Calculating the height of a geostationary satellite of Earth

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SUMMARY

The discussion focuses on calculating the height of a geostationary satellite above Earth. Using the gravitational constant (6.667 x 10^-11 nm^2/kgm^2), Earth's mass (6 x 10^24 kg), and Earth's radius (6400 km), the initial calculation for the radius (r) yielded 53,583.6 km, leading to a height (h) of 47,183 km. However, the correct radius for a geostationary orbit should be approximately 36,000 km. The confusion arose from misinterpreting the orbital velocity and the relationship between radius and velocity.

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  • Understanding of gravitational force and constants
  • Familiarity with orbital mechanics and geostationary satellites
  • Proficiency in algebra and unit conversion
  • Knowledge of the equation GM/r = v^2
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kokodile
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Homework Statement


Calculate the height of a geo-stationary satellite of earth.

Gravitational force of earth=6.667 x 10^-11 nm^2/kgm^2
Mass of earth=6x10^24 kgm
Radius of earth=6400 km
V=86400

Homework Equations


GM/r=v^2
r=R+h

The Attempt at a Solution


I plugged everything into the equation and got 53,583.6 for r. Then since I need only height of the satellite to earth, I subtracted 6400 km from 53,583.6 and I got 47,183 for h. However, my professor said the answer for r should be around 36,000 km. Here is what I did.

(6.667x10^-11)(6x10^24)/r=86400^2

From multiplying, I got 4x10^14/r=7464960000

From this, I got r=53,583.6
Then h=47,183.

Are my calculations incorrect?
 
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Looks to me like you calculated the number of seconds in a day and set that as a velocity in m/s. Have I misunderstood your working?
 
haruspex said:
Looks to me like you calculated the number of seconds in a day and set that as a velocity in m/s. Have I misunderstood your working?
I just realized that as well. I can't believe I've been staring at this problem for half an hour without seeing this. Thanks
 
kokodile said:
I just realized that as well. I can't believe I've been staring at this problem for half an hour without seeing this. Thanks
That's why it's important to carry the units through your calculations. If you don't know what the numbers represent, how can anyone else know?
 
Thansk everyone
 
Last edited:
haruspex said:
Looks to me like you calculated the number of seconds in a day and set that as a velocity in m/s. Have I misunderstood your working?

I still can't figure this problem out. If I need to find the velocity to calculate r, how can I do that when I need r to calculate velocity?
 
kokodile said:
I still can't figure this problem out. If I need to find the velocity to calculate r, how can I do that when I need r to calculate velocity?
How are the radius and velocity related for a geostationary satellite?
 
haruspex said:
How are the radius and velocity related for a geostationary satellite?

Well the larger the radius, the smaller the velocity of the satellite will be.
 
kokodile said:
Well the larger the radius, the smaller the velocity of the satellite will be.
And yet remain geostationary?
 
  • #10
haruspex said:
And yet remain geostationary?
Yes. So it has the same velocity as Earth. But I'm still confused as to how to find the velocity of earth. Even if I use the equation 2pir/T, I get .465 m/s. And when I plug that into the equation GM/r=v^2, I get a very large number that must be incorect.
 
  • #11
kokodile said:
So it has the same velocity as Earth.
No, not the same velocity. What must be the same?
 
  • #12
haruspex said:
No, not the same velocity. What must be the same?
The time it takes for one orbit?
 
  • #13
kokodile said:
The time it takes for one orbit?
Well, not quite right. The time for one orbit of the satellite must equal what?
From that, write an equation relating radius of orbit to velocity.
 

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