How to solve this problem in sinusoidal motion? (Satellite orbiting the Earth)

AI Thread Summary
The discussion focuses on solving a problem related to sinusoidal motion in the context of a satellite orbiting Earth. The original poster calculated the angular frequency and frequency, finding a very small frequency of approximately 10^-5 s^-1, which raised concerns about potential errors in their solution. Participants suggested comparing the calculated altitude for geosynchronous orbit with known values and emphasized checking units for consistency. They noted that while the formula used might be incorrect, the numerical result aligns with expected outcomes. Additionally, they prompted a reevaluation of the expected acceleration at a radius of 42,000 km compared to the acceleration at Earth's surface.
Safinaz
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Homework Statement
A satellite orbits the Earth at a speed of 3100 m/s in an orbit that passes through the north and south points and has a radius of ##4.2 \times10^7## m. Consider a point moving in a straight line along the north-south axis of the Earth and passing through its center such that its speed is always equal to the component of the satellite's velocity in the north-south direction. Find a) the frequency of the point's motion, b) the acceleration of the point at the two end points of the motion, and c) its maximum speed.
Relevant Equations
The angular frequency : ## \omega= \frac{v}{x}##

The frequency ## f= 2 \pi \omega ##

For a SHM: ## a = - \omega^2 x##

The maximum velocity: ## v = - v_{max} sin ( 2 \pi f t ) ##
I tried to solve by calculating:

## \omega = \frac{v}{x} = 3100/ (4.2 \times10^7) ##,

Which makes the frequency so small:
## f = 2 \pi \times 3100/ (4.2 \times10^7) \sim 10^{-5} ~ s^{-1} ##

And also the acceleration and velocity. So I think there is a mistake in my solution.

Any help to solve this problem accurately is appreciated.
 
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Safinaz said:
Which makes the frequency so small:
## f = 2 \pi \times 3100/ (4.2 \times10^7) \sim 10^{-5} ~ s^{-1} ##
You could look up the altitude for geosynchronous orbit and compare that to ##4.2 \times 10^7## meters.

What is the frequency for geosynchronous orbit?
 
Safinaz said:
The frequency ## f= 2 \pi \omega ##
Check the units: radians/cycle x radians/second =?
Safinaz said:
Which makes the frequency so small:
## f = 2 \pi \times 3100/ (4.2 \times10^7) \sim 10^{-5} ~ s^{-1} ##
What does that make the orbital period to be? Is that unreasonable?
 
haruspex said:
Check the units: radians/cycle x radians/second =?
While OP has written down the wrong formula, their numerical result corresponds to the correct one.
 
Safinaz said:
And also the acceleration and velocity. So I think there is a mistake in my solution.
Here is another reality check for your solution. Why do you think your acceleration is small? What do you expect it to be at a radius of 42000 km given that it is about 10 m/s^2 at the Earth radius 6700 km?
 
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