## August 31, 2004

### One Hundred Interesting Mathematical Calculations, et Cetera: Number 26: Orbiting the Earth

Orbiting the earth. From PBS:

TeacherSource . Math . Orbiting the Earth | PBS: For a spacecraft, satellite, or other body (such as the moon) that orbits the earth, an equation links 1) how high above the earth the object is, and 2) the period or time it takes the object to complete one orbit. This equation is

Period = 2 (pi) x square root (a3/mu)

The Period is the time in seconds it takes to complete one orbit.

a is called the "semi major axis." Calculate this by taking the average of the altitude when the object is closest to the earth with the altitude when it is farthest from the earth and then add the radius of the earth. Do this calculation in kilometers.

Mu is the gravitational constant of the earth. This number is 398,601 km3/sec2.

For your information:
Miles x 1.609 gives you kilometers.
Kilometers x 0.6215 gives you miles.
Perigee: for an object orbiting the earth, this is the point closest to the center of the earth.
Apogee: for an object orbiting the earth, this is the point farthest from the center of the earth.

6. John Glenns actual orbit in the Mercury spacecraft ranged from 159 km (perigee) to 265 km (apogee) above the earth. Calculate the time it took Glenn to complete one orbit using this method. The radius of the earth is approximately 6382 kilometers.

7. On the space shuttle mission, John Glenn and the rest of the crew will be about 300 miles or 482 kilometers above the earth. Assuming a circular orbit, how long will it take them to complete on orbit?

Have you heard of people who have dishes in their yard or on their houses to receive TV channels? These dishes work off of satellites that are in Geosynchronous orbits. Geosynchronous means that from our vantage point on earth, the satellite seems to remain in the same place in the sky. These are circular orbits in which the time it takes to complete one orbit is the same as the 24 hours it takes the earth to complete one revolution. So people who use these dishes dont have to go out and adjust the dish.

8. Using the formula above, figure out how high a satellite would need to be for a geosynchronous orbit. Remember that a includes the radius of the earth. Convert your answer to miles when you are done.

Another interesting problem--but harder--would be deriving Period = 2 (pi) x square root (a3/mu)

Posted by DeLong at August 31, 2004 11:45 AM | TrackBack
Comments

Upon completion of these exercises, send your work and a CV to any publication which employs Gregg Easterbrook.

Posted by: P O'Neill at August 31, 2004 12:54 PM

Gee, Celestial Mechanics. Deriving all of this from first principals.

With a central mass

acceleration = G M / r^2

but, for a circular orbit

acceleration = Omega^2 r

(omega = 2 PI / Period)

Thus

G M / r^2 = Omega^2 r

or

G M = Omega^2 r^3 (Kepler's Law)

Thus

Period^2 = r^3 (2 PI)^2 / GM
or
Period = 2 PI SQRT(r^3 / GM)

(GM is the "mu" in the post, and for a circular orbit r = a)

Everything else is just arithmetic.

Posted by: Marshall Eubanks at August 31, 2004 01:21 PM

Oh, and don't forget that one complete sidereal rotation of earth is about 23 hours 56 minutes. The last 4 minutes (1/365th of a day) are spent catching up with the sun, so to speak.

Posted by: joe at August 31, 2004 01:51 PM

Oops, I see you mentioned that in the previous post. Never mind.

Posted by: joe at August 31, 2004 01:56 PM

Isn't this the calculation Newton once did, roughly, that, when it give him a nearly correct value for the time it takes the moon to circle the earth, made him realize that it is earth's gravity that keeps the moon in its orbit?

Posted by: Bruce Garrett at August 31, 2004 02:15 PM

For some more fun with this formula, here are some altitudes for various Shuttle missions.
http://hypertextbook.com/facts/2000/AndresMok.shtml

And here are the apogee and perigee for the Space Station's nominal orbit.
http://tinyurl.com/6kkca

Posted by: melior at August 31, 2004 03:25 PM

But assuming a circular orbit is cheating since the formula is true for an elliptical orbit as well!!!!!!!!!!!!!!

Posted by: davidf at August 31, 2004 05:02 PM

Another physics phun phact, and this one explains WHY Easterbrook is so wrong:

The Earth is a very good approximation to a sphere. With just first-year college physics and calculus (Easterbrook, are you listening?), it's easy to demonstrate that a sphere (for objects outside it) has the same gravitational properties of a point mass -- a speck. By extension, any two spheres of the same mass (again, as long as the orbiting object is outside r) have identical gravitational properties.

Keeping mass the same, you could bloat the earth to double its size, and the only thing that would change is your commute. No need to adjust your satellite dish -- or your satellite.

Posted by: Dragonchild at August 31, 2004 05:35 PM

Yes, Newton (and Halley and Hooke) did basically
the same calculation. IIRC, the first time (with the apple) Newton
got one of his constants wrong, and so got the wrong local acceleration, so he dropped this line of thought, but the second time Halley was with him, corrected the error, and he got the correct number and was encouraged to write and publish Principia.

Here is another fun fact - if a body has average
density rho and radius R, then it has mass M = 4 rho R^3 / 3

So, an orbit that skims the surface has

Period = 2 PI SQRT(R^3 / GM)
or
Period = 2 PI SQRT(3 R^3 / 4 G rho R^3)
or
Period = 2 PI SQRT(3 / 4 G rho)

So, a "low" orbit takes 90 minutes (more or less) for any size body with the Earth's mean density (~ 5.5 gm / cm^3).

A low Mars orbit or a low Venus or a low Mercury orbit - all about the same period - and, if you took your bowling ball (about the same density) into space, a speck of dust would also orbit it
every 90 minutes, if nothing disturbed it.

All, to my taste, much more interesting than the RNC.

Posted by: Marshall Eubanks at August 31, 2004 06:51 PM

All I can say to all of you is ...wow.

Posted by: victoria bond at August 31, 2004 08:54 PM

The circular orbit case is dead easy, as Marshall Eubanks indicates above. [IIRC, we covered this in my first high-school mechanics class and I suspect one could do so even earlier.] As far as I know, though, the elliptical orbit requires second-semester calculus (and is usually only taught in multivariable).

Posted by: Anarch at September 1, 2004 12:34 AM

yes, the reference to 'semi-major axis' makes it inclusive of an elliptical orbit. Kepler's law also specifically works for elliptical orbits and reduces to the simpler r^2 for a circle, which is a general case of an ellipse where a=b.

Posted by: Suresh Krishnamoorthy at September 1, 2004 09:58 AM