Distance of geosynchronous satellite from earth

In summary: Hi TSny,I'm not so sure about this, but should I have calculated the orbital distance as x + radius of earth, or x+6416000?You should calculate the orbital distance as x+6416000.
  • #1
hangontoyourego
2
0
Hello, I had a bit of trouble figuring out this problem:

1. Homework Statement


Given the following, determine the distance in miles above the Earth's surface of a geosynchronous satellite.

MEarth=5.98E24 kg
REarth=4,000 miles
1 mile=1604 m

Homework Equations



Fg=(Gm1m2)/r2

FC=(mv2)/r

The Attempt at a Solution



((6.67E-11m^3/kg x s^2)(mSatellite)(5.98E24kg))/(6416000m+x)^2 = (mSatellitevSatellite^2)/(6416000m+x)

For v of the satellite, I said the velocity is equal to the distance of the orbit divided by 86,400 seconds, so I have:

(3.98866E14)/(4.1165056E13+12832000x+x^2) = (((6.2832x)/(86400))^2)/(6416000+x)

and then

http://www4b.wolframalpha.com/Calculate/MSP/MSP94291ci3c60ee01fbifd00000i565236dhh0d08b?MSPStoreType=image/gif&s=2&w=476.&h=66.

Then, I found that x equaled 40,216,400 meters or 11,658 miles. Is this correct?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Hello, and welcome to PF!

Your expression for the distance traveled during one orbit is not quite correct. See if you can spot the problem.

It would be much nicer if you first solved the equations symbolically and then plug in numbers. I would also recommend that you first solve (in symbols) for the radius r and then find x.
 
  • #3
Hi TSny,

I'm not so sure about this, but should I have calculated the orbital distance as x + radius of earth, or x+6416000?

So,

((G)(MEarth))/(REarth+x)2 = ((x+REarth)^2)/(REarth+x)

((G)(MEarth))/(REarth+x)2 = (x+REarth)

((G)(MEarth)) = (x+REarth)3

((G)(MEarth))(1/3) = (x+REarth)3)(1/3)

which, substituting the numbers in, would be:

((6.67E-11 m3/kg⋅s2)(5.98E24 kg))(1/3) = x + 6416000

73610.93588 = x + 6416000

This can't be right though, since it yields a negative answer, right? Or do you take the absolute value of x and the answer is 6342389.064/1604 or 3,954 miles?
 
  • #4
hangontoyourego said:
Hi TSny,

I'm not so sure about this, but should I have calculated the orbital distance as x + radius of earth, or x+6416000?

So,

((G)(MEarth))/(REarth+x)2 = ((x+REarth)^2)/(REarth+x)

What's the deal with this equation?

You basically have written GM/r2 = r. That doesn't make sense. What happened to the orbital period of the satellite?
This can't be right though, since it yields a negative answer, right? Or do you take the absolute value of x and the answer is 6342389.064/1604 or 3,954 miles?

Only a desperate man takes the absolute value of something which is wrong, hoping to make it right.

Draw a sketch of the problem, and then apply your formulas. What's the formula for the velocity of an object traveling a circular path of radius R and period T?

What's the R for a satellite in a geosynchronous orbit? Remember, the satellite must orbit the entire earth, you know, center and all.
 
  • #5
hangontoyourego said:
Hi TSny,

I'm not so sure about this, but should I have calculated the orbital distance as x + radius of earth, or x+6416000?

Yes, the radius of the circular orbit is r = R +x where R is the radius of the earth. So, the distance traveled in one orbit is the circumference of a circle with radius r. Note that the relation r = R + x means that you can easily solve for x if you first solve for r.

Your attempt at a solution in the first post shows that you have the right idea.

If you let T be the time for one orbit, how would you write v in terms of the symbols r and T?

You were correct to set the gravitational force equal to mv2/r where m is the mass of the satellite. So, in symbols, you have

GMm/r2 = mv2/r where M is the mass of the earth.

Combine this with your expression for v in terms of r and T and see if you can solve symbolically for r.
 

1. What is the distance of a geosynchronous satellite from Earth?

The distance of a geosynchronous satellite from Earth is approximately 35,786 kilometers (22,236 miles). This distance is known as the geostationary orbit, which allows the satellite to maintain a fixed position above a specific point on Earth's equator.

2. How is the distance of a geosynchronous satellite from Earth maintained?

The distance of a geosynchronous satellite from Earth is maintained through a balance of the satellite's velocity and the gravitational pull of Earth. The satellite's orbital speed is synchronized with the rotation of the Earth, allowing it to maintain a relatively fixed position above a specific point on the equator.

3. What are the benefits of having a geosynchronous satellite at a specific distance from Earth?

Having a geosynchronous satellite at a specific distance from Earth allows for continuous communication and observation of a specific location on Earth's surface. This is useful for applications such as weather forecasting, satellite television, and GPS navigation.

4. Can the distance of a geosynchronous satellite from Earth change?

Yes, the distance of a geosynchronous satellite from Earth can change due to various factors such as atmospheric drag, solar pressure, and gravitational forces from other celestial bodies. However, these changes are minimal and can be corrected through small adjustments to the satellite's orbit.

5. How does the distance of a geosynchronous satellite from Earth affect its lifespan?

The distance of a geosynchronous satellite from Earth does not significantly affect its lifespan. However, the satellite's orbit does gradually decay over time due to atmospheric drag, which can lead to a decrease in its operational lifespan. To counteract this, satellites are equipped with thrusters to make small corrections to their orbit and maintain their position in the geostationary orbit.

Similar threads

  • Introductory Physics Homework Help
2
Replies
39
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
5K
  • Introductory Physics Homework Help
Replies
16
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
22
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
18K
Back
Top