Calculus in the velocity and acceleration of satellites.

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Homework Help Overview

The discussion revolves around the application of calculus to understand the velocity and acceleration of satellites in relation to their distance from Earth. The original poster is exploring how to incorporate calculus into their project, particularly in the context of orbital mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive acceleration from the orbital speed equation but encounters discrepancies with accepted formulas. They question the role of the radius in their calculations and consider the relevance of calculus in their project.

Discussion Status

Participants are actively engaging with the original poster's attempts, providing clarifications about the nature of velocity and acceleration as vectors. Some suggest that the differentiation approach may not be appropriate for finding acceleration directly from speed. There is an ongoing exploration of the relationship between these concepts and the implications of using calculus.

Contextual Notes

The original poster mentions their limited experience with calculus and vectors, which may affect their understanding of the problem. They also express a desire to relate their findings to practical measurements of satellite distances from Earth.

VinnyO

Homework Statement


I am working on a project dealing with the velocity and acceleration of satellites based on their distance from Earth. I was recommended to include some calculus in this project. Originally I thought I could just take the derivative of the orbital speed equation to find acceleration, given the mass of the Earth and the gravitational constant would be constants in the situation. When I tested this hypothesis with the accepted acceleration equation the answers came out different. I am fairly sure this is because the remaining variable is R (the radius of the orbit) as opposed to time, but perhaps I am making a different mistake. So at this point I am in search for a way to make calculus relevant in this situation.

Homework Equations


v = SQRT(G*M/R)
a = G*M/R^2

The Attempt at a Solution


I also have been looking for a way to calculate the distance of objects like the ISS from Earth based on observations/ measurements I can take myself (from Earth), but I assume any solution to this would be related to trigonometry or geometry.
 
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VinnyO said:
take the derivative of the orbital speed equation to find acceleration
Remember that velocity and acceleration are vectors. Differentiating a speed (a scalar) does not give acceleration.
Please post your algebra.
 
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Remember that acceleration is change in velocity with respect to change in time
 
Moved to Calculus & Beyond section
 
haruspex said:
Remember that velocity and acceleration are vectors. Differentiating a speed (a scalar) does not give acceleration.
Please post your algebra.
haruspex said:
Remember that velocity and acceleration are vectors. Differentiating a speed (a scalar) does not give acceleration.
Please post your algebra.
Thanks for your reply!
Currently I have only taken one year of calculus in which I have not learned vectors (the report is for math in the program I am in and because I have taken calculus I was encouraged to utilize that skill). However with a bit of research on the basics of such I realize how obvious my mistake is. With that in mind my first though was that perhaps I could find velocity given the acceleration formula with some integration. But anyways in short this was my math:
I took an online example where R (the radius of the orbit) was 6.47 x 10^6 m. When plugged into the acceleration formula, a = (6.673 x 10^-11 N m^2/kg^2) • (5.98 x 10^24 kg) / (6.47 x 106 m)^2, a (acceleration) was equal to 9.53 meters per second squared.
To test my idea to use calculus I took the orbital speed equation, v = SQRT(G*M/R), and plugged in the gravitational constant, 6.673 x 10-11 N m^2/kg^2, for G and the mass of Earth, 5.98 x 10^24 kg, for M
I than simplified this to v=998806037 / 50*R^1/2
The derivative of this was a= -998806037 / 100R^3/2
I finally just plugged in the same R value into the new potential acceleration equation, a= -998806037 / 100(6.47x10^6)^3/2, and got a=6.07x10^-4 which was not equal to the answer produced by the equation so I knew I made some wrong assumption.
 
VinnyO said:
The derivative of this was a= -998806037 / 100R^3/2
That is the derivative wrt radius, dv/dR. Acceleration is the derivative of velocity wrt time, dv/dt.
 

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