How Far is an Object Orbiting Earth if Its Period is 32 Days?

AI Thread Summary
An object with a mass of 7.1x10^23 kg orbits Earth with a period of 32 days, and the gravitational force acting on it can be expressed using Newton's law of gravitation. The discussion revolves around calculating the distance from Earth to the object by isolating the variable r in the gravitational force equation. Participants emphasize the importance of converting the orbital period from days to seconds and correcting errors in the velocity equation. After several attempts and clarifications, the correct approach involves using the relationship between gravitational force, centripetal acceleration, and orbital period to derive the final equation for r. Ultimately, the correct distance is found to be approximately 4.26e8 meters.
Digdug12
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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 don't 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.
 
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Hint: Apply Newton's 2nd law. How is the object accelerating?
 
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:
Digdug12 said:
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.
 
ok so i setup it up to look like this:
Fg=m*r*(32/2pi)2
Then i substituted it back into the original equation and i got:
G*M*Me/M*(32/2pi)2=r
and this comes out to 1.87e10
the correct answer is supposed to be 4.26e8
 
Digdug12 said:
ok so i setup it up to look like this:
Fg=m*r*(32/2pi)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.
 
ahh thanks for catching that Doc. Al :D I'm sorry i still haven't solved it, I am 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?
 
Digdug12 said:
is the mass in Fg=M*ac the mass of the object, right?
Right.

Show me your final equation.
 
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
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
oooo, sweet. thanks, I got it now :D thanks a bunch
 
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