Orbit changes, Kepler's Third Law

  • Thread starter jcook735
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  • #1
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Tidal effects in the Earth-Moon system are causing the Moon's orbital period to increase at a current rate of about 35 ms per century. Assuming the Moon's orbit around the Earth is circular, to what rate of change in the Earth-Moon distance does this correspond? Hint: differentiate Kepler's third law.



Homework Equations


I know that I need to use T2 over r3 = T2 over r3


The Attempt at a Solution



I tried converting the moon's period (27.3 days) to ms and putting the 35 ms into ms per day into ms per century, but that yielded no results, since the number was so small the ms/day didnt change the moon's period at all in my calculator, maybe im supposed to do this one by hand? This method doesn't seem feasible sicne i will have to be squaring things, can someone help me?
 

Answers and Replies

  • #2
gneill
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Perhaps you might begin by setting aside the calculator and performing the suggested differentiation (symbolically). Differentiate Kepler's third law with respect to time to find a relationship between the rate of increase of period and the rate of increase of the orbit's semi-major axis.
 
  • #3
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Isnt the differentiate of Kepler's Third law my relevant equation?
 
  • #4
D H
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Isnt the differentiate of Kepler's Third law my relevant equation?
Where have you calculated the derivative of Kepler's third law with respect to time?
 
  • #5
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Oh is that T2=4(pi)2r3 / GM?
 
  • #6
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the only other thing i could think of is 2T over r cubed but that doesnt seem right..
 
  • #7
D H
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T is the period, not time. You have been given DT/dt.

Given a value for dT/dt, and given Kepler's third law, what is da/dt?
 

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