I'm doing a project concerning space travel to Mars, and I need some help with calculations.
This is the description of the assignment:
After completing Assignment 1 we realized that our small spacecraft needs to travel at over 7 km/s to keep orbiting rather than spiraling down towards the Earth. This is fast! At this speed it needs less than 2 hours to circle the entire planet. But how about the force of friction? Wouldn’t this force slow our craft and eventually send it into a spiral anyway? If eventually we get tired of making circles around Earth and attempt to make a run for the Red Planet, what would I need to do? Can I just point my craft in the right direction and let it go?
Besides answering the above questions, this assignment will concentrate on how can we arrive in the 350Km above ground orbit to begin with. Assume that we lunch the spacecraft from the surface of Earth at 85 degrees with the horizontal. Decide how long would you need to accelerate at a constant acceleration - which humans can survive - in order to achieve the cut-off velocity. As discussed in class, when this cut-off velocity is achieved we will turn off the engines and our spacecraft will travel and engage eventually in the 350 km orbit at the speed calculated during Assignment 1. At the cut-off point, our spacecraft becomes a projectile and behaves just like Newton’s cannon. Please consider the progressive diminishing of the Earth’s gravity with altitude and make reasonable approximations as needed.
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I already got the speed at 7,700 km/s. I also have a general picture of the situation...but it's the math that concerns me. Help please?
Thank you very much!
Why do you think you need 7700 km/s? The earth satellite orbital speed needed is about 1000 times less. You need a hgher speed to get to Mars, but not THAT much higher.
The most fuel-efficient way to get to Mars is NOT to aim at it and try to get there as fast as you can. You need to make a carefully timed maneuver, adding orbital velocity to your spacecraft in a tangential direction that adds enough kinetic energy to create an elliptical orbit with perihelion at the earth's orbital distance and aphelion at Mars' orbital distance. Timing is needed to make sure Mars is there when you finally arrive. (Ideally, you should try to arrive at Mars when it is closest to the sun, since its orbit is quite eccentric).
These are some hints to help you complete your project. Good luck wth it.
The trip time to Mars depends on the configuration of the planets at launch, the size of the rocket, and the weight of the payload. It could be as long as 970 days or as quick as 240 days. The following will explain this in more detail. The following was based on the conditions several years ago.
The time for a space probe, launched from Earth, to reach the planet Mars, or any planet for that matter, is a function of the location of the planets relative to one another at the time of launch, the final velocity of departure from earth orbit, and the velocity direction, at burnout of the rocket stage. The exact phasing and distances of the planets from each other, dictates the required magnitude and direction of the velocity. There are fundamentally two extremes to examining the time required to travel to a planet in a direct planet to planet flight. One requires a minimal expenditure of rocket energy but results in the longest trip time. The other requires a huge expenditure of rocket energy but results in a shorter trip time, relatively speaking. The minimum energy, one way trip time, is ~259 days (assuming the average radii of the earth's and Mar's orbits) while the fast track approach could get you there as fast as 70 days or less, depending on the final burnout velocity of the rocket stage. It is also possible to launch at a non-optimum phasing of the planets and take longer than the 259 days, such as the recent Mars Global Surveyor, which took ~10 months to arrive at Mars in September 1997. The Pathfinder spacecraft, traveled on a fast track trajectory, reaching Mars in ~7 months, on July 4, 1997. The following will hopefully shed some additional light on the subject for you.
The minimum energy approach for a probe to reach any planet on its own is by means of the Hohmann Transfer Orbit. By minimum energy, I mean the lowest possible final velocity of the probe as it departs its earth orbit. The Hohmann Transfer Orbit (HTO) is an elliptical orbit that is tangent to both of the orbits of the planets between which the transfer is to be made. In other words, a probe placed into a heliocentric orbit about the Sun would leave the influence of the earth with a velocity vector tangent to the earth's orbit and arrive at the destination planet's orbit with a velocity vector tangent to its orbit. One of the focii of this elliptical orbit is the Sun and the orbit is tangent to both the Earth's orbit and the target planet's orbit.
Lets explore what it takes to send a space probe to our planet in question. Lets assume a launch vehicle has already placed our probe and its auxiliary rocket stage in a circular, 250 mile high, low Earth orbit (LEO) with the required orbital velocity of 25,155 feet per second, fps.
By definition, a probe being launched on a journey to another planet, must be given sufficient velocity to escape the gravitational pull of Earth. A probe that is given exactly escape velocity, will depart the Earth on a parabolic trajectory, and just barely escape the gravitational field. This means that its velocity will be approaching zero as its distance from the center of the Earth approaches infinity. If however, we give our probe more than minimal escape velocity, it will end up on a hyperbolic trajectory and with some finite residual velocity as it approaches infinity, or the edge of the sphere of influence. This residual velocity that the probe retains is called the "hyperbolic excess velocity." When added to the velocity of the Earth in its orbit about the Sun, the result is the heliocentric velocity required to place the probe on the correct Hohmann transfer trajectory to rendezvous with Saturn.
