Determine Altitude from Constant Speed and Acceleration of Satellite S

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Homework Help Overview

The problem involves determining the altitude of a satellite traveling in a circular orbit around the Earth, given its constant speed and centripetal acceleration. The context includes the Earth's diameter as a reference for calculations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between speed, acceleration, and radius in circular motion. Questions arise regarding the correct interpretation of acceleration and the units of speed. There is an exploration of how to relate the radius of the orbit to the altitude above the Earth's surface.

Discussion Status

Some participants have provided guidance on the equations involved and clarified the relationship between the radius and altitude. There is an acknowledgment of previous misunderstandings regarding the calculations, but no consensus has been reached on the final altitude value.

Contextual Notes

Participants note the importance of using the correct radius (Earth's radius) in calculations and the implications of unit conversions for speed. There is a recognition of the need to clarify assumptions about the definitions of variables involved.

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Homework Statement



The satellite S travels around the Earth in a circular path with a constant speed of 20 Mm/h. If the acceleration is 2.5m/s2, determine the altitude h. Assume the Earth's diameter to be 12 713 km.

i5akx3.png


Homework Equations



[tex]a_{n}= \frac{v^{2}}{\rho}[/tex]

The Attempt at a Solution



I thought that [tex]a_{t}= 0[/tex] because speed is constant and [tex]a_{n}= 20 Mm/h[/tex] and I just solved the above equation for [tex]\rho[/tex] but that came out to a negative number so that's not right.
 
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KillerZ said:

Homework Statement



The satellite S travels around the Earth in a circular path with a constant speed of 20 Mm/h. If the acceleration is 2.5m/s2, determine the altitude h. Assume the Earth's diameter to be 12 713 km.

i5akx3.png


Homework Equations



[tex]a_{n}= \frac{v^{2}}{\rho}[/tex]

The Attempt at a Solution



I thought that [tex]a_{t}= 0[/tex] because speed is constant and [tex]a_{n}= 20 Mm/h[/tex] and I just solved the above equation for [tex]\rho[/tex] but that came out to a negative number so that's not right.
Why are you using the tangential speed as the acceleration? The acceleration is, as you have stated, [itex]a = v^2/r[/itex]. But there is another expression for a as well, since the acceleration is provided by ...? Write the equation for that acceleration. With those two equations you should be able to solve for the two unknowns, a and r.

AM
 
Ops that was a typo. I meant [tex]a_{n}= 2.5 m/s^{2}[/tex]
 
KillerZ said:
Ops that was a typo. I meant [tex]a_{n}= 2.5 m/s^{2}[/tex]
Ok. I misread the question too. You are given the acceleration. What units must v have in your equation [itex]a_n = v^2/r[/itex]?

AM
 
v should be m/s I think. Which I calculated 20 Mm/h = 5555.556 m/s
 
KillerZ said:
v should be m/s I think. Which I calculated 20 Mm/h = 5555.556 m/s
So what is r? How is r related to h?

AM
 
I am assuming r would be to the center of the Earth and h is r - the Earth's radius.
 
Ok I think I got this:

[tex]v = 20 Mm/h = 5555.6 m/s[/tex]

[tex]a_{n} = 2.5 m/s^{2}[/tex]

[tex]a_{n}= \frac{v^{2}}{\rho}[/tex]

[tex]\rho= \frac{v^{2}}{a_{n}}[/tex]

[tex]\rho= \frac{5555.6}{2.5} = 12345679.01 m[/tex]


[tex]h = \rho - Earth's radius = 12345679.01 - 6356500 = 5989179.01 = 5989.18 km[/tex]
 
that looks better now, I think you were getting a negative number because you were subtracting the diameter instead of the radius.
 
  • #10
Ya that's what I was doing.
 

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