Satellite period, which equation?

In summary, the equations for a spherical orbit state that T(secs)^2 = (4(pi^2)(6,500,000^3 m))/((6.67*10^-7 N*m^2/kg^2)(5.98*10^24 kg))
  • #1
izforgoat
16
0

Homework Statement



I'm looking for the period of an orbiting object a certain height from the Earth's surface, I am given this height. So I have the total radius of 6,5OO,OOO m, g = 9.81 m/s^2 and the mass of Earth = 5.98*10^24 kg

Please note that for this problem G is another constant than what it usually is.

Homework Equations



Here is where I am confused I do not know whether to use the T = (2[tex]\pi[/tex]r)/[tex]\sqrt{gr}[/tex]

or the T[tex]^{2}[/tex] = (4[tex]\pi^{2}r^{3}[/tex])/(GM[tex]_{earth}[/tex])

where G = 6.67*1O^-7

The Attempt at a Solution



K. So when I use the method of going with T = (2[tex]\pi[/tex]r)/[tex]\sqrt{gr}[/tex]

I get about 5114 seconds for the period.

when I use T[tex]^{2}[/tex] = (4[tex]\pi^{2}r^{3}[/tex])/(GM[tex]_{earth}[/tex])
I get 52.13584223 seconds, which doesn't logically seem right but since G is different I don't know.

does anyone know what the right method is?
 
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  • #3
And your value for G is wrong. Though it's hard to really say until you put units on it.
 
  • #4
thx a lot for that clarification. Still it doesn't make sense that its period would be 52 seconds here's my work.

T(secs)^2 = (4(pi^2)(6,500,000^3 m))/((6.67*10^-7 N*m^2/kg^2)(5.98*10^24 kg))

are the units for T in seconds? was it alright that I changed km to m for the radius?
 
  • #5
Dick said:
And your value for G is wrong. Though it's hard to really say until you put units on it.

I'm using a different value for this problem
 
  • #6
G=6.67*10^(-11)*N*m^2/kg^2. Note the exponent.
 
  • #7
izforgoat said:
I'm using a different value for this problem

Is it an 'alternative universe' problem? Why would you use a different value for G? It's a 'universal constant'.
 
  • #8
Dick said:
Is it an 'alternative universe' problem? Why would you use a different value for G? It's a 'universal constant'.

you could say that. But either way I don't think it would have much difference for this equation than the plug and chug. Right now I want to know if I am calculating everything else right. I assume I am.
 
  • #9
The reason you getting 52 seconds is because you are putting in a value of G that is 10000 times too large. Other than that you are doing fine.
 
  • #10
thank you
 

1. What is the equation for calculating the period of a satellite?

The equation for calculating the period of a satellite is:

T = 2π√(a^3/GM)

where T is the period in seconds, a is the semi-major axis of the satellite's orbit in meters, G is the gravitational constant (6.67 x 10^-11 m^3/kg/s^2), and M is the mass of the central body (such as the Earth) in kilograms.

2. How is the period of a satellite related to its orbital altitude?

The period of a satellite is directly related to its orbital altitude. This is because the period of a satellite is determined by the distance between the satellite and the central body, as well as the mass of the central body. As the altitude of a satellite increases, the distance between the satellite and the central body also increases, resulting in a longer period of orbit.

3. Can the equation for satellite period be used for any satellite orbit?

The equation for satellite period can be used for any satellite orbit, as long as the orbit is circular. For non-circular orbits, a more complex equation is needed to calculate the period, taking into account the eccentricity of the orbit.

4. How does the mass of the central body affect the satellite period?

The mass of the central body has a direct influence on the period of a satellite. As the mass of the central body increases, the gravitational force also increases, resulting in a shorter period of orbit for the satellite.

5. How accurate is the equation for calculating satellite period?

The equation for calculating satellite period is highly accurate for circular orbits. However, for non-circular orbits, it may not provide an exact value for the period due to the influence of the satellite's eccentricity. In such cases, a more complex equation is needed to calculate the period accurately.

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