

Subject:
Orbital mechanics
Category: Science > Physics Asked by: halejrbga List Price: $5.00 
Posted:
09 Dec 2002 17:55 PST
Expires: 08 Jan 2003 17:55 PST Question ID: 122122 
If a satellite is in a circular orbit above Earth at an altitude of 4000 miles, how fast is it moving and how long would it take to make one revolution around the Earth? 

Subject:
Re: Orbital mechanics
Answered By: haversianga on 09 Dec 2002 19:16 PST Rated: 
Interesting question, halejrbga! First, let's tackle the speed, since it's the hard part of your question. The diameter of the earth is about 12.7 million meters, or 6.35 million meters diameter, plus about 6.5 million meters above the surface gives not quite 13 million meters from the earth's center of mass to the satellite. To achieve a stable orbit, the gravitational force pulling the satellite towards the earth must equal the centripetal force pushing it away. If I remember my physics right: M1*M2*G/r^2 = gravitational attraction M1v^2/r) = centripetal force We factor out M1 (the mass of the satellite), and r, leaving and are left with M2 * G / r = v^2, or M2 kg * 6.673 * 10^11 m^3 s^2 kg^1 / 13EE6 m = v^2 m^2 s^2. (meters, kilograms, and seconds). The mass of the Earth is about 6EE24 kg. Checking the units, they cancel (the left hand side gives m^2/s^2, as does the right). Performing the square root, we find v = ~5550 m/s (about 12.5 thousand miles per hour). Now, as for one revolution around the earth. The circumference of a circle is pi * diameter. The diameter of the circle the satellite sweeps out is about 26 million meters, so the circumference is about 81.5 million meters. 81.5 million meters / 5550 meters / second = 15000 seconds = about 4 hours. This makes intuitive sense since 4 hours is one sixth of 24 hours, and a geosynchronous orbit, which takes 24 hours, is 6 times as high. After all those units, we get "go twice as high, and it takes twice as long". I hope whatever you need this for is as interesting as finding the answer was! Haversian 
halejrbga
rated this answer:
A great answer and a great service for those of us who never took physics. 

Subject:
Re: Orbital mechanics
From: javitga on 17 Dec 2002 13:34 PST 
Please note that the above answer assumed that the body is orbiting Earth in a circular path ON THE SAME PLANE AS THE EQUATOR. Currently an Earth orbit with radius 4000 miles is only useful for earth sciences and astronomy. Its period of about 4 hours make it virtually useless for communication purposes and its height makes it useless for Earth observation. For example, there were two satellites called Lageos1 and Lageos2 which were deployed in 1976 and 1992 respectively. These tiny satellites orbited our planet at an altitude of nearly 4000 miles. The lageos satellites are passive vehicles covered with retroreflectors designed to reflect laser beams transmitted from ground stations. By measuring the time between transmission of the beam and reception of the reflected signal from the satellite, stations could precisely measure the distance between themselves and the satellite. Regards. 
Subject:
Re: Orbital mechanics
From: racecarga on 05 Mar 2003 16:05 PST 
"Go twice as high and it takes twice as long" is incorrect. The period of a satellite is proportional to the radius of the orbit to the power 1.5. So "go 4 times as high, takes 8 times as long" is correct (if you measure height as the distance from the center of the earth), though not very useful. If you measure height from the surface of the earth, the period of the satellite is not proportional to any power of the height. It is just happenstance that the period of a satellite at 1/6 the altitude above the earth's surface as a geosynchronous satellite is about 1/6 of a day. Actually, such a satellite would orbit 6.32 times each day. 
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