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Satellite orbital radius Question

  • #1
I am a first-year physics/chemistry major in First-term Calculus-based Physics. Normally, I have no trouble with my physics assignments, but this problem...man, i have no idea even where to start, even after analyzing it for two hours straight! Any help would be appreciated!

5. You are an astronaut in the space shuttle pursuing a satellite in need of repair. You find yourself in a circular orbit of the same radius as the satellite, but 20 km behind it.

(a) How long will it take to reach the satellite if you reduce your orbital radius by 0.9 km?

(b) By how much must your reduce your orbital radius to catch up in 7.3 hours?
km
 

Answers and Replies

  • #2
For circular orbit,

[tex]a_\textrm{centripetal} = \frac{v^2}{r}[/tex]

correct? Now, this centripetal acceleration must be provided by gravity (what else is going to produce it?), so acceleration due to gravity is

[tex]a_\textrm{gravity} = G\frac{M_\textrm{Earth}}{r^2}[/tex]

So centripetal acceleration and acceleration due to gravity will be equal. Now we have an equation relating velocity and radius. Can you take it from here?

The second problem works the same way, just backwards.

cookiemonster
 
  • #3
Oh, I forgot! I know that this relationship needs to be used somehow, but I still can't seem to see how:

(mv^2)/r = Gm*(m of earth)/(r^2)
 
  • #4
don't i need to know the original radius? or does that not really matter?
 
  • #5
It seems to me that you do, but everybody's working me so hard that I'm running around like a madman and I could very well be wrong. Is the initial orbital radius given? Do they want you to solve it symbolically?

See where you can take that equation.

cookiemonster
 
  • #6
The initial orbital radius isn't given, but they still expect me to come up with a numerical answer...lol, soon I'm probably just going to start guessing, hope I come up with the right answer!
 
  • #7
*clip*
This all was misleading, so forget it.

cookiemonster
 
Last edited:
  • #8
All right. Here we go.

[tex]\frac{1}{v_i^2} - \frac{1}{v_f^2} = \frac{r_0 - r_0 + .9 km}{GM}[/tex]

Take it from there.

cookiemonster
 
  • #9
ok...i figured out what i was supposed to do...
[tex]\frac{mv^2}{r} = \frac{GmM}{r^2}[/tex]

m cancels out, leaving:
[tex]\frac{v^2}{r} = \frac{GM}{r^2}[/tex]

solving for v gives:
[tex]v = \sqrt{GM}*r^{-1/2}[/tex]

taking the derivative gives us [tex]dv[/tex] in terms of [tex]r[/tex] and [tex]dr[/tex]:
[tex]dv = \sqrt{GM}*r^{-3/2}*dr[/tex]

Now, I guess I was supposed to assume that the satellite was a low-orbiting body, thus the orbital radius is nearly equal to the radius of the earth.
Thus, [tex]r = 6.38 * 10^3 km[/tex]
[tex]M = 5.97 * 10^{24} kg[/tex]
[tex]G = 6.67 * 10^{-11} Nm^2kg^{-2}[/tex]
[tex]dr = 0.9 km[/tex]

Solving gives:
[tex]dv = 2.02 m/s[/tex]

Because [tex]dr[/tex] is so small in comparison with the orbital radius, the effect it has on the rest of the solution is negligible.

Because all aspects of the two orbiting bodies are effectively equal except for the velocities, we can now use:

[tex]t = \frac{d}{v}[/tex]
[tex]d = 20 km[/tex]
[tex]v = 2.02 m/s[/tex]

From this, we obtain [tex]t = 9.90 hrs[/tex]
Part B uses these equations, but in reverse.
 
  • #10
angelical_kitten said:
Now, I guess I was supposed to assume that the satellite was a low-orbiting body, thus the orbital radius is nearly equal to the radius of the earth.
Personally, I think that's a poor assumption on the part of the problem writers. But that's just me.

And, just for your future use, the dv/dr method yields the same results as substituting in

[tex]v_f = v_i + \Delta v[/tex]

but requires less algebra. It's pretty intuitive to see how this is true, since we know in the limit delta v = dv.

cookiemonster
 
Last edited:

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