Newton's Variation of Kepler's third law

  • Thread starter Jess048
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Calculate the period of the Earth's moon if the radius of orbit was twice the actual value of 3.9 x 10^8 m.

a. 1.13 x 10^6 s or 13 days
b. 2.3 x 10^6 s or 26 days
c. 5.14 x 10^6 s or 59 days
d. 6.85 x10^6 s or 79 days

So far I got: v=velocity; T=time; h= hieght from surface
v=?
T=?
h=?
re= 6.38 x 10^6
me= 5.97 x 10^24
G= 6.67 x 10^-11

Another question I have is, if the radius of earth is doubled but the mass remains the same what would happen to the acceleration due to gravity. Would I divide 9.80 by 2, multiply by 2, or multiply by 1/4.
 

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  • #2
HallsofIvy
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You titled this "Newton's variation on Kepler's third law"! How about using Kepler's third law? It states that the period of an orbit, squared, is proportional to the cube of the radius of the orbit: p2~ r3. Of course, the proportionality depends on the mass of the object being orbited (that was "Newton's variation") and so differs for objects orbiting the sum, objects orbiting the earth, etc. But here, we are talking about orbiting the earth only. Use the present period and radius of the moon to find the constant of proportionality: [itex]k= \frac{p^2}{r^3}[/itex] then use the new radius (twice the old) with [itex]p^2= kr^3[/itex] to find the new period. Actually, with a little thought you don't need to actually find k: r is doubled so p2 is multiplied by ___ and so p is multiplied by ___
 

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