Orbital Period for GPS

In summary, the GPS satellites orbit at an altitude of 2.0×10^7 m and have an orbital period of approximately 11 hours and 31 minutes. To calculate the orbital period, the equation T = 2pi sqrt( r^3 / GM) can be used, where G is the gravitational constant (6.67*10^-11), M is the mass of the Earth (5.97*10^24), and r is the distance from the center of the Earth to the satellite (6.37*10^6 + 2*10^7). It is important to ensure that all values are in the correct units (m/s/kg) in order to
  • #1
rsfancy
8
0
GPS (Global Positioning System) satellites orbit at an altitude of 2.0×10^7 m.

Find the orbital period.(hours)I use the equation T=(2pi/sqrt(GM))r^(3/2).

Shouldnt that be the correct way of doing it? I know I am supposed to add the altitude to the radius of the earth, G is known, M of Earth is known. I just don't understand where I am going wrong with this. It should be a simple problem yet I am not able to get the correct answer. is it possible I am doing something as silly as not converting something somewhere?

If anyone could offer some help, it would be greatly appreciated.
 
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  • #2
What's the problem you are having?
Lookup the values for 'G' and 'M' (of the earth)
Add the Earth's radius to the altitude and plug into the equation.

Make sure everythign is in the correct units m/s/kg

The equation is normally written T = 2pi sqrt( r^3 / GM), might be simpler to use.
 
  • #3
I used the formula you provided and I still got the answer wrong.I can't imagine I am capable of messing this up this bad.

G=6.67*10^-11
M=5.97*10^24
R=6.37*10^6 + 2*10^7

the answer I got was =.001617844
Rounding to two significant figures= .0016

This is driving me insane.I know that whatever it is I am doing wrong,is going to end up being something small and infinitely stupid.
 
  • #4
I think a bit of finger trouble on the calculator.
It's always worth being able to do a rough approx just using the rules of exponents for large number calculations. Remember to multiply simply add the exp, to divide subtract.

ps. the answer isn't exact becaue I only did a few decimal places - but you see the idea

GM = 6.7E-11 * 6E24 = 6.7*6 E(24-11) = 40 E13 m^3 s^-2
r = 6.4E6 + 20E6 = 26E6 m
r^2 = 26*26*26 E(6+6+6) = 17500E18 m^3

sqrt( r^3/GM) = sqrt( 17500E18 / 40E13 ) = sqrt(437E5) = 6600
t = 2pi*6600 = 41500 seconds =11 h
 

What is the Orbital Period for GPS?

The Orbital Period for GPS (Global Positioning System) refers to the amount of time it takes for a GPS satellite to complete one full orbit around the Earth.

How long is the Orbital Period for GPS satellites?

The Orbital Period for GPS satellites is approximately 12 hours. This means that each satellite completes two orbits in a single day.

Why is the Orbital Period for GPS satellites important?

The Orbital Period for GPS satellites is important because it determines how often a satellite passes over a specific location on Earth. This allows for more accurate and consistent data collection for GPS devices.

How is the Orbital Period for GPS calculated?

The Orbital Period for GPS is calculated using Kepler's third law of planetary motion, which states that the square of the orbital period is proportional to the cube of the semi-major axis (distance from the satellite to the Earth's center). This calculation takes into account the mass of the Earth and the satellite, as well as the gravitational constant.

Has the Orbital Period for GPS changed over time?

No, the Orbital Period for GPS has remained consistent since the first GPS satellite was launched in 1978. However, satellites may be adjusted or replaced over time to maintain accuracy and functionality.

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