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Derived equation...initial speed...

by 7randomapples
Tags: initial velocity, inward force
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7randomapples
#1
Dec11-09, 11:17 AM
P: 7
1. The problem statement, all variables and given/known data
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. Relevant equations
-Derived equation provided:
Finward = Fg
m1 v^2 / r = Gm1m2/r
v^2 = Gm2/r
v= (square root of) Gm2/r


3. 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?
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cepheid
#2
Dec11-09, 11:24 AM
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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.
dr_k
#3
Dec11-09, 11:59 AM
P: 69
Quote Quote by 7randomapples View Post
...
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?

7randomapples
#4
Dec11-09, 12:07 PM
P: 7
Derived equation...initial speed...

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?
cepheid
#5
Dec11-09, 12:14 PM
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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.
7randomapples
#6
Dec11-09, 12:23 PM
P: 7
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?
cepheid
#7
Dec11-09, 12:30 PM
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It would give you the orbital speed of the satellite (which is constant in magnitude).


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