Suppose you are living on a straight line. There are several satellites at height 20,000 kilometers and you get readings saying that satellite 1 is directly above the point x 1 ± 10 − 10 and is at a distance h 1 = 21 , 000 ± 10 − 2 from you, satellite 2 is directly above x 2 ± 10 − 10 and at a distance h 2 = 52 , 000 ± 10 − 2 . Where are you ( x 0 ) and to what accuracy? Hint: Consider separately the cases x 1 &lt; x 2 and x 2 &gt; x 1 .My results:Actually we have two equations: ( x 1 − x 0 ) 2 + ( 2 ⋅ 10 4 ) 2 = h 1 2 ( x 2 − x 0 ) 2 + ( 2 ⋅ 10 4 ) 2 = h 2 2 We can express x 0 from them as: x 0 = x 1 + x 2 2 + ( h 1 − h 2 ) ( h 1 + h 2 ) 2 ( x 2 − x 1 ) But how to compute accuracy of measurement? I'm troubling with x 2 − x 1 in denominator.

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Suppose you are living on a straight line. There are several satellites at height 20,000 kilometers and you get readings saying that satellite 1 is directly above the point       x    1    ±      10          −      10       and is at a distance       h    1    =  21  ,  000  ±      10          −      2       from you, satellite 2 is directly above       x    2    ±      10          −      10       and at a distance       h    2    =  52  ,  000  ±      10          −      2      . Where are you (      x    0  ) and to what accuracy? Hint: Consider separately the cases       x    1    &amp;lt;      x    2   and       x    2    &amp;gt;      x    1  .My results:Actually we have two equations:  (      x    1    −      x    0        )    2    +  (  2  ⋅      10    4        )    2    =      h    1    2    (      x    2    −      x    0        )    2    +  (  2  ⋅      10    4        )    2    =      h    2    2  We can express       x    0   from them as:      x    0    =                    x        1            +              x        2              2    +            (              h        1            −              h        2            )      (              h        1            +              h        2            )              2      (              x        2            −              x        1            )      But how to compute accuracy of measurement? I&#039;m troubling with       x    2    −      x    1   in denominator.

tennispopj8 · Accepted Answer

Your expression for       x    0   is incorrect. Subtracting the first equation from the second (which also needs to be amended, as given in my comment), we obtain:  (      x    2    −      x    1    )  (      x    2    +      x    +    −  2      x    0    )  =      h    2    2    −      h    1    2  Which means:      x    0    =                    x        1            +              x        2              2    −                    h        2        2            −              h        1        2                    2      (              x        2            −              x        1            )      Note that errors in measurement are much smaller than the measurements themselves. Thus, taking a differential, which can be approximated as the error, we get:  d      x    0    =            d              x        1            +      d              x        2              2    −      (                  (                  x          2                −                  x          1                )        (        2                  h          2                d                  h          2                −        2                  h          1                d                  h          1                )        −        (                  h          2          2                −                  h          1          2                )        (        d                  x          2                −        d                  x          1                )                    2        (                  x          2                −                  x          1                          )          2                      )  Here,   d      x    1    =      10          −      10        =  d      x    2   and   d      h    2    =      10          −      2        =  d      h    1  .The second differential is merely a consequence of   d  (      v    u    )  =            v      d      u      −      u      d      v              u      2      .Hence error in       x    0  , as well as       x    0   itself, is calculated.

Suppose you are living on a straight line. There are several satellites at height 20,000 kilometers

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