• Support PF! Buy your school textbooks, materials and every day products Here!

Gravity Question

  • Thread starter Poisonous
  • Start date
  • #1
26
0
The problem statement, all variables and known data:

You launch an object into a circular orbit around a spherical, smooth rock, 30 cm in diameter. The rock also has the same mass density as Earth, ρ. How long does it take the object to orbit the rock one time?

Attempt at a solution:

I know you should start by finding the mass of the rock, which turns out to be:
(4/3)pi*r3

I'm not sure what to do after that. Any help?
 
Last edited:

Answers and Replies

  • #2
Doc Al
Mentor
44,913
1,171
Apply Newton's 2nd law to solve for the period. What force acts on the object? What is its acceleration?

What radius of orbit are you mean to assume? Is the object supposed to be sliding along the surface of the rock?
 
  • #3
26
0
Yeah, the problem says the object is in orbit just barely above the surface, so I think you can assume that it's the same as if it were sliding without friction.

The force that acts on the object is gravity. The acceleration due to gravity is given by the formula -GM/r2, where I can plug in the M from above.

I'm not sure how to relate that to F=ma.
 
  • #4
Doc Al
Mentor
44,913
1,171
Yeah, the problem says the object is in orbit just barely above the surface, so I think you can assume that it's the same as if it were sliding without friction.
Good. So we have the radius of the orbit.

The force that acts on the object is gravity.
Good.
The acceleration due to gravity is given by the formula -GM/r2, where I can plug in the M from above.
That's true, but unhelpful.

I'm not sure how to relate that to F=ma.
Hint: What kind of motion does the object undergo? Write the generic formula for that sort of acceleration. Plug that into F=ma.
 
  • #5
26
0
The object is moving in uniform circular motion. So, F=m*v2/r

You know r and the force is gravity, but do you know m? Or does that cancel out with the force of gravity formula?

I assume it does, so then v = [tex]\sqrt{GM/r}[/tex]

or

v = [tex]\sqrt{4/3*G*pi*rho*r^2}[/tex] which has me a little worried, since the problem has a note that says we shouldn't have to do any arithmetic at all to solve it.
 
Last edited:
  • #6
Doc Al
Mentor
44,913
1,171
Note that you are asked to find the period, so express v in terms of the period.

It's not obvious to me how you can get a numerical answer without doing any arithmetic. (Perhaps I just need more coffee.) You can certainly express the answer in terms of ρ.
 
  • #7
ehild
Homework Helper
15,478
1,854
Compare it with an object orbiting just above the surface of the Earth.

ehild
 
  • #8
26
0
Doc Al: So I could do it with v*omega instead of v2/r

ehild: I was thinking something like that, but how..?
 
  • #9
ideasrule
Homework Helper
2,266
0
v = [tex]\sqrt{4/3*G*pi*rho*r^2}[/tex] which has me a little worried, since the problem has a note that says we shouldn't have to do any arithmetic at all to solve it.
Don't be discouraged; keep going. How do you get the period from the speed? Remember that this is a circular orbit, so you can easily calculate the distance that the satellite travels. Once you get an expression for the period, you'll realize why you don't need any arithmetic to solve the problem.
 
  • #10
ideasrule
Homework Helper
2,266
0
Doc Al: So I could do it with v*omega instead of v2/r

ehild: I was thinking something like that, but how..?
What's v*omega? Anyhow, you don't need to bring omega into the picture.
 
  • #11
Doc Al
Mentor
44,913
1,171
Doc Al: So I could do it with v*omega instead of v2/r
Don't bother using ω²r, if that's what you're thinking. Just replace v with distance per period.
 
  • #12
ehild
Homework Helper
15,478
1,854
Just write out the equations acceleration due to gravity = centripetal acceleration , both for Earth and for the rock. Find out how the angular velocities ω are related. Do not forget that the densities are equal. And you know the minimum speed of a satellite which can orbit around the Earth .


ehild
 
Last edited:
  • #13
26
0
Since it's circular, the period is T = 2*pi*r/v, but that is not making it immediately obvious why I shouldn't have to do arithmetic.
 
  • #14
Doc Al
Mentor
44,913
1,171
Doc Al:
The angular velocities are the same. On the Earth, it is sqrt(g/R)
That's certainly true, but you still need some arithmetic to figure out the period.
 
  • #15
Doc Al
Mentor
44,913
1,171
Since it's circular, the period is T = 2*pi*r/v,
Good. Express v in terms of T and use that in your early equation. Then you can solve for the period.
but that is not making it immediately obvious why I shouldn't have to do arithmetic.
Unless they want you to express the answer symbolically, you'll need a little arithmetic.
 
  • #16
93
0
You are almost there. Carry out the isolation for T in general terms and you will observe that the radius term cancels out. Thus it really doesn't matter how big the rock is at all. If the density of the rock is the same as the earth's, then the orbital period at the surface would also be the same. If you knew this orbital period for the earth, then you would not have to do any arithmetic (although I doubt anybody would know).
 
  • #17
26
0
Actually, I do know the orbital period on earth's surface, since it is in the packet this question is in. It is given as 5069 seconds. But, the problem is that the answer I'm getting is not 5069.

So, after the algebra, you get:

T^2 = (2pi)/(4/3*G*rho)

solving for T yields: 11709.99

Did I do something wrong?
 
  • #18
Doc Al
Mentor
44,913
1,171
Actually, I do know the orbital period on earth's surface, since it is in the packet this question is in.
Ah, so that's why you don't need any arithmetic! :wink:

It is given as 5069 seconds. But, the problem is that the answer I'm getting is not 5069.

So, after the algebra, you get:

T^2 = (2pi)/(4/3*G*rho)
That's not quite right. Redo your algebra.

But the main thing to realize is that the period is independent of the radius.
 
  • #19
26
0
Hmm, I am supposed to be solving:

GMm/r2 = m * v2/r

plugging in and simplifying to:

G(4/3*π*r*ρ)= (2πr/T)2/r

right?

Because that simplifies all the way to: T^2 = (3pi)/(G*ρ) which doesn't give 5069.
 
  • #20
Doc Al
Mentor
44,913
1,171
Hmm, I am supposed to be solving:

GMm/r2 = m * v2/r

plugging in and simplifying to:

G(4/3*π*r*ρ)= (2πr/T)2/r

right?

Because that simplifies all the way to: T^2 = (3pi)/(G*ρ)
Good.
which doesn't give 5069.
What did you use for G and ρ? (Show how you calculated ρ.)
 
  • #21
93
0
Because that simplifies all the way to: T^2 = (3pi)/(G*ρ) which doesn't give 5069.
I hope you rooted whatever number you got. Not undermining your intelligence. Just checking all the possibilities.
 
  • #22
26
0
Ah, I forgot to convert density to kg/m^3. Thanks for the help. I get about 5060 now.
 
  • #23
ideasrule
Homework Helper
2,266
0
A useful thing to remember: the Space Shuttle, which orbits only 300 km above the surface, orbits in 90 minutes. When I read the part about not needing any arithmetic, I thought you were expected to know this or something.
 

Related Threads on Gravity Question

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
0
Views
746
  • Last Post
Replies
6
Views
9K
  • Last Post
Replies
5
Views
584
  • Last Post
Replies
14
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
3
Views
844
  • Last Post
Replies
6
Views
1K
Top