Radius of Orbits

  • Thread starter Ryo124
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  • #26
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For G i put 6.67 x 10^11 G = 6.67 x 10^11
For Me i put 6 x 10^24 kg Me = 6.0 x 10^24 kg
For T i put 28800 sec T = 28800 sec

What am I doing wrong? Are my steps correct? What are you getting for your answer?
 
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  • #27
D H
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One error: you didn't square the entire expression, [itex]2\pi r/T[/itex]. However, correcting this will not fix the 1014 problem. As Dick noted, please show exactly what numbers you used.
 
  • #28
nrqed
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For G i put 6.67 x 10^11 G = 6.67 x 10^11
For Me i put 6 x 10^24 kg Me = 6.0 x 10^24 kg
For T i put 28800 sec T = 28800 sec
G is 6.67 times 10 to the minus 11
 
  • #29
D H
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G is 6.67 times 10 to the minus 11
Even more correctly, 6.67x10-11 m3/kg/s2. It is a good idea to always explicitly represent units.
 
  • #30
nrqed
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Even more correctly, 6.67x10-11 m3/kg/s2. It is a good idea to always explicitly represent units.
You are right. I just wanted to emphasize the problem with the exponent.
 
  • #31
D H
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You're right too. That is a very big problem.
 
  • #32
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Alright, I changed it and got 2.56 x 10^7 m for the radius of the orbit. . . Can someone work the problem out and see if this is the correct answer?

Also, no one told me if my final equation was correct... [2(pi)^2]r^3 = GmeT^2
Is this equation correct?
 
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  • #33
nrqed
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Alright, I changed it and got 2.56 x 10^7 m for the radius of the orbit. . . Can someone work the problem out and see if this is the correct answer?

Also, no one told me if my final equation was correct... [2(pi)^2]r^3 = GmeT^2
Is this equation correct?
I get [itex] 2.03 \times 10^7 m [/itex] You forgot to square the 2. It's (2 pi)^2
 
  • #34
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OK. I did it and got 2.04 x10^7 m. I did you just round up in your calculation?
 
  • #35
nrqed
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OK. I did it and got 2.04 x10^7 m. I did you just round up in your calculation?
I keep getting 2.03 x 10^7 m. I did not round off.
(I mean I rounded off my final answer but not in the intermediate steps)
 
  • #36
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OK. Weird, I did it again and got 2.03 x 10^7. Cool.

Thanks for your help and patience everyone, I guess I made this problem harder than it needed to be.
 
  • #37
D H
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Since you only used two decimal places of accuracy for G and Me, it would be better to say 2.0x107 meters in this case.

As an aside, it is often better to use the planet gravitational constant (the product of universal gravitational constant and the planet mass rolled into one constant) rather than the product of universal gravitational constant and the planet mass as separate values. What's the difference? Accuracy. In the case of the Earth, we know the product G Me to 9 places of accuracy: 398,600.4418±0.0008 km3/s2. In comparison, we only know G to 4 decimal places.
 

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