Orbits of Satellites-Proportions and Escape Speed

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  • #1
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Homework Statement



A missile is launched upward with a speed that is half the escape speed. What height (in radii of Earth) will it reach?

a. R/4
b. R/3
c. R/2
d. R
e. 2R

Homework Equations



escape speed=√(2GM/r)

The Attempt at a Solution



r is inversely proportional to the square of the speed so if v is now 2v then r will be r/4
the answer key says its supposed to be r/3 :s
 

Answers and Replies

  • #2
gneill
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How does v become 2v? When the rocket reaches is maximum height its velocity will be zero.

Why not look at the specific mechanical energy, [itex]\xi[/itex] (the sum of the kinetic energy and potential energy per unit mass) which is always constant for any body in orbit (or free-fall)?

[tex] \xi= \frac{v^2}{2} - \frac{\mu}{r} [/tex]

Also note that height is not the same thing as radius :wink:
 
  • #3
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i meant v/2

can we use the formula

v=√(GM*((1/r(earth))+(1/r(height)))??
 
  • #4
Actually the problem doesn't make sense to me either. Maybe I have already forgotten my mechanic stuff.

the v becomes v/2, that means KE/4
KE/4 = PE/4, my equation is (GMm/r²)*r*1/4 that is GMm/4r.
The rocket reaches 3R height and velocity becomes zero and starts to fall back to Earth surface.

So what have I done wrong?
 
  • #5
gneill
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i meant v/2
Okay, why does the velocity become v/2 ?
can we use the formula

v=√(GM*((1/r(earth))+(1/r(height)))??
I don't recognize that formula. Where does it come from?
 
  • #6
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Because the speed is half the escape speed and he formula is the escape speed formula but usualy the height is set to infinity so we get the formula from the beggining but in his case we are being asked for he height so we need it
 
  • #7
gneill
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Because the speed is half the escape speed and he formula is the escape speed formula but usualy the height is set to infinity so we get the formula from the beggining but in his case we are being asked for he height so we need it
Ah. So your v/2 is the initial velocity, Vesc/2. Got it.

Since [itex] V_{esc} = \sqrt{\frac{2 \mu}{r_e}}[/itex] you can plug 1/2 of that into the total mechanical energy formula I gave above to find the constant energy [itex]\xi[/itex] at launch time. With that energy you can use the same formula again to find r when the velocity becomes zero. Note that the total mechanical energy formula is just a statement of the conservation of energy, in this case kinetic and potential energy.
 
  • #8
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[tex] \xi= \frac{v^2}{2} - \frac{\mu}{r} [/tex]
what is μ in this formula??
 
  • #9
gneill
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what is μ in this formula??
It's the gravitational parameter for the Earth: μ = GM
 
  • #10
D H
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what is μ in this formula??
It's the gravitational parameter for the Earth: μ = GM
Just to add: The gravitational parameter for some object μ is observable in and of itself if that object has sub-objects in orbit about it. In many cases, scientists know μ to a very high degree of precision (e.g., eleven decimal places for the Sun). In comparison, G is arguably the least known of all key constants in physics (only four or so decimal places). In fact, the primary evidence for many astronomical objects' mass is μ/G for that object.

Bottom line: It's not just a matter of convenience to lump GM into a single parameter μ. In many cases, using μ rather than GM is conceptually more precise (G and M are known to four places, μ to nine or more) and more correct (M is just μ/G).
 
  • #11
Actually nevermind, I see how now.

KE - PE(earth) = 0 for escaping gravity.
now KE is 1/4 because v is v/2.
So
PE(earth) - KE = 3PE/4 (the potential energy at the point where KE has taken the Satellite to.)

That is 3*GMm/ 4r
In other word it is 4r/3.
That is r/3 above Earth's radius.
 
Last edited:

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