Consider a moon that is 1.15 x 10^6 km from the planet it is orbiting .

  • Thread starter vipson231
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Consider a moon that is 1.15 x 10^6 km from the planet it is orbiting.....

Homework Statement



Consider a moon that is 1.15 x 10^6 km from the planet it is orbiting. This moon makes one complete orbit every 27 days. What is the orbital period of a satellite orbiting the planet at a distance of 4.10 x 10^6 km?

a)96.3 d
b)181.8 d
c)63 d
d)4.01 d
I was given a formula T^2/r^3 = k constant

So i set up the equation (27)^2/(1.15 x 10^6)^3= T^2/ (4.10 x 10^6)^3 and got T^2= 3.30 x 10^22 and take square root of that which gives me T=1.87151521 x 10^11. However my answer is not a choice above. Did i do something wrong with my calculation?
 

Answers and Replies

  • #2
lewando
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...Did i do something wrong with my calculation?
Yes. T2 is not equal to 3.3x1022.
 
  • #3
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:O Thx I'll go back and see what I did wrong
 
  • #4
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Oh I made a careless computation error I input all the values at once in my calculator and got 181.75 ≈181.8.
(27^2)(4.10x10^6)^3/ (1.15 x 10^6)^3 = 33035.85699

radical of 33035.85699 = 181.75 = 181.8 Cheers! =)
 

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