You are observing the motion of Mars across the night sky from Earth and note that it occasionally appears to reverse direction and loop backwards in its path. Summarise Kepler’s 3 laws of planetary motion and use them to explain why this phenomenon occurs?
Kepler’s 3 laws can be classified into the “Law of orbit”, “Law of areas” and the “Law of periods. The first law, (Law of orbits) states that planets travel in an elliptical path with the sun at one focus. The second law, (Law of areas) states that planets, or anything else, moving in orbit “sweep out” equal areas in equal time.
A diagram from the website “http://oneminuteastronomer.com” explaining this is found to the right. The last law, (law of periods) states that the square of the period of a planet is proportional to the cube of the semi-major axis of its orbit (average distance between the Sun and the planet).
The phenomenon mentioned in the question above is called the retrograde effect. It is an optical illusion that makes it look like mars decides to stop and do a loop every 2 years (this is from the perspective of an observer from earth). A picture from Nasa describes this effect:
Mars appears to do this because our earth travels around the sun twice in the same time that mars does once (This is due to the third law (the semi-major axis of earth is smaller than that of mars, this means we’re closer to the sun)). For this to happen, we must overlap mars; while we’re catching up to mars it appears to slow down a stop (relative motion) and then start moving again. Furthermore, our orbit is on a slight different plane to Mars’ and therefor the loop illusion can be seen.
Explain and define the terms in Newton’s Universal law of gravitational attraction. Why does it lead to planetary motion that we observe?
Newton’s law of gravitational attraction states that: “ Fg = – (GMm) / (r2) ”.
“Fg” is the force of gravitation attraction. “G” is the gravitation constant, its value is approximately 6.674×10−11 N⋅m2/kg2, the negative sign that appears before it is just convention as the force is attraction. “Mm” is two masses being multiplied together, the little “m” is the mass of the orbiting object whereas the big “M” is the mass of the object that the first object is orbiting. These masses show that two masses generate an attractive force between themselves. “r2” is the radius of the two objects in the orbit squared. This can be changed to find the gravitational energy “ Eg = – (GMm) / (r) ”.
Objects in orbit all have this gravitation force/energy as they themselves have a mass, the “r” or “r2” term’s value will change if the orbit is not circular (due to properties of ellipsis), this means that the total gravitational energy will go down, but it is converted into kinetic energy (0.5*m*V2) which is why planets speed up when their radii increases.
Given that there are 365 days in 1 Earth year, and the mean distance of Earth
from sun is 1 AU (Astronomical unit), look up the mean Mars distance from the
sun and calculate the period of its orbit around the sun in days.
The mean mars distance from the sun is “227,940,000 km” according to NASA. This is 1.52368 (5.d.p) AU. This AU is a relationship between the mean distance of Earth
from sun and the mean distance between mars and the sun. This relationship is devoted by “d”, and an equation we can use is “d3 = t2” (Kepler’s third law) (when using the sun as a point of origin (huge mass)) where “t” is the ratio factor for the time taken in days. Therefore, “t = (1.523683)0.5” which equals 1.8808 (5.d.p); multiplying this value by the time taken for earth to go around the sun will give you the value for how long mars takes in earth days, this value (by taking 365 as an earth year) equals: “1.8808*365” = 686 days.
This is Mars’ period, taking earth’s year to a more precise value would give a more precise value of Mars’ period.
State Newton’s equation for motion of a planet in orbit about the sun and define
each of the terms. What 4 types of planetary motion paths does this equation
describe, and what is the name for this general class of curve? What is the
physical meaning of the special case for ε = 1.
The Newtonian equation for motion of a planet in orbit around the sun is as followed:
1/r= GM/(r_0^2 v_0^2 ) (1+ε cosθ )
It can be arranged to make ε the subject:
ε= ⌈(r_0 v_0^2)/GM-1⌉
The equation above has many terms; “r” is the radius (distance that the planet is away from the sun at any time,) whereas “r0” is the distance at closest approach (smallest radius); “G” is the gravitational constant (6.67*10-11 ) and “M” is the mass of the sun; “V0” is the maximum velocity of the planet in its orbit (it will be at the point of closest approach); “ε” is the eccentricity of the orbit (a measure of how elliptical) and “cosθ” is the cosine function of the position angle that the planet is in it’s orbit.
The 4 types of planetary motion paths are: Circular, Elliptical, Parabola, Hyperbola. These are all conic sections by nature (general class of curve) and assuming “ε = 1” is referring to eccentricity, the special case is a parabolic trajectory (Kepler orbit). While moving away from a source, it is called parabolic escape trajectory, otherwise its known as the parabolic capture trajectory.
