ps01sol - “ 7 € ’3 % „ C    9  Q ‘ $3 g@ # %  „...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: “ 7 € ’3 % „ C    9  Q ‘ $3 g@ # %  „ ‰ ˆ ‡ † 3 … SQ C  9  9' B & f !! V &!  ( 9    ) & 9    & e 0 &€ €)S  & 9 QBƒx€ & 9‚€SyQ „ & 0 B  Q !  ' & 9   X w v u G F D hx Q % S H  €Sy ax S w v u G F D S H A@ 2 F t s D C & f !! V &!  ( 9   ) & 9  0 ) B   ( ' & '  & 9    &  & 9   9   & e & V r & e  & Y 0 ) ! ) & &  qp U a $ 3 #   & ) i '  9  ) & 6   ( ' & '  ) & 6 B   b  V  ! !    9 V )  0 &  ' & ! e & ) & f ) ' 0 ) a U h S $ 3 g@ # '  ' 0 &  )  &!  ( 9    ) & 9  0 ) B    ) & &  ( c!! B  ' & '   & 6 B!  Y & 9 8 a U WS $3 $@ g & f !! V & '    & 6 B!  Y e ) 0 )  (  &    ' & 9   &  V   c  ) & 0 & 9    # d@ $  & '    c  ) & 0 & 9   b a U WS $3 R@ " 2 1 H D % `    &  V   & 6 B!  Y & 9   X 7 U W $3 "@ R 2 P  B 0  9   & & 9     0 &  B & V &  & 9 V Q U TS $3 R@ 4 % Q P I " H A@ % G F E D C    9  9   & & 9    & '   B  !    & 9    A@   & !   &   6   &  &  & 9 8 7 6 4 5 $ 43 2 1 '   '   0 )  (  &    '  & &  $ " " % $ # " ! !         ¢  ©¤ ¢ ¨ § ¦ ¥¤ £ ¢ ¡ V U U TS 9 R D 9 6 3 9 Q 6 TS 9 R D 2 6 3 0 2 Q 6!!$ %       %    &     )$  " $  $      $     ¨ !  &    )  &  PB  ' ¨ $  ¨  ¨      H  & "    " $     H  ( ! ! ¨ ) &    &  I @ 9 6 7 2 6   ! ! $ %  H  & "    $  (     ' $  %  ¨ ! ¨A &   $  ) $     G  ) ¨      $ ! ! ¨    (  !  )  ! ! $ %          ) $      F © ' 8 8 9 6 7 2 65 6 E D 9 6 7 2 65 4 C 9 3 9 B 7 2 B 0   " $  &  ' ' ¨  )$  $ (! & ¨  $!  ¨  & ' $ !!$ %  ' ¨  $  &$   9  ¨  9  2 5 98 9 6 7 65 4 C 29 3 9 B 0 98$ 26$ #7¨ 6 &4¨C %29 32B    % " ! ¨  $   " $ & ¨     )  &       %    (   !!  A¨ &   ' ¨  $     ©  @  ¨   '  $8 2 6 7 9 65 4 3 9 1 7 2 1    @   )$   ¨ !  &     & ¨ 9 6  ¨ 2 6  &   % 8 9 6 7 65 4 3 9 1 8 2 6 7 65 4 3 2 1 0   " $  &  ' ' ¨  )$   $ % (! & ¨  $!   ¨  & ' $ !!$ %   & ¨ % %   " $  $   $ # ¨ ¨ " ! ¨                   © ¨ § ¦ ¥¤ £ ¢ ¡ Problem 3: Radians and Estimation (see section 1.11 in course notes) a) Hold a dime at arm’s length. What angle in radians is subtended by the diameter of the dime? When I hold a dime at arm’s length, the distance from my eye to my fingertips is x ≅ 7.1× 101 cm (see figure below.) The diameter of the dime is y ≅ 1.8 cm . The radian measure of θ is given by θ = s r ≅ y x = 1.8 cm 7.1× 101 cm = 2.5 ×10−2 rad . b) Using your result from part a), estimate the length of the infinite corridor at MIT. In order to do this, choose a reference height at one end of the corridor and estimate its height. Then go to the other end of the corridor and measure what fraction of the diameter of the dime corresponds to your reference height. You can now calculate the length of the corridor by using similar triangles. The length is published by MIT. Can you find the published value? I used a doorway at the end of the infinite corridor (building 8) as a reference height which I measured to be h = 9.0 ft = 2.7 m . I measured the size of this height from the other end of the corridor compared to the dime and found it was approximately 1 2 of the diameter of the dime or y ≅ 1cm . Therefore the angular diameter of the door is approximately θ ≅ y x = 1cm 7.1×101 cm =1.4 ×10−2 rad ≅ 1× 10−2 rad . The length of the infinite corridor is then d ≅ h θ ≅ ( 2.7 m 1×10−2 rad ) ≅ 2.7 × 102 m . In the MIT Bulletin 1999-2000, pp.10 , the published result for the length of the infinite corridor is 8.25 × 102 ft = 2.51× 102 m . So our estimate is ( 2.7 ×102 m ) ( 2.7 × 102 m ) ≅ 1.1 of the actual value. c) Now use your dime and wait until the moon is out (the moon is new on Sept. 14 so you need to make a measurement as quickly as possible on the quarter moon) to