ALGEBRA

PROBLEM 6.

PROBLEM 6. In Chapter 9, Problem 7, we show that the sidereal period (in seconds) can be computed by the formula P = 2pi(a^3/GM)^.5, where a is the average radius of orbit from the center of the body about which the satellite is in motion, G is the constant of universal gravitation, and M is the mass of the body about which the satellite orbits.

a. Find the sidereal period of the High Energy Astronomy Observatory (HEAO) satellites, which have an average altitude above Earth of 430 km. The radius of the Earth averages 6370 km, and the value of the product GM for Earth is 3.99 x 10^14 m^3/sec^2.

Solution: The radius of orbit is the sum of the radius of Earth and the average altitude of the satellite:


a = 6370 km + 430 km = 6.80 x 10^6 m.

Then the sidereal period in seconds is


P = 2(3.142)[(6.80 x 10^6)^3/(3.99 x 10^14)]^.5
P = (6.284)(6.80)[(6.80/3.99)]^.5 x 10^2 = 5580 s

The sidereal period then is 93.0 minutes, or 1.55 hours.

b. Compute the synodic period of the HEAO satellites, given that their orbits are posigrade.

Solution: In Fig. 3.1, let x be the position of the observer (assumed on the equator) when the satellite is directly overhead and let y be the observer's position one synodic period later, due to the rotation of Earth.

If we can find the angular distance A, we shall be able to use it to find the synodic period. In one synodic period the observer traveled an angular distance A, and the satellite traveled an angular distance 360 + A, measuring the angular distance in degrees. The observer travels 360 in 24 hours, or 1 in 24/360 hours, so it takes the observer (24/360) (A) hours to travel the angular distance A. From part (a), the sidereal period is 1.55 hours. It takes the satellite 1.55/360 hours to travel 1, and the time that elapses between successive viewings over the observer is therefore ( 360 ) (360 + A) hours.


24/360 A = 1.55/360 (360 + A)
24 A = (1.55)(360) + 1.55 A
22.45 A = 558
A = 24.9 degrees

Problem 6.b.

b. Compute the synodic period of the HEAO satellites, given that their orbits are posigrade.

Solution: In Fig. 3.1 (below), let x be the position of the observer (assumed on the equator) when the satellite is directly overhead and let y be the observer's position one synodic period later, due to the rotation of Earth.

ALGEBRA PROBLEM 6

If we can find the angular distance A, we shall be able to use it to find the synodic period. In one synodic period the observer traveled an angular distance A, and the satellite traveled an angular distance 360 + A, measuring the angular distance in degrees. The observer travels 360 in 24 hours, or 1 in 24/360 hours, so it takes the observer (24/360) (A) hours to travel the angular distance A. From part (a), the sidereal period is 1.55 hours. It takes the satellite 1.55/360 hours to travel 1, and the time that elapses between successive viewings over the observer is therefore (1.55/360 )(360 + A) hours.


(24/360) A = (1.55/360) (360 + A)
24 A = (1.55)(360) + 1.55A
22.45 A = 558
A = 24.9 degrees

So the synodic period is (1.55/360) (360 + 24.9) = 1.66 hours = 99.6 minutes. We observe that the synodic period is 6.6 minutes longer than the sidereal period.

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