Derived equation initial speed

AI Thread Summary
To calculate the initial speed for a satellite in a circular orbit at 1000 km above Earth, the correct formula is v = √(Gm2/r), where r is the distance from the center of the Earth to the satellite. The radius r should be the sum of Earth's radius (approximately 6.37 x 10^6 m) and the altitude (1000 m), resulting in r = 6.37 x 10^6 + 1000 m. Using the gravitational constant G = 6.67 x 10^-11 and Earth's mass m2 = 5.98 x 10^24, the calculation yields an orbital speed of approximately 8.9 x 10^5 m/s. This speed represents the orbital velocity required for the satellite to maintain its circular path.
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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