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!spacestation.gif!

By navigating to the site [http://spaceflight.nasa.gov/realdata/tracking/] you can obtain tracking information giving the altitude and speed of the International Space Station (ISS).  Suppose that the information displayed on the site was as shown in the screen capture above.  By making the assumption that the space station's orbit is a circle with its center at the center of the earth, find the approximate magnitude of the acceleration experienced by the space station as a result of the gravitational pull of the earth.

h4. Solution

*System:*  The ISS will be treated as a [point particle, subject to an external ].

*Interactions:* External influence from the earth (gravity).

*Model:*  [Uniform Circular Motion].

*Approach:*  We know from the Law of Interaction that the acceleration experienced by the ISS must satisfy:

{latex}\begin{large}\[ a = \frac{v^{2}}{r} \] \end{large}{latex}

We know that _v_ = 7699.59 m/s, but _r_ requires some thought.  The altitude of 346450 m is not the full radius of the orbit, it is only the height of the ISS above the surface of the earth.  To find the full orbital radius, we must add on the earth's radius.  The earth's radius can be found on the web or in a number of books to be _R_~e~ = 6,380,000 m.  Thus, the orbital radius is _r_ = 6,730,000 m.  With this determined, we find:

{latex}\begin{large} \[ a = \mbox{8.81 m/s}^{2} \] \end{large}{latex}

{note}Earth's gravity is *not* insignificant at level of the ISS's orbit.  If it was, the ISS would just fly off into space!{note}