Calculate the period of the transfer orbit to Mars in days, assuming a minimum energy trajectory, and circular Mars and Earth orbits.
The same system can be used for this question as was used in question 3. In this question, I used the formula: “d3 = t2 “. The equation for working out d is as followed:
“d = a = (R1 + R2)/2”. from earlier, we know R1 = 1.5236 (4.d.p) and R2 = 1, therefor:
“ a = (1.5236 + 1) / 2 = 1.2618 “. Therefor: “1.26183 = t2”. This means that t = 1.417 (3.d.p) years which equals 517 days (taking a year to be 365 days). Dividing this by two gives you 259 days, which is half of the orbit which is the transfer orbit.
Determine, with the aid of two sketches and a description showing Mars / Earth orbits and the outbound and inbound transfer orbit, the minimum duration in years for an Earth Mars return mission using Hohmann transfers. Remember that both Earth and Mars are moving during the transfer, and your crew will need to wait on Mars for planetary realignment before returning. It is recommended you break your working into 4 parts, the outbound journey, the wait on Mars, the return journey, and the summation of the parts. Don’t forget to convert your units!
The Mars stay time mention above is evaluated by working out the speed at which Mars travels around the sun “(Period of mars – Period of earth) / (Period of Mars)”. This ratio is used to work out how many days until earth and Mars are aligned again, this is how long you would have to wait on mars.
List and briefly describe 3 potential hazards that space travellers will face on an interplanetary journey of the duration you have calculated above.
For the duration mentioned above, many hazards could potentially occur. For instance, there are tonnes of different space debris (whether it be man-made stage parts or an asteroid field) hurling through space at incredibly fast velocity that on impact with a spacecraft or space suit would generate tremendous force and be disastrous and deadly to the travelling space explorers. Another possibility is that the astronauts could experience psychological dangers as many have done in past due to depression, lack of gravity and loneliness. Another hazardous risk is that space is full of highly charged partials (radiation) that on impact with humans can be deadly, it destroys cells and can cause cancer. These are just a few examples of the hazards.
Calculate the velocity change needed for a Hyperbolic departure from Earth onto minimum energy Mars transfer orbit. Assume the following: • an Earth velocity around the sun of 29.8km/s). • A 500km circular equatorial parking orbit around Earth (for which the circular velocity is 7.62km/s) The overall velocity change to go from low Earth to low Mars orbit is 6.07km/s. Explain briefly difference between the Earth escape number you worked out above, and this new figure of 6.07km/s? Given the figures above, wow would you then calculate the velocity changes for the return trajectory and hence the overall round trip delta-V (not including descent from Mars orbit to Mars surface and back again)?
To reach the Mars surface from a circular Mars orbit of altitude 500km, the orbital velocity must be mitigated. Describe potential approaches for soft landing on the Mars surface, and approaches to reducing the velocity change that the propulsion system must deliver. Discuss the range of options to reduce the required landing velocity and then to achieve a soft landing on the surface of Mars.
To safely land on the surface of mars would require lots of careful planning. The first thing that would be requited is a heat shield. Entering the atmosphere at high speeds will result in a lot of friction which in turn results in a lot of heat; heat shields take excess heat away from what needs to land (what happens to this heat depends on the heat shield) and hence protecting it. Heat shields can also lower the velocity of the lander by up to 90%, a parachute should then be used to slow it down even more.
The lander can still be at 100 miles per hour at this stage so it would not be a soft landing. A correct landing system must be deployed to make sure the lander lands softly, for small and medium sized rockets a combination of retro rockets (a small auxiliary rocket fired in the direction of travel) and airbags are used. Impact is usually 30 miles per hour and the lander will bounce to a stop; soft landing.
Larger landers will use more retro rockets which would also be more powerful. Instead of airbags, landing legs will be available to use as the lander would be descending at only 6 miles per hour; soft landing.
For much larger Landers, retro rockets would not be enough, instead a Jetpack system would be used to slow descent to two miles per hour. The lander would then be lowered down from the jetpack by cables so that the lander can land on its wheels; soft landing.
If the total landing velocity delta-V (Descend from Mars orbit to Mars surface) is 3.4km/s, and the same delta-V is required to go from the surface to low Mars orbit before escape, determine the overall delta-V requirement for the mission?
This is the equation for the total velocity change for a mission where V1 equals
“ ”. We are told that and that V1 = 3.4km/s;
this means that = 3.4 + 3.4 = 6.8km/s.
However, this is only the total change in velocity for leaving and entering Mars; the actual total would include the change in velocity for getting to Mars and back, we’re are given this in question 8. Therefore:
= 6.8km/s + 6.07km/s + 6.07km/s = 18.94km/s
How might the transfer time for a minimum energy / Hohmann trajectory be improved to reduce the amount of time the astronauts spend in interplanetary space? What are the disadvantages of such an approach?
