Calculus 2 math project: Calculating orbital work

[AndrewC]

1. My calculus 2 math teacher has us doing a project, involving calculus, on any topic we choose. I chose the topic of calculating work, as in Force x Distance. Specifically calculating work done by lifting a mass into orbit, or to escape velocity. A method is used in my textbook for calculating the work to go from Earth's surface to an orbital altitude using an integral from a to b, and also to escape velocity with the upper limit of the integral being infinity. I wanted to use this same method but instead to calculate the work to go from a low Earth orbit to a low Mars orbit. I'm specifically thinking of the Hermes orbital spacecraft from the Martian movie. I want to calculate how much work it would require to get to mars.

Homework Equations

:[/B]

F=Gm1m2/R^2

Ke=(1/2)mv^2

The Attempt at a Solution

I'm no expert in orbital mechanics so I mostly had to guess on the method I used. My assumption was that once a spacecraft leaves Earth's gravity well, at escape velocity, it will simply coast until it is within the vicinity of Mars. Another assumption is that any orbit the craft is in besides about the sun would be perfectly circular (I realize how impossible this is). This means I wasn't concerned with the path the spacecraft would take. My only concern was how much it would need to use its propulsion system, and thus how much work would be done.

First calculation:
The work required to build the hermes spacecraft in a 1000km orbit around Earth. I couldn't find the actual orbital altitude, so I guessed. Obviously in reality the craft would have been built in sections, with multiple rocket launches, but the total work required would be the same as if it had all been launched at once. I also estimated its mass to be comparable to the ISS, as they are similar in size and structure.

F=Gm1m2/R^2 Definite integral from a to b yields:

Gm1m2(1/a-1/b)
use mass of Earth = 5.98x10^24, gravitational constant = 6.67x10^-11, and mass of spacecraft = 420,000 kg.
let a = radius of Earth = 6.37x10^6 meters, let b = orbital height + Earth radius = 6.37x10^6+10^6 meters.

Plugging in we get 3.56836596×10^12 Joules. This is where i noticed my first problem. If you plug the calculated kinetic energy into Ke=(1/2)mv^2 and solve for v the velocity comes out much to small, it apparently only works for the kinetic energy at escape velocity:

I then calculated the same integral but instead letting a= orbital altitude + Earth radius, b = infinity, and setting up an improper integral. This gave me 2.273x10^13 J. Adding those two together, and setting (1/2)mv^2 equal to the total Joules gives the escape velocity 11,190.75 m/s.

The reason I did the part above as two integrals is the hermes spacecraft would have only needed to use its propulsion system from its orbital height to escape velocity.

Ultimately to find the total Kinetic energy I calculated the kinetic energy the spacecraft would have at Mars escape velocity, which = 5.279x10^12 J. In order to achieve this; after the craft was close to the edge of Mars' sphere of influence it would have to shed 2.10199x10^12 J. That brings it down to the kinetic energy at Mars escape velocity. Tt would require a further reduction of 4.6007x10^12 J to get down to a low 500km Mars orbit. I calculated the energy reduction using the same method as above except with Mars mass and radius.

So the final kinetic energy is:

(kinetic energy from escaping earth) - (kinetic energy slowing down before mars) - (kinetic energy to low Mars orbit) = (2.62984x10^13J) - (2.10199x10^13) - (4.6007x10^12) = 6.783x10^11J at 500km Mars orbit.
The total energy used would be adding 2.273x10^13 + 2.10199x10^13 + 6.783x10^11 = 4.83566x10^13J

Sorry there is so much writing, but I'd really appreciate any feedback on this. If any clarification is needed I apologize I'm quite new to using this forum.

 

[Ray Vickson]

1. My calculus 2 math teacher has us doing a project, involving calculus, on any topic we choose. I chose the topic of calculating work, as in Force x Distance. Specifically calculating work done by lifting a mass into orbit, or to escape velocity. A method is used in my textbook for calculating the work to go from Earth's surface to an orbital altitude using an integral from a to b, and also to escape velocity with the upper limit of the integral being infinity. I wanted to use this same method but instead to calculate the work to go from a low Earth orbit to a low Mars orbit. I'm specifically thinking of the Hermes orbital spacecraft from the Martian movie. I want to calculate how much work it would require to get to mars.

