Geosynchronous Satellites: Distance & Formula Analysis

In summary: Substituting in the values for R_e, h, and GM_e gives: $$v = R_e\sqrt{\frac{g m}{R_e + h}} = R_e\sqrt{\frac{g m}{(R_e+h)^2}}$$ This means that the period is 24 hours if the geosynchronous satellite is within 5000km of the Earth's surface.
  • #1
guyvsdcsniper
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Homework Statement
A satellite orbits Earth at altitudes of 5000 km. What is its speed with respect to the center of Earth? The radius of the Earth is 6380 km and its mass is 5.98 x 1024 kg.
Relevant Equations
V=sqrt(G*me/r)
V=2pir/T
My work is attached. I did get the answer right using the formula V=√(G*me/r).

Im just confused as to why I can't use V=2pi*r/T to also get the right answer? With this equation I would assume T=86400s.

I was able to use both equations to get the correct answer for a problem relating to geosynchronous satellite. The satellite for this problem was 200km from the Earth's surface, which is significantly closer than 5000km.

Does distance play a factor and if so, how?

I appreciate any and all responses.
 

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  • #2
quittingthecult said:
With this equation I would assume T=86400s.
What makes you think the period would be 24 hours?
 
  • #3
haruspex said:
What makes you think the period would be 24 hours?
I was just assuming that based off the geosynchronous satellite I solved. I was completely wrong. I used the value I got from my answer and solved for T and the period was 12078s.

I totally jumped the gun asking the question here without really thinking about my question. Sorry.
 
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  • #4
quittingthecult said:
I was just assuming that based off the geosynchronous satellite I solved. I was completely wrong. I used the value I got from my answer and solved for T and the period was 12078s.

I totally jumped the gun asking the question here without really thinking about my question. Sorry.
No problem.
 
  • #5
quittingthecult said:
I was just assuming that based off the geosynchronous satellite I solved. I was completely wrong. I used the value I got from my answer and solved for T and the period was 12078s.

I totally jumped the gun asking the question here without really thinking about my question. Sorry.
You can also get the answer without the mass of the Earth or ##G##, but using the known surface gravity. The magnitude of the force on the satellite is: $$F = \frac{GM_e m}{r^2} = \frac{mR_e^2}{r^2}\frac{GM_e}{R_e^2} = \frac{mR_e^2}{r^2}g$$ And the centripetal force is $$F_c = \frac{mv^2}{r}$$ Equating these two gives: $$v = R_e\sqrt{\frac g r} = R_e\sqrt{\frac g {R_e + h}}$$
 

FAQ: Geosynchronous Satellites: Distance & Formula Analysis

What is a geosynchronous satellite?

A geosynchronous satellite is a satellite that orbits the Earth at the same speed and direction as the Earth's rotation, allowing it to maintain a fixed position relative to a specific location on Earth.

How far away is a geosynchronous satellite?

A geosynchronous satellite is typically located at an altitude of approximately 22,236 miles (35,786 kilometers) above the Earth's surface.

What is the formula for calculating the distance of a geosynchronous satellite?

The formula for calculating the distance of a geosynchronous satellite is: distance = radius of Earth + altitude of satellite

How long does it take for a geosynchronous satellite to complete one orbit?

A geosynchronous satellite takes approximately 24 hours to complete one orbit around the Earth, which is the same amount of time it takes for the Earth to complete one rotation.

What are some common uses for geosynchronous satellites?

Geosynchronous satellites are commonly used for telecommunications, weather forecasting, and navigation systems. They are also used for monitoring the Earth's surface, such as for tracking weather patterns or detecting changes in the environment.

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