# Th1 Solution - 36 th International Physics Olympiad...

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Unformatted text preview: 36 th International Physics Olympiad. Salamanca (España) 2005 Th 1 Solution Page 1 of 5 R.S.E.F. Th1 AN ILL FATED SATELLITE SOLUTION 1.1 and 1.2 ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎨ ⎧ ⋅ = ⇒ = ⋅ = ⇒ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⇒ ⎪ ⎪ ⎪ ⎪ ⎭ ⎪ ⎪ ⎪ ⎪ ⎬ ⎫ = = = m/s 10 07 3 m 10 22 4 4 2 3 7 3 1 2 2 2 2 2 2 . v r g R v . r T R g r R GM g T r v r v m r m M G T / T T T T π π 1.3 ⇒ = = 2 2 v m v R g v m r L T 2 v R g m L T = ⇒ − = − = − = 2 2 2 2 2 2 1 2 1 2 1 mv mv r m R g mv r m M G mv E T T 2 2 1 mv E − = 2.1 The value of the semi-latus- rectum l is obtained taking into account that the orbital angular momentum is the same in both orbits. That is ⇒ = = = = 2 2 2 2 2 4 2 2 2 2 1 r v R g m R g v R g m m M G L l T T T T r l = The eccentricity value is 3 2 2 2 2 2 1 m M G L E T + = ε where E is the new satellite mechanical energy ( ) 2 2 2 2 2 2 1 2 1 2 1 2 1 mv v m E v m r m M G v v m E T − ∆ = + ∆ = − ∆ + = that is ( ) 1 2 1 1 2 1 2 2 2 2 2 − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = β ∆ mv v v mv E Combining both, one gets β ε = This is an elliptical trajectory because 1 < = β ε . 36 th International Physics Olympiad. Salamanca (España) 2005 Th 1 Solution Page 2 of 5 R.S.E.F....
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Th1 Solution - 36 th International Physics Olympiad...

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