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

In summary, the distance from the Earth to the object can be found using the equation r = (G*M*Me*T^2/4*pi^2)^(1/3), where G is the gravitational constant, M is the mass of the object, Me is the mass of the Earth, and T is the period of rotation. It is important to remember to convert the period from days to seconds before plugging it into the equation.
  • #1
Digdug12
15
0

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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  • #2
Hint: Apply Newton's 2nd law. How is the object accelerating?
 
  • #3
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
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.
 
  • #5
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
Digdug12 said:
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
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?
 
  • #8
Digdug12 said:
is the mass in Fg=M*ac the mass of the object, right?
Right.

Show me your final equation.
 
  • #9
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
 

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

1. What is rotational motion?

Rotational motion is the movement of an object around an axis or center point. It involves the rotation of an object along a circular path.

2. What is angular velocity?

Angular velocity is the rate of change of angular displacement over time. It is measured in radians per second.

3. How is rotational motion different from linear motion?

Rotational motion involves movement along a curved path, while linear motion involves movement along a straight path. Additionally, rotational motion involves angular displacement, velocity, and acceleration, while linear motion involves displacement, velocity, and acceleration along a straight line.

4. What is torque?

Torque is a measure of the force that can cause an object to rotate around an axis. It is calculated by multiplying the force applied to the object by the distance from the axis of rotation.

5. How can rotational motion problems be solved?

To solve rotational motion problems, you can use equations such as Newton's Second Law for rotational motion, torque equations, and conservation of angular momentum. It is important to identify the given information, determine the unknown variables, and choose the appropriate equation to solve for the unknowns.

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