# Circular Orbit and Period

1. Feb 6, 2008

### cstout

1. The problem statement, all variables and given/known data
The Hubble Space Telescope orbits Earth 615 km above Earth's surface. What is the period of the telescope's orbit?

2. Relevant equations
T = 2(pi)r/v

3. The attempt at a solution

T = 2(pi)(6.37x10^6)/9.8

2. Feb 6, 2008

### mgb_phys

Remember the radius of the orbit is measured from the centre of the earth.

3. Feb 6, 2008

### cstout

Isn't 6.37 × 106 m the radius of the earth, or do I use another measurement.

4. Feb 6, 2008

### Littlepig

So, the radius of the orbit is radius of earth? That means the satellite is orbiting in surface, right?

5. Feb 6, 2008

### cstout

it orbit 615km above the Earths surface

6. Feb 6, 2008

### mgb_phys

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. Feb 6, 2008

### Artus

Remember to express total distance (Earth's radium + orbital distance) in meters to mantain a coherence among unities.

8. Feb 6, 2008

### mgb_phys

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. Feb 6, 2008

### cstout

how would you use Kepler's 3rd law in relation to this question though, you aren't given some of the information needed.

10. Feb 6, 2008

### belliott4488

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. Feb 6, 2008

### mgb_phys

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. Feb 6, 2008

### cstout

Ok, I finally got it, the answer is 1.61 hrs. Thanks for the help

13. Feb 6, 2008

### mgb_phys

That sounds about right, 90 minutes is pretty typical for LEO

14. Jan 29, 2009

### joeiii63

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: Jan 29, 2009
15. Feb 2, 2009

### Victorian91

I don't think so..