According to the book “Multiple Gravity assists – interplanetary trajectories” by A.V. Labunsky, the transfer time for a minimum energy Hohmann trajectory can be reduced by getting into the gravitational field of Ganymede (which is Jupiter’s largest moon) as fast as possible and using it’s angular momentum to “slingshot” around the moon and leave it’s field with a path trajectory for Mars (Or where Mars would be when you cross your paths). This means you don’t have to wait the two years for mars to be in perfect alignment. This method does have disadvantages however as you would need more propellant to travel a further distance and at a faster velocity so current technology would not be efficient enough; the cost would be too great. You would also have to make sure that Ganymede will be in the right position when you get there.
A rough sketch of the trajectory can be seen below:
The same thing can be done with Venus; this would lower the total mission time by around 400 days. Another way to improve on this value is to use a hyperbolic path (opposed to a circular path) with much higher change in velocities. This would require more thrust which requires more energy; however, this would result in a trajectory that is no longer a minimum energy transfer.
Briefly describe two possible approaches, one making use of Mars’ atmosphere and the other the planet Venus, that would reduce the overall delta-V and / or duration requirement for the mission? Your answers should make use of the terms ‘energy’ and ‘momentum’.
One possible approach for reducing the overall change in velocity is making use of Mars’ atmosphere. This requires using a spacecraft that has a larger enough surface area to provide a large enough atmospheric drag that counteracts the force acting on the spacecraft due to gravity resulting in less velocity which also results in less kinetic energy (0.5mv2). There would be less of an overall impulse created (due to Newton’s third law) meaning the spacecraft would have less momentum. This means propulsion methods do not need to be used to slow the spacecraft down.
Another method in using a gravity assist. These are also useful as they allow the increase or decrease in velocity with needing any propellant. They work as planets retain most of the solar system’s angular momentum which can be utilized to accelerate the spacecraft. By starting in orbit on earth, heading towards Venus (when in correct alignment) would increase the spacecraft’s velocity as it’s distance from the sun would be decreasing. The velocity would be increased even more if the spacecraft entered Venus’ orbit and then left it again in a “slingshot manoeuvre”. Angular momentum is added to the spacecraft due to this manoeuvre, hence the kinetic energy of the spacecraft would increase. The spacecraft could then do the same to the earth to increase in velocity even more before reaching Mars at a much higher velocity than it would be if it went straight there from earth’s orbit.
Both these methods reduce the need for thrusters to reach high speed, this means that a means of propulsion with a lower specific impulse (or effective exhaust velocity) can be used without dropping the kinetic energy or momentum of the spacecraft; this also means you can carry less propellant, as you don’t need as much thrust.
Diagrams of these methods can be seen below:
The rocket equation is “delta-V = ve ln (M0 / Mf)”, as the overall ratio of (M0 / Mf) and ve will be lowered, the overall delta-V will also be lowered. This also means that the duration requirement of mission would decrease.
Assuming the overall delta-V for the mission can be reduced to 10.93km/s by use of various techniques, use the rocket equation to calculate the overall rocket propellant requirement per unit of payload. It is recommended you calculate the mass ratio first. Remember also that Initial mass = (structure + payload + propellant) and final mass = (structure + payload). Assume the propulsion stages use LOX / LH2 propellants with Isp 450s, that the payload mass is 47t (based on the Apollo lunar mission), and that the structural mass is about 5% of the propellant.
The rocket equation is “delta-V = ve ln (M0 / Mf)”, this can be re-written as
“delta-V = Isp * g0 * ln (M0 / Mf)”. Delta-V can be reduced to 10.93km/s, the propulsion system uses a specific impulse of 450s, and the payload mass is 47 tonnes (which is 47000 Kg). We also know that the structural mass is 5% of the propellant. Know all this; we can deduce that the Mass ratio is equal to: “(propellant mass + 47000Kg + (propellant mass / 20)) / ((propellant mass / 20) + 47000Kg)”. We now know every part of the rocket equation apart from the propellant mass. The working or this is all followed:
Assume each astronaut requires 5-10kg of essential consumables (water, air, food) per day in life support. Assuming a crew of 6, and a spacecraft empty (dry) mass of 47t, estimate the overall payload mass, including astronauts & consumables? What role would recycling play in changing this figure? And what impact would this have on the overall mission propellant? You will need to use your answer to the previous question.