2. Homework Equations :

F=Gm1m2/R^2

Ke=(1/2)mv^23. The Attempt at a Solution

I'm no expert in orbital mechanics so I mostly had to guess on the method I used. My assumption was that once a spacecraft leaves Earth's gravity well, at escape velocity, it will simply coast until it is within the vicinity of Mars. Another assumption is that any orbit the craft is in besides about the sun would be perfectly circular (I realize how impossible this is). This means I wasn't concerned with the path the spacecraft would take. My only concern was how much it would need to use its propulsion system, and thus how much work would be done.

First calculation:
The work required to build the hermes spacecraft in a 1000km orbit around Earth. I couldn't find the actual orbital altitude, so I guessed. Obviously in reality the craft would have been built in sections, with multiple rocket launches, but the total work required would be the same as if it had all been launched at once. I also estimated its mass to be comparable to the ISS, as they are similar in size and structure.

F=Gm1m2/R^2 Definite integral from a to b yields:

Gm1m2(1/a-1/b)
use mass of Earth = 5.98x10^24, gravitational constant = 6.67x10^-11, and mass of spacecraft = 420,000 kg.
let a = radius of Earth = 6.37x10^6 meters, let b = orbital height + Earth radius = 6.37x10^6+10^6 meters.

Plugging in we get 3.56836596×10^12 Joules. This is where i noticed my first problem. If you plug the calculated kinetic energy into Ke=(1/2)mv^2 and solve for v the velocity comes out much to small, it apparently only works for the kinetic energy at escape velocity:

I then calculated the same integral but instead letting a= orbital altitude + Earth radius, b = infinity, and setting up an improper integral. This gave me 2.273x10^13 J. Adding those two together, and setting (1/2)mv^2 equal to the total Joules gives the escape velocity 11,190.75 m/s.

The reason I did the part above as two integrals is the hermes spacecraft would have only needed to use its propulsion system from its orbital height to escape velocity.

Ultimately to find the total Kinetic energy I calculated the kinetic energy the spacecraft would have at Mars escape velocity, which = 5.279x10^12 J. In order to achieve this; after the craft was close to the edge of Mars' sphere of influence it would have to shed 2.10199x10^12 J. That brings it down to the kinetic energy at Mars escape velocity. Tt would require a further reduction of 4.6007x10^12 J to get down to a low 500km Mars orbit. I calculated the energy reduction using the same method as above except with Mars mass and radius.

So the final kinetic energy is:

(kinetic energy from escaping earth) - (kinetic energy slowing down before mars) - (kinetic energy to low Mars orbit) = (2.62984x10^13J) - (2.10199x10^13) - (4.6007x10^12) = 6.783x10^11J at 500km Mars orbit.
The total energy used would be adding 2.273x10^13 + 2.10199x10^13 + 6.783x10^11 = 4.83566x10^13J

Sorry there is so much writing, but I'd really appreciate any feedback on this. If any clarification is needed I apologize I'm quite new to using this forum.

Please avoid posting your message in bold font---it looks like you are yelling at us. I have edited out the bolding in the above, but I cannot do it in your original message---only you can do that. Just go to the start of your message (in Edit), outline the parts you want to "unbold" (by using the mouse to do the selection), then "unclick" the B button at the top of the input panel. I realize that since you are new here you probably did not intend your whole message to come out in bold font, but now that you are aware of the possibility, you should always preview your message to make sure it is as you want it.
Mod note: I edited the original post, as well as this one to get the quotes as they should be.

 

[AndrewC]

Hi sorry it took me so long to respond. Okay I saw it was in bold and I knew it was wrong. I think it was so late at night and I was pretty tired so I didn't take the time to edit it carefully enough, sorry about that. I don't see an option anywhere to edit the original post?