Lecture5_Oct03_2007

# Lecture5_Oct03_2007 - R E C A P The Solar System Newtons...

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Unformatted text preview: R E C A P The Solar System Newtons and Keplers Laws Revisted Deriving Keplers Third Law It is far better to grasp the universe as it really is than to persist in delusion, however satisfying and reassuring. &quot;The brain is like a muscle. When it is in use we feel very good. Understanding is joyous.&quot; Carl Sagan R E C A P Newtons and Keplers Laws Revisted Deriving Keplers Third Law M 1 M 2 R 1 R 2 V 1 V 2 R E C A P Newtons and Keplers Laws Revisted Deriving Keplers Third Law This equation, M 1 R 1 = M 2 R 2 , defines the center of mass and its position in the simple system we set up. The smaller mass is at a greater distance from the center of mass, and vice versa for the larger mass. Recall that we already established that M 1 R 2 V 2 2 M 2 R 1 V 1 2 and so having established a relationship between R and M above we may get the following additional equality M 1 V 2 M 2 V 1 = = Newtons and Keplers Laws Revisted Deriving Keplers Third Law Now, further consider the force between the objects. Lets consider object 2 the less massive object and work from that perspective (note that the other is equally valid though). Using both Huygens rule, and Newtons Universal Law of Gravity, we note that the force acting on object 2 is- GM 1 M 2- M 2 V 2 2 (R 1 +R 2 ) 2 R 2 F = = now lets do some further manipulation and re-arranging GM 1 M 2 M 2 V 2 2 4 R 2 2 (R 1 +R 2 ) 2 R 2 4 R 2 2 = GM 1 M 2 M 2 R 2 4 2 V 2 2 (R 1 +R 2 ) 2 4 2 R 2 2 = Note that the circled term on the far right is 1/P 2 Newtons and Keplers Laws Revisted Deriving Keplers Third Law The total separation between objects, R , is simply R 1 +R 2 , and so we get (after canceling M 2 as well) that GM 1 4 R 2 2 R 2 P 2 = Now, rewrite the separation, R , as follows R = R 2 (1+R 1 /R 2 ) And similarly, note that the total mass of the system, M , can be similarly defined, so that M 1 = M/(1+M 2 /M 1 ) Recall from earlier the equivalence of mass and distance ratios in this system; the parenthetical expressions above are in fact equal, and hence M 1 / R 2 =M/R Newtons and Keplers Laws Revisted Deriving Keplers Third Law Substituting in the expression for M 1 , we get A further final manipulation yields GR 2 (M/R) 4 R 2 2 R 2 P 2 = Now, cancel R 2 , and rearrange a bit, and get GM 4 2 R 3 P 2 = P 2 = 4 2 R 3 GM This is Keplers Third Law expressed in a manner similar to the text, but including the terms which make the units work properly. Note that nowhere in this derivation have we specified the bodies ; this formula works in any binary system! (Or to a good approximation, in any multiple system, like the Solar System, where there is one dominant body and many secondary bodies.) The Solar System Basic Structure of the Solar System The Sun The sun is a mass of incandescent gas A gigantic nuclear furnace Where hydrogen is built into helium At a temperature of millions of degrees Yo ho, it's hot, the sun is not...
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## This note was uploaded on 04/07/2008 for the course PHSC 119 taught by Professor Gladders during the Fall '08 term at UChicago.

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Lecture5_Oct03_2007 - R E C A P The Solar System Newtons...

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