try an estimate the angular diameter of the moon. Once you have this estimate, what additional information would you need in order to estimate the mass of the moon? Make some estimates regarding these additional quantities and then estimate the mass of the moon. Look up the actual value and compare it with your estimate. How did you do? I measured the moon and found the diameter of the moon corresponded to approximately 1 2 of the diameter of the dime or y ≅ 1cm . Note that this means that if you stand at one end of the infinite corridor and look at the sun or moon it will entirely fill the doorway. Try it out. There is one day a year when the sun sets exactly in front of the infinite corridor. Therefore the angular diameter of the moon is θ ≅ y x = 1cm 7.1×101 cm =1.4 ×10−2 rad ≅ 1× 10−2 rad . I estimate the distance from the earth to the moon to be approximately Re ,m ≅ 3 × 108 m . The diameter of the moon is then D = Re,mθ ≅ ( 3 × 108 m ) (1.4 ×10−2 rad ) = 4 × 106 m . The volume of the moon is Vm = ( 4 3) π ( D 2 ) = (π 6 ) D 3 ≅ (π 6 ) ( 4 × 106 m ) ≅ 3 × 1019 m3 . 3 3 I estimate the density of the moon to be ρ m ≅ 5 × 103 kg − m −3 , similar to the density of rocks found on Earth. The mass of the moon is then mm = ρ mVm = ρ m (π 6 ) D 3 ≅ ( 5 × 103 kg − m −3 ) (π 6 ) ( 4 × 106 m ) ≅ 1.7 × 1023 kg ≅ 1× 1023 kg 3 . The accepted standard value for the mass of the moon is 7.35 × 1022 kg . So our estimate was 1×1023 kg 7 × 1022 kg ≅ 1.4 times the accepted value. Not bad!!! E 2 8 7 s 3 54 6 3 G U 3 I  !  ¨ "% & ¨ $    # $  (¨     % $     2 4 r eq e 3 iph ` g 3 G W  A     " ¨ # R  !  " ¨ $ & ¨ # R   "  "   " #  % D      2 Y f 4 8 e 3 5 4 V 6 d 3 UW %  (  %    2 &  S%  " ¨ %  A "% # ) 7 3 `5 4 V 6 c U 3 b %  " ¨  "   " ¨ # R  %  "  "   " # %    D    A "% # )  $   " $ ¨ !   "      "  A ¨ "% # B %    % $  ` " % a " U 3 G V Y X 54 W ©   " ¨ $   ! ! ¨ $ ¨ !   " % V &   `%   % $    " % UT &  S% " ¨ % %   " ¨ # R          " $ ¨ !   "   %    % $    "    Q © A 2 $¨ !  "   %    %$   "     % " ¨  " P I H GF " ¨ "     ¨ "  A   %   " !  #   "   " &  ¨ "     "! % $  " E D    A    $   " $ ¨ !   "      " "   ! C #% "% # B %  % $    "   &   %   " @ 2 %   " !  #   "   "    "  A % "     "!  % $  "   "  A    $   "      " %   % $    "   &   %   "  " ¨  " %   %  % ¨ "@ 2  % $    " &  ¨ $ ¨ !   " $    )  "      " !  #    %  ' ¨ 1   "     ¨    " %    " % "   %  $  $  ¨ $    " $  (   $  9 2 8 76 3 54 3 1 " ¨ % "$ ¨ "% $ ¨ !   "      $    " " ¨ % "$ ¨ "%   % $     2 01   "  )     "!  $ &   "    ¨  %  !  (¨    %  ' ¨  " ¨  & $  ! !   " %    %  $ # "!           © ¨ § ¦ ¥¤ £ ¢ ¡ S ©S "    %" ! ¨     " "  # & " ' !" ¨        !  Q H R ! ¨ " !   ! ¨ "  # &" ' !"      !  Q ¨   ! ¨  ! ¨ " !  "  !   ! (E C  P I HG  F¨EC     B A©   D5 D 765 C  ¨  !  ! ¨ © 1 !    "    %" ! ¨    8    $  ¨  !  ¨  !& " 76 5 @ 9 " ¨ 4&& ¨   ¨   ¨ ( 3    $ 0      ¨ " " ( 0    $ 0    !   ! (       & ¨  '   !    © 2 1 "   %" ! ¨         ¨   " " # 0 ( ) !   ! (        % "  '        &   %    $ "  "    #  ! ¨  " !           ¨         © ¨ § ¦ ¥¤ £ ¢ ¡ e d 9 W 876 V 5 4 V Q U QW Y U 9 876 53 Q c 9 6 E D 534 2 W H C 0 2 b 4 V3 C )     " a   & !       !  A $  © !   B   ©  ! ( ! $  %  & IW H U CQW Y U IH 2 `QW Y U I H 2 W H )  B Q  $     ! "  $ (  ©  ©&   $  #     ©   X FW VP 1 U T 2 S )    #  ! "  $ (  ©   %    © (  ©    B  F9©6E D R5 4F PQ1 2 I H  B   $    © (  © B " B G  B  ©  ©&   $@ 98#6 54  © ¨  3 2 C 0)   !  !   B  &    !  A $  ©       ©   © # 7 3 2 1 0 )     (    ©   ©  ( ! %   " ' !   &  $    &  $  %  $  #  "  ©  !     ©          © ¨ § ¦ ¥¤ £ ¢ ¡ ...
View Full Document

Ask a homework question - tutors are online