Help understanding basic astronautics?

[snickers1]

Hello,

I am currently studying some introductory concepts involving orbit. Can anyone help me with the equation I would use to solve the following problem:

Assume you wish to place in orbit a satellite that always remains directly above the same point on the Earth's equator. What velocity and altitude must the satellite have at the instant of burnout of the rocket booster?

Thanks for any suggestions!

 

[kungfuscious]

If you want to have something in orbit above the same place in the equator, this is a special orbit called 'geostationary'. It only works around the equator, and it's done by putting something in orbit such that its own orbital period is the same as that of the Earth rotating on its own axis. So, since the Earth takes 24 hours to rotate, so should your orbital period be 24 hours.

T = 24 hrs. Convert that to seconds, and you get 24hrs*3600sec/hr = 86400sec

Orbital mechanics in your case is actually pretty easy. It's just a matter of thinking about the forces acting on your satellite when it is in orbit. In this case, it's going around in a circle, so you have the centripetal force being caused by the gravitational force of attraction between the satellite and the earth.

Remember that centripetal force Fc = (mv²)/r (where m = mass of satellite, v = speed of satellite in meters per second, and r = radius of orbit ABOVE THE CENTER OF THE EARTH)

Gravitational force is given by: Fg = (GMm)/(r^2) (where G = gravitational constant = 6.67x10^-11, M = mass of the earth, which is 5.98x10²⁴ kg, and r and m were defined above).

Set the two forces equal to each other: (mv²)/r = (GMm)/r²

You see that the mass of the satellite cancels out, and you can move the r around to get:

GM = rv²

Now you need to find both the velocity and the radius, which seems impossible with only one equation. (you need to know r and v). However, you can use the fact that the orbital period is 24 hours. You'll just need to convert your speed v into something more manageable.

Remember that speed = distance traveled / time taken. Since we assume it goes in a circle (this is a simplification - orbits are actually ellipses), then the distance traveled in one orbit is the same as the circumference of a circle: dist = 2*pi*r. You can also say that the time taken is the orbital period, which was stated earlier.

Then, you can say that v = 2*pi*r / T

Remember that we were working on the equation GM = rv²

Now replace v to get: GM = (4*pi²*r³)/T²

Then you can solve for r: r³ = GMT²/(4*pi²).

You solve for r by finding the cube root of the right side and put in the values for G and M and T and finally get:

r = 4.2x10⁷m. BUT, this is just the height above the center of the earth. If you want to know how high it is above the SURFACE of the earth, you have to subtract the radius of the earth, which is 6378000m, so you get 35 873km above the earth, or 35 873 000 m above the surface of the earth. Phew! (you can check on wikipedia for geostationary orbit height, and this matches)

Then you can find the orbital speed. Remember that we have figured out that v = 2*pi*r / T Now you just plug in your answer for r (the one that is the radius above the center of the earth), and put in T and you get:

v = 3073 meters / sec = 11061km/hour

I hope that helps!