Analyzing Relativistic Effects on Low-Flying Satellites

AI Thread Summary
The discussion focuses on calculating the relativistic factor gamma for a low-flying satellite traveling at 8000 m/s. The user finds that both direct calculation and Taylor series expansion yield nearly identical results, leading to confusion about the utility of the two methods. It is clarified that while the results appear the same within the limits of the calculator's precision, they are not truly identical, especially at lower velocities. The conversation emphasizes the importance of understanding the differences in results when using approximations versus exact calculations. Ultimately, the discussion highlights the nuances of relativistic effects in practical applications.
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Homework Statement


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A low flying Earth satellite travels at about 8000m/s. For the satellite, the relativistic factor $$\gamma = \frac{1}{\sqrt{1-\beta^2}}$$ where $$\beta = \frac{v}{c}$$ is close to 1 because v<<c. Estimate by how much gamma actually deviates from 1 by expanding gamma in a taylor series and evaluating all terms up to second order.

Homework Equations

The Attempt at a Solution


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To be sure, I plugged in 8000 for v in gamma originally. That is, I evaluated $$\gamma = \frac{1}{\sqrt{1-\beta^2}}$$ with $$\beta = \frac{8000}{3\times10^8}$$ When I put this gamma into my TI-89 and used 12-point-float, it came back with 1.00000000036. This makes sense, since the satellite is moving so slow compared to the speed of light. However, when I did the taylor expansion, I came up with $$\gamma \approx 1 + \frac{\beta^2}{2}$$ and when I put in the aforementioned beta value, i came up with the EXACT same value as before. So my question is a semantic one I suppose: what is the question wanting me to do? How can I get a more useful number when both "techniques" yield the same number?
 
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mch said:

Homework Statement


[/B]
A low flying Earth satellite travels at about 8000m/s. For the satellite, the relativistic factor $$\gamma = \frac{1}{\sqrt{1-\beta^2}}$$ where $$\beta = \frac{v}{c}$$ is close to 1 because v<<c. Estimate by how much gamma actually deviates from 1 by expanding gamma in a taylor series and evaluating all terms up to second order.

Homework Equations

The Attempt at a Solution


[/B]
To be sure, I plugged in 8000 for v in gamma originally. That is, I evaluated $$\gamma = \frac{1}{\sqrt{1-\beta^2}}$$ with $$\beta = \frac{8000}{3\times10^8}$$ When I put this gamma into my TI-89 and used 12-point-float, it came back with 1.00000000036. This makes sense, since the satellite is moving so slow compared to the speed of light. However, when I did the taylor expansion, I came up with $$\gamma \approx 1 + \frac{\beta^2}{2}$$ and when I put in the aforementioned beta value, i came up with the EXACT same value as before. So my question is a semantic one I suppose: what is the question wanting me to do? How can I get a more useful number when both "techniques" yield the same number?
They're not exactly the same value. They're only the same to the number of digits that your calculator has provided. If it could provide more digits (without roundoff), the numbers would be different. How would you calculator do if the velocity was only 1 meter per second? How would your approximate equation do?

Chet
 
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