ANSWERS to EVSC494/794 PROBLEM SET 1

1. Assume the Moon is in a circular orbit around the Earth. Calculate its period of rotation, T, in days.

The period of rotation of a satellite in circular orbit is just the distance traveled in one revolution divided by the velocity of rotation. In the first lecture on orbits it was shown that the gravitational force acting on a satellite in orbit is balanced by the centrifugal force which is a fuction of its mass times its velocity squared divided by the radius of the orbit. By substituting for the definition of the orbital velocity (the distance traveled in one revolution divided by the period, or 2p r/T) in the centrifugal force, and rearranging we came up with the following:

(Note this expression describes Kepler’s third law, as applied to satellites of the earth, that is, the ratio of the period squared to the radius cubed is a constant.)

The radius of rotation in this equation is the distance from the Earth to the Moon (384,400km) plus the radius of the earth (6371km). Recall that we have to assume the mass of the earth is a point mass, and the radius is therefore defined from the center of the earth.

Recall the definition of a Newton, which is just a force, and has the units of mass times acceleration or simply (kg m/sec²)

You should be able to inspect this answer and recognize this as something you know, the period of the moon is indeed about 28 days, there are 28 days between any given phase of the moon (eg., full moon to full moon or new moon to new moon).

2. What is the magnitude of the acceleration that holds the moon in its orbit? Express your answer in m/sec² .

We know that F=ma, and for our satellites in orbit, the force is defined by the gravitational attraction between the two masses (the mass of the earth and in this case, the mass of the moon).

Note this is a rather small acceleration.

3. If NOAA has a satellite that orbits the earth every 102 minutes, and we assume the orbit is circular, how high above the surface of the earth is the satellite located?

T=102 minutes=102*60=6120 seconds

T²=3.745x10⁷ seconds

If the radius is calculated to be 7230km, then the distance of this satellite above the surface of the earth is just 7230km-6371km=859km. You can check the NOAASIS web site to see that in fact the NOAA polar orbiter’s have characteristics similar to this with periods of about 102 minutes and altitudes above the surface of the earth of about 860km.

4. Use equation 1.3 to determine the distance above the earth of a satellite in geosychronous orbit (an orbit in which the satellite has the same angular velocity as the Earth: 7.292115x10^-5 rad sec^-1. Recall that the angular velocity is simply 2p/T where T is the period of rotation.

Given an angular velocity equal to that of the Earth, the period is just 86164 seconds, and we set up the following equation:

Again, we subtracted the radius of the earth to get the height of the geosynchronous satellite above the surface of the earth.

5. Go to the NOAASIS website and get the value of the semi-major axis of the GOES-8 satellite. How does it compare to the answer in number 4 above. Why are they different?

The NOAASIS site reports the semi-major axis of the GOES-8 to be 42,160.98180km. This is different from the answer to the last question because we calculated the distance of the satellite above the earth. However, the semi-major axis is analogous to r in equation 1.3, it is the distance from a point-mass, and we have made the assumption that this mass is concentrated at the center of the earth. Therefore, we can compare this semi-major axis with the value we calculated for r in the previous question. Now we see that our calculation differs by ~5km. This difference results from the fact that we assumed a perfectly circular obrit, which is not true for the real geosynchronous satellite GOES-8.