Derived equation initial speed

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Homework Help Overview

The discussion revolves around calculating the initial speed required for a satellite to maintain a circular orbit at an altitude of 1000 km above the Earth. The problem involves applying Newton's law of gravitation and centripetal force equations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the derived equation for orbital speed and question the correct interpretation of the distance variable 'r' in the context of gravitational force and centripetal acceleration.

Discussion Status

There is an ongoing examination of the calculations and assumptions regarding the distance from the center of the Earth to the satellite. Some participants suggest clarifying the definition of 'r' and whether the calculated speed represents the initial velocity or orbital speed.

Contextual Notes

Participants are navigating through the implications of using the correct radius for calculations, with specific attention to the altitude of the satellite and the radius of the Earth. There is a focus on ensuring that the derived equations are applied correctly in the context of the problem.

7randomapples
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Homework Statement


Use the equation derived about to calculate the exact value of the initial speed that will produce a circular orbit of the satellite at an altitude of 1000 km above earth.2. Homework Equations
-Derived equation provided:
Finward = Fg
m1 v^2 / r = Gm1m2/r
v^2 = Gm2/r
v= (square root of) Gm2/r

The Attempt at a Solution


Finward = Fg
m1 v^2 / r = Gm1m2/r
v^2 = Gm2/r
v = (square root of) Gm2/r
v = (square root of) (6.67x10^-11)(5.98x10^24)/1000
v = (square root of) 3.98...x10^11
I didn't round of 3.98...on my calculator; I saved the whole number and calculated the square. This gave me:
v = 631558.39
v = 6.3x10^5 m/s ?

Is this initial velocity though? Or did I miss a step?
 
Last edited:
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In Newton's law of gravitation, r is the distance from the *centre* of mass 1 to the *centre* of mass 2. This is true for the equation for centripetal force as well -- r is the distance from the object to the centre of its orbit. In other words, you have not used the correct distance for r.
 
7randomapples said:
...
v = 6.3x10^5 m/s ?

Is this initial velocity though? Or did I miss a step?

Draw a picture, and ask, what is r?
http://img705.imageshack.us/img705/790/fma3.jpg
 
Last edited by a moderator:
Finward = Fg
m1 v^2 / r = Gm1m2/r
v^2 = Gm2/r
v = (square root of) Gm2/r
v = (square root of) (6.67x10^-11)(5.98x10^24)/(1000/2)
v = (square root of) (6.67x10^-11)(5.98x10^24)/500
v = (square root of) 7.977...
I didn't round of 7.977...on my calculator; I saved the whole number and calculated the square. This gave me:
v = 893158.4406
v = 8.9x10^5 m/s ?

Is this the initial velocity?
 
Where do you get 1000/2 from? This is not correct. I think you should read my previous post again. The distance 'r' has a very specific definition here.
 
Finward = Fg
m1 v^2 / r = Gm1m2/r
v^2 = Gm2/r
v = (square root of) Gm2/r

G= gravitational constant = 6.67x10^-11
m2= mass of Earth = 5.98x10^24
r= radius of Earth + altitude of satellite above Earth = 6.37x10^6 + 1000 ?

If everything else is right, would this equation give me initial velocity or average?
 
It would give you the orbital speed of the satellite (which is constant in magnitude).
 

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