It is worth noting at this point that it is somewhat naive to talk about a space probe reaching infinity and escaping a gravitational field completely. It is somewhat realistic, however, to say that once a probe has reached a great distance (on the order of 500,000 to one million miles) from Earth, it has, for all intensive purposes,
escaped. At these distances, it has already slowed down to very near its hyperbolic excess velocity. It has therefore become convenient to define an imaginary sphere surrounding every gravitational body as the body's "sphere of influence", SOI. When a space probe passes through this SOI, it is said to have truly escaped. Over
the years, it has become difficult to get any two people to agree on exactly where the SOI should be located but, nevertheless, the ficticious boundry is widely used in preliminary lunar and nterplanetary trajectory studies.
As stated earlier, a probe that departs from a 250 mile high LEO with the minimal escape velocity of 36,605 feet per second will end up at the edge of the Earth's SOI with near zero velocity and remain there forever. We must then give our probe a final Earth departure velocity in excess of minimal escape velocity. Lets back into this required velocity in the following manner.
The final heliocentric velocity required by our probe for a Hohmann transfer to Mars is ~107,350 fps, relative to the Sun. Since the probe, at the beginning of its journey, picks up and retains the velocity of the Earth about the Sun (97,700 fps), we can subtract this from our required heliocentric velocity to determine our required hyperbolic excess velocity. Thus, 107,350 - 97,700 = 9,650 fps, the hyperbolic excess velocity required by our probe at the edge of the Earth's SOI. To achieve this hyperbolic excess velocity, on the hyperbolic escape trajectory from our 250 mile parking orbit, requires a final rocket burnout velocity of 36,889 fps, relative to the Earth. The velocity vector must nominally be perpendicular to the radius from the Sun, through the center of the Earth and parallel to the Earth's direction around the Sun. Since our circular parking orbit velocity is already 25,155 fps, we need therefore only add a velocity increment (deltaV) of 11,734 fps with our auxiliary rocket stage burn.
Our launch vehicle has already delivered us to our 250 mile LEO. The auxiliary stage now fires, imparting its impulsive velocity change of 11,734 fps to push us out of the circular orbit onto our required hyperbolic orbit from which it escapes the Earth altogether. The journey would take ~259 days to travel the 180 degrees around the Sun and arrive in the vicinity of Mars. Using the more exact perihelion and aphelion distances for Mars, ~128,416,000 miles and ~154,868,000 miles, and 93,000,000 miles for the earth's distance from the sun, the Hohmann Transfer trip times would be ~238 days and ~281 days respectively. Obviously, the phasing of the planets is important for this type of transfer and the launch must occur at a carefully predetermined time such that both Mars, and the probe, will reach the heliocentric rendezvous vicinity at the same time.
The disadvantage of this type of trajectory is that when the spacecraft has reached Mars, the earth is too far past the position from which a reverse Hohmann Transfer trajectory could return the spacecraft back to earth. So the spacecraft would either stay in this elliptical trajectory until such time as the earth might meet up with it in the future or it could be placed into a permanent orbit around Mars. Obviously such a mission would be unmanned.
If it was desirable to place a manned spacecraft into a Martian orbit after a Hohmann Transfer trip to Mars, and eventually return it to Earth, they would have to remain there for 450 days until the planets were in the correct reverse configuration that would allow a return Hohmann Transfer trajectory back to Earth. Needless to
say, such a mission would require an abundance of life support equipment and supplies and last a total of 970 days, or 2 years, 7 months and 24 days.
If bigger rockets were available to deliver the spacecraft out of earth orbit, it could reach Mars in shorter times during what is called opposition class missions. The planets are considered in opposition when the Sun, Earth, and Mars are all on a straight line and the Sun and Mars are on opposite sides of the Earth. One version of such a mission might take a total of 240 days. The outbound trip time would take 120 days with the spacecraft reaching Mars at opposition, passing Mars by and returning to Earth in the reverse of the trajectory it took to get to Mars in the first place. This would take a bigger delivery rocket than that required to send the spacecraft on a Hohmann Transfer trajectory. Another version, also taking a total of 240 days, is one where the spacecraft goes into Martian orbit for 30 days and then returns to Earth. The outbound trip in this case would require 105 days and the spacecraft would reach Mars 15 days before opposition, go into orbit around Mars, wait 30 days until 15 days after opposition, and depart Mars on a 105 day return trip. This mission, though taking the same 240 days to complete, would require a bigger rocket yet as the trip time had to be reduced to 105 days. Clearly, there are other, more expensive, choices of departure velocities for even shorter trip times. The choice is primarily a function of how much money one is willing to spend on a launch vehicle and the launch site hardware.
