 = 1320000 − 120000 = 1200000 kg 𝑚𝑣0 1320 ∆𝑣 = 𝐼𝑠𝑝 𝑔 𝑙𝑛[𝑀𝑅] = 𝐼𝑠𝑝 𝑔 𝑙𝑛 [ ] = 350(9.81) ln = 8233 m/s 𝑚𝑣𝑏𝑢𝑟𝑛𝑡 120 𝑀𝑝 𝑀𝑝 1200000 𝑚̇𝑝 = 𝑡 so 𝑡𝑏𝑢𝑟𝑛 = 𝑚̇ = 1500 = 800 seconds. 𝑏𝑢𝑟𝑛 𝑝 𝐹 = 𝑚̇𝑝 𝐼𝑠𝑝 𝑔 = 1500(350)9.81 = 5150250 N. (b) Taking the EM Hall effect rocket engine: 𝑀𝑝 = 𝑚𝑣𝑜 − 𝑚𝑣𝑏𝑢𝑟𝑛𝑡 = 1320000 − 120000 = 1200000 kg 𝑚𝑣0 1320 ∆𝑣 = 𝐼𝑠𝑝 𝑔 𝑙𝑛[𝑀𝑅] = 𝐼𝑠𝑝 𝑔 𝑙𝑛 [𝑚 ] = 4000(9.81) ln 120 = 94093 m/s 𝑣𝑏𝑢𝑟𝑛𝑡 𝑀𝑝 𝑀 1200000 𝑚̇𝑝 = 𝑡 so 𝑡𝑏𝑢𝑟𝑛 = 𝑚̇𝑝 = 0.1 = 139 days. 𝑏𝑢𝑟𝑛 𝑝 𝐹 = 𝑚̇𝑝 𝐼𝑠𝑝 𝑔 = 0.1(4000)9.81 = 392.4 N. So you need the chemical rocket to blast off and fight against gravity and reach the circular stable orbit around the Earth but to transfer to another planet you need long burn, high specific impulse engines which will continuously fire to give long term improvements in velocity.