| Composition Setup |
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By navigating to the site
http://spaceflight.nasa.gov/realdata/tracking/
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you
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can
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obtain
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tracking
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information
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giving
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the
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altitude
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and
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speed
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of
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the
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International
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Space
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Station
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(ISS).
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Suppose
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that
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the
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information
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displayed
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on
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the
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site
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was
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as
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shown
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in
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the
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screen
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capture
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above.
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By
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making
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the
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assumption
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that
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the
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space
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station's
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orbit
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is
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a
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circle
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with
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its
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center
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at
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the
...
center
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of
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the
...
earth,
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| Excerpt |
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find |
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the |
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approximate |
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magnitude |
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of |
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the |
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acceleration |
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experienced |
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by |
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the |
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space |
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station |
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as |
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a |
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result |
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of |
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the |
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gravitational |
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pull |
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of |
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the |
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earth. |
Solution
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Based upon the assumption of uniform circular motion, the acceleration of the ISS must satisfy:
| Latex |
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{excerpt} h4. Solution *System:* The ISS will be treated as a [point particle]. *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
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know
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that
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v
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=
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7699.59
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m/s,
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but
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r
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requires
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some
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thought.
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The
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altitude
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of
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346450
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m
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is
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not
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the
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full
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radius
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of
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the
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orbit,
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it
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is
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only
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the
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height
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of
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the
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ISS
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above
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the
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surface
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of
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the
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earth.
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To
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find
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the
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full
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orbital
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radius,
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we
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must
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add
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on
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the
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earth's
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radius.
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The
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earth's
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radius
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can
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be
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found
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on
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the
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web
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or
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in
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a
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number
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of
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books
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to
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be
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R
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e =
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6,380,000
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m.
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Thus,
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the
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orbital
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radius
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is
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r
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=
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6,730,000
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m.
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With
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this
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determined,
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we
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find:
| Latex |
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}\begin{large} \[ a = \mbox{8.81 m/s}^{2} \] \end{large}{latex} {note} |
| Note |
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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} |
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