Now all of the above assumes the probe is totally on its own for the entire journey. Another approach to reaching another planet is to launch the probe toward an intermediary planet, say Venus for instance, and let Venus alter the trajectory of the probe through what is typically referred to as the sling shot or swingby approach. The use of the term slingshot is somewhat misleading however as it tends to imply that the intermediate planet throws, or hurls, the probe away from it toward another planet with added energy or velocity relative to the turning planet, which it does not. In reality, a probe that enters the sphere of influence of another planets gravitational field, will be pulled in toward the planet along a hyperbolic trajectory to some predetermined minimum altitude, called the perigee altitude, increasing its velocity all the way, and once passing this minimum altitude point, will continue out and away from the planet on the exact mirror image trajectory that it came in on. When the probe again reaches the edge of the planets sphere of influence, it will be traveling at the exact same speed, relative to the turning planet, that it entered the sphere of influence with in the first place, but, its direction will have been changed such that it is now heading toward our destination planet, with the added velocity of the turning planet's orbital velocity. This hyperbolic encounter, in addition to altering the direction and heliocentric velocity of the probe, also alters the eccentricity and and energy of the probe's orbit with respect to the Sun. This is where the term slingshot came into use as the energy level of the probe is changed with respect to the Sun but not the passive turning planet. Another option in utilizing a planet to change direction is to add some velocity to the probe when it passes through its perigee altitude by means of another auxilliary rocket stage. Under this type of scenario, the exiting trajectory of the probe will be different from the entering trajectory.
The utilization of the swingby method to alter the trajectory of a probe can often be accomplished with less energy than the HTO method, though not always. The key advantage of the swingby approach is usually the reduction of time for the probe to reach its destination planet. The locations of all the planets concerned, and the control of the probe throughout, become much more critical for the success of such a mission.
Got to be kidding me one guy 7700 value but doesnt let alone nobody reviews this nutters response
The response is nutter because there are no data to make the attempt at a semi valid question contemplative
To solve this problem, you will need to use the principles of physics, specifically the laws of motion and gravity. Here's a step-by-step guide to help you with the calculations:
1. Understand the problem:
- The spacecraft needs to achieve a speed of 7 km/s to maintain a stable orbit around Earth and avoid spiraling down.
- You need to determine how long the spacecraft should accelerate at a constant acceleration to reach this speed.
- The spacecraft will be launched from the surface of Earth at an angle of 85 degrees with the horizontal.
- Consider the progressive diminishing of Earth's gravity with altitude.
2. Define the variables:
- Let's denote the time needed to accelerate as 't'.
- The acceleration required for humans to survive will be denoted as 'a'.
- The cut-off velocity needed to achieve for orbit is 'v'.
- The angle of launch from the horizontal is given as 85 degrees.
3. Find the initial velocity:
- Assuming the spacecraft starts from rest, the initial velocity can be calculated using the equation: v₀ = at, where 'v₀' is the initial velocity.
- Rearrange the equation to solve for 'v₀': v₀ = at. Since the acceleration is constant, you can use the average acceleration for this calculation.
- Substitute the given angle of launch as 85 degrees and convert it to radians (85*(π/180)).
4. Calculate the time needed to accelerate:
- Rearrange the equation v = u + at to solve for time 't'.
- Substitute the initial velocity 'v₀' as calculated in step 3, and the required velocity 'v' as 7 km/s.
- Solve for 't' using the rearranged equation.
- This will give you the time needed to accelerate to the required speed.
5. Take into account Earth's diminishing gravity:
- As the spacecraft moves away from Earth's surface, the force of gravity decreases. You'll need to make reasonable approximations for this.
- Consider using the formula for the gravitational force: F = G * (m₁ * m₂) / r², where 'G' is the gravitational constant, 'm₁' and 'm₂' are the masses of two objects, and 'r' is the distance between them.
- Since the spacecraft is relatively small compared to Earth, you can approximate the force of gravity as decreasing linearly with altitude.
6. Perform the calculations:
- Substitute the value of 't' obtained from step 4 into the equation v = u + at to calculate the final velocity 'v'.
- Calculate the distance 's' traveled during acceleration using the formula s = ut + (1/2)at².
- Determine the altitude at the cut-off velocity using the distance traveled.
Remember to use consistent units throughout your calculations and be mindful of any approximations made. Good luck with your project!