If we assume that an average of 7.5 kg ((5+10)/2) of essentials are consumed each day by life support for each astronaut, which have an average mass of 77kg (Estimated mass of Neil Armstrong) each. How long it takes to get to Mars will depend on the speed you are travelling, NASA is developing technology that should be able to get to Mars in around 40 days, but for this purpose I will take the result from question 6 (973 days). This means that the overall payload mass equals:
47000kg + (((7.5kg * 973 days) + 77Kg of astronaut) * 6 astronauts) = 91247Kg of payload.
With recycling of the essential consumables, this number would go down by a big amount as you wouldn’t need 7.5 kg for each astronaut each day. Assuming all air and water can be completely recycled and re-used (100% efficiency) only the mass of food would be needed each day. This would change the equation to:
47000kg + (5kg of air and water + (2.5kg of food * 973 days)) * 6 astronauts = 61925Kg.
This means you would need less mission propellant as payload mass has decreased.
Estimate how many Ariane 5 launches would be required to launch just the propellant you have calculated as needed the Mars mission elements (propellant and life support consumables as calculated, spacecraft mass as per Apollo at 47t) to Low Earth orbit, assuming 48degree inclination and 500km altitude? Why would your answer probably underestimate the total mission mass?
The Ariane 5 evolution storable variant of the rocket has a launch capacity of 21000kg payload to LEO. This is from NASA’s database about the Ariane 5 but the L/V per graph gives 20000Kg payload. Our total payload is 91247Kg (without recycling). Therefore “91247kg / 20000kg = 4.56 launches” You cannot have a fraction of a launch so this is equivalent of 5 launches to put the payload into space.
(With recycling considered this equates to 3 launches)
To put the total propellant calculated into space you would need (1124481/20000=56.2) 57 launches.
This is probably an under estimation as payload mass might needed to be changed or extra propellant might be needed to get the craft to dock correctly. It might also be an underestimation if you do not include the payload mass (essentials).
Discuss quantitatively at least 3 impacts of Mars mission design on solar array performance if you are to adapt arrays originally designed for (low) Earth orbit.
Solar arrays originally designed for low earth orbit wouldn’t be very useful the closer it gets to Mars; this is for a variety of reasons. One of these reasons is that the solar intensity (the number of photons from the sun in a given area) is inversely proportional to the square of the distance of the sun. This means that further away from the sun, the less photons, the less photons, the smaller the amount of electricity that a solar array can produce. The solar arrays, while in low earth orbit, may be able to produce enough energy to keep a space craft running but as the space craft gets further away from the sun (to Mars), the energy produced may not be enough. Therefore, the surface area of these arrays would need to adjusted and increased to make sure they can provide enough electricity for the core systems.
The other main problem is that Mars is a very dusty planet, lots of this space dirt can get on the solar array surface and reduce its efficiently by up to 80 percent (and even more). This is said to be the cause of past mission failures such as Phoenix’s descent. This problem is worsened as Mars is plagued by dust storms, the only way to avoid this is to create a self-cleaning system or create the surface out of something that the dust won’t stick to. Debates are going on with scientists to determine the best source of energy for future mars’ colonies.
In the recent Matt Damon film ‘the Martian’, the astronauts travelled between Earth and Mars orbit in the Hermes spacecraft. What 3 aspects of this space vehicle were unique compared to all other interplanetary spacecraft?
“Hermes” is the name of a spacecraft in the film Martian, opposed to the current ways of getting something to Mars, the spacecraft is a re-usable “Taxi” in a sense. The spacecraft is theoretical but it’s background science is real. Individual missions dock with “Hermes”, this allows Hermes to never have the need to land and just travel between Mars and Earth repeatedly (Or stay LEO). To do this, Hermes uses a more advanced concept than that was conceived by Walter Hohmann in 1925. He proposed that every 26 months Earth and Mars are in the right position for an Orbit between them to be travelled at a low change in velocity (delta-V). Earth’s and Mars’ positions relative to launch and arrival can be seen below:
https://briankoberlein.com – visited 22/11/2016,
This is called the Hohmann orbit, it is not as simple as it seems as each orbit would have a slightly different orientation and Earth and Mars are not quite in the same plane. Therefore, this large spacecraft would need a thrusting force to adjust its orbit. It would need this thrust very gradually so a chemical source would generate too much thrust for not a long enough period. Hermes’s transfer is closer to a Brachistochrone trajectory (curve) Ion drives would be the more optimised as they generate a constant small thrust by accelerating charged particles (the novel of the film states that the ship moves at 2mms-2). Current interplanetary spacecraft does not travel in this Hohmann orbit and they are not re-usable (a re-usable technology is currently being researched and designed); current technology does not provide ion drives that are powerful enough to move a big space craft such as Hermes.
As a side note, Hermes