Derived equation initial speed

In summary: To find the initial velocity, you would need to take into account the direction of the velocity vector, which would depend on the starting position of the satellite. In summary, to calculate the initial speed for a satellite to maintain a circular orbit at an altitude of 1000 km above Earth, one must use the derived equation Finward = Fg and the values for the gravitational constant, Earth's mass, and the distance between the satellite and Earth's center (radius of Earth + altitude). This will give the orbital speed of the satellite, but to find the initial velocity, the direction of the velocity vector must also be taken into account.
  • #1
7randomapples
7
0

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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  • #2
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.
 
  • #3
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:
  • #4
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?
 
  • #5
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.
 
  • #6
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?
 
  • #7
It would give you the orbital speed of the satellite (which is constant in magnitude).
 

Related to Derived equation initial speed

What is the derived equation for initial speed?

The derived equation for initial speed is:
v0 = v - at, where v is the final speed, a is the acceleration, and t is the time.

What is initial speed?

Initial speed, denoted as v0, is the speed of an object at the beginning of its motion.

How do you calculate initial speed using the derived equation?

To calculate initial speed using the derived equation, plug in the values for final speed, acceleration, and time. Then, solve for v0.

What is the difference between initial speed and final speed?

The main difference between initial speed and final speed is that initial speed is the speed at the beginning of an object's motion, while final speed is the speed at the end of its motion. Additionally, initial speed can be zero, while final speed cannot be zero unless the object has come to a complete stop.

Can the derived equation for initial speed be used for any type of motion?

Yes, the derived equation for initial speed can be used for any type of motion, as long as the acceleration remains constant. This includes linear, circular, and projectile motion.

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