Circular Orbit and Period

  • Thread starter cstout
  • Start date
  • #1
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Homework Statement


The Hubble Space Telescope orbits Earth 615 km above Earth's surface. What is the period of the telescope's orbit?


Homework Equations


T = 2(pi)r/v


The Attempt at a Solution



T = 2(pi)(6.37x10^6)/9.8
 

Answers and Replies

  • #2
mgb_phys
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Remember the radius of the orbit is measured from the centre of the earth.
 
  • #3
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Isn't 6.37 × 106 m the radius of the earth, or do I use another measurement.
 
  • #4
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Isn't 6.37 × 106 m the radius of the earth, or do I use another measurement.

So, the radius of the orbit is radius of earth? That means the satellite is orbiting in surface, right?
 
  • #5
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it orbit 615km above the Earths surface
 
  • #6
mgb_phys
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The orbital radius is the distance from the centre of the Earth to the satelite.
Hence the radius of the Earth PLUS the altitude = approx 7000km.
 
  • #7
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Remember to express total distance (Earth's radium + orbital distance) in meters to mantain a coherence among unities.
 
  • #8
mgb_phys
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And since you aren't given the speed you also need Kepler's 3rd law to work out the period. Not sure where you got the 9.8 from in your original equation - but it's wrong!
 
  • #9
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how would you use Kepler's 3rd law in relation to this question though, you aren't given some of the information needed.
 
  • #10
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Nope - you have everything you need. What information do you think you need for Kepler's 3rd Law that you aren't given?

And by the way, mgb_phys - I'll be cstout was using the value for the approximate acceleration due to gravity at the Earth's surface, i.e. 9.8 m/s^2, which you'd just have to call a pretty bad guess.
 
  • #11
mgb_phys
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Remember the simplified form of the law doesn't need the mass of the orbital body
, you can assume that G(M+m) is pretty much equal to GM. Hubble only weighs about 10tons!
 
  • #12
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Ok, I finally got it, the answer is 1.61 hrs. Thanks for the help
 
  • #13
mgb_phys
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That sounds about right, 90 minutes is pretty typical for LEO
 
  • #14
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Try using this formula: 4(pi)² r³/(M+m)G = P²

Where:
r = radius between the centers of mass of both bodies measured in meters
M = the larger mass in kilograms (5.9742 X 10^24 for the earth)
m = the smaller mass in kilograms (this mass is usually so much smaller than the
larger mass that it can almost always be disregarded)
G = 6.67259 X 10-¹¹
P = the period of orbit in seconds

This formula yields an answer of approx. 5,811 seconds, which is about 96.85 minutes.
 
Last edited:
  • #15
I don't think so..

Since the radius,r of the telescope is just about 10% of the radius of the Earth.
Then it can be ignored..

Just use R for the radius of Earth.

Hence, Period of revolution of telescope
= 2Pi Square root R/g

=84.43min
 

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