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Rotational Motion Problem

  • Thread starter Digdug12
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  • #1
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Homework Statement


An object of mass 7.1x1023 circles the earth and is attracted to it with a force whose magnitude is given be Gmem/r2. If the period of rotation is 32 days, what is the distance from the earth to the object? Here G=6.67x10-11 Nm2/kg2, me=6x1024 kg.


Homework Equations


Fg=G*m*Me/r2
T=2pi/Omega

The Attempt at a Solution


I used the first equation to isolate r, the distance from the earth to the object to get sqrt((G(m*Me)/Fg)=r
But this is where I get stuck, i dont see how to find the other unknown, Fg I'm guessing it has to do something with the period of 23 days, but I cannot connect the two in my head.
 

Answers and Replies

  • #2
Doc Al
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Hint: Apply Newton's 2nd law. How is the object accelerating?
 
  • #3
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I set it up so that Fg=Mac, since Ac=V2/r, but the radius is still in the equation. If substituted it back in i would have another unknown, velocity. What am i missing here? is there another force?
edit:
Should i use v=r*omega, Period=2pi/omega, and substitute v=r*(28/2pi) into the Fg=M(V^2/r)?

edit2:
i got the final equation GMeM/M(28/2pi)^2=r but its off by roughly 1x10^2., is there a problem with the period? I see that i used 28 days instead of 32 but i tried it with 32 also and it was still off, do i need to convert days into seconds or something?
 
Last edited:
  • #4
Doc Al
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edit:
Should i use v=r*omega, Period=2pi/omega,
Good.
and substitute v=r*(28/2pi) into the Fg=M(V^2/r)?
You messed that step up a bit. Redo.
 
  • #5
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ok so i setup it up to look like this:
Fg=m*r*(32[tex]/2pi[/tex])2
Then i substituted it back into the original equation and i got:
G*M*Me/M*(32[tex]/2pi[/tex])2=r
and this comes out to 1.87e10
the correct answer is supposed to be 4.26e8
 
  • #6
Doc Al
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ok so i setup it up to look like this:
Fg=m*r*(32[tex]/2pi[/tex])2
Still not right. For some reason, you are inverting things. v = r*(2pi/T), not r*(T/2pi). Redo your equation for v from post #3.
 
  • #7
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ahh thanks for catching that Doc. Al :D I'm sorry i still havent solved it, im really getting frustrated haha... i made the change in the equation but my answer is still off, is the mass in Fg=M*ac the mass of the object, right?
 
  • #8
Doc Al
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is the mass in Fg=M*ac the mass of the object, right?
Right.

Show me your final equation.
 
  • #9
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for the first few tries i canceled one r on each side of the equation, it came out to be
r=G*M*Me/M*(2pi/32)^2

then i realized i was cancelling a r with a 1/r, so i carried the r^2 over and it became r^3, so i took the cube root of the answer, tried cancelling the M's, and still nothing.
 
  • #10
Doc Al
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The M's will cancel and you will need to take a cube root. What's the final version of your equation for r³?

Be sure to convert the period from days to seconds! (D'oh!)
 
  • #11
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oooo, sweet. thanks, I got it now :D thanks a bunch
 

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