Circular motion in planets and satellites

In summary: The radius of the spaceship?R. Therefore, the velocity is 6.67*10^-11 m/s.This is not the answer to the question.The question asks to solve for the length of the time interval the spaceship would have to complete one rotation on its axis, so isn't that finding the period, not the acceleration?You are correct, but to do so we need the (tangential) velocity at which it rotates. (I think. Somebody PLEASE correct me if I am wrong before I make it worse lol).PLEASE correct me
  • #1
4
0

Homework Statement



A spaceship in outerspace has a donut shaped with a 500 m outer radius. The inhabitants stand with their heads towards the center and their feet on an outside rim. Over what time interval with the spaceship have to complete one rotation on its axis to make a bathroom scale have the same reading for the person in space as when on Earth's surface?

Homework Equations



Possibly G*m1*m2/r^2

G= 6.67*10^-11 N*m^2

Or

4*pi^2*R/ T^2

The Attempt at a Solution



Honestly I'm completely lost here. I'm not sure which formula to use to be able to solve for this.
 
Physics news on Phys.org
  • #2
hplover17 said:
Possibly G*m1*m2/r^2
Hmmm ---- possibly. What are m1 and m2 here?
 
  • #3
Correct me if I'm wrong, but you're looking for the period in which it rotates so that the acceleration is equal to the acceleration due to Earth's gravity?
 
  • #4
Gravity will be negligible. There is another effect that the scale will detect. Hint: the inhabitants rotate together with the space station.
 
  • #5
Bystander said:
Hmmm ---- possibly. What are m1 and m2 here?

There were no values given for any mass, so I'm not sure.
 
  • #6
Mass doesn't matter.. It will cancel out.
 
  • #7
If you want to use Gm1m2/r2, it will give you the acceleration at Earth's surface provided you use Earth's mass for one of the mass values.
hplover17 said:
to make a bathroom scale have the same reading for the person in space as when on Earth's surface?
Or, you can use previous knowledge of that value.
hplover17 said:
what time interval
hplover17 said:
one rotation on its axis
What do those two parts of the question bring to mind?
 
  • #8
Bystander said:
If you want to use Gm1m2/r2, it will give you the acceleration at Earth's surface provided you use Earth's mass for one of the mass values.

Or, you can use previous knowledge of that value.

What do those two parts of the question bring to mind?
Do we have to find the period here? If so, how do we do that in this problem? Do we set something equal to mv^2/r?
 
  • #9
hplover17 said:
Do we set something equal to mv^2/r?
Well, that would be equal to the force created, but we only need the acceleration, correct? And as Force = (Acceleration)(Mass), then without (Mass), v2/r = ?
 
  • #10
mrnike992 said:
Well, that would be equal to the force created, but we only need the acceleration, correct? And as Force = (Acceleration)(Mass), then without (Mass), v2/r = ?

The question asks to solve for the length of the time interval the spaceship would have to complete one rotation on its axis, so isn't that finding the period, not the acceleration?
 
  • #11
You are correct, but to do so we need the (tangential) velocity at which it rotates. (I think. Somebody PLEASE correct me if I am wrong before I make it worse lol).
 
  • #12
mrnike992 said:
PLEASE correct me
For all participants: is there a relationship between period (or angular velocity) and acceleration?
 
  • #13
My train of thought is that we can use the acceleration to find tangential velocity, and then use that and the circumference of the spaceship to find the time, or the period. Probably wrong, that's just what I'm going with.
 
  • #14
I'll restate my last question: What is the relationship between angular velocity and acceleration?
 
  • #15
The question is the period of rotation of the spaceship about its axis not its period along its orbit.

Over what time interval with the spaceship have to complete one rotation on its axis
 
  • #16
Are these not the same?
 
  • #17
But I'm going to step out now, definitely going to keep an eye on this post though. Sorry op, not trying to hijack your post lol
 
  • #18
mrnike992 said:
My train of thought is that we can use the acceleration to find tangential velocity, and then use that and the circumference of the spaceship to find the time, or the period. Probably wrong, that's just what I'm going with.
Instead of the tangential velocity at the outside of the spaceship, determine the angular velocity from the expected centripetal acceleration.
 
  • #19
hplover17 said:
A spaceship in outerspace
There are no orbits involved at all.
 
  • #20
hplover17 said:

Homework Statement



A spaceship in outerspace has a donut shaped with a 500 m outer radius. The inhabitants stand with their heads towards the center and their feet on an outside rim. Over what time interval with the spaceship have to complete one rotation on its axis to make a bathroom scale have the same reading for the person in space as when on Earth's surface?

Homework Equations



Possibly G*m1*m2/r^2

G= 6.67*10^-11 N*m^2

Or

4*pi^2*R/ T^2

The Attempt at a Solution



Honestly I'm completely lost here. I'm not sure which formula to use to be able to solve for this.

Set the radial acceleration of the ship equal to the gravity of earth, solve using v^2 / r = Ar to find the velocity, then divide the parameter by the velocity.
 
  • #21
Sal_Cubed said:
Set the radial acceleration of the ship equal to the gravity of earth, solve using v^2 / r = Ar to find the velocity, then divide the parameter by the velocity.
:welcome:

Note that this homework thread is eight years old!
 

1. What is circular motion in planets and satellites?

Circular motion in planets and satellites refers to the path that these objects follow as they orbit around a central body, such as a star or a planet. This motion is characterized by a constant distance from the central body and a uniform speed.

2. What causes circular motion in planets and satellites?

The gravitational force between the central body and the orbiting object is what causes circular motion. This force acts as a centripetal force, pulling the object towards the center and keeping it in its circular path.

3. How is circular motion different from elliptical motion?

Circular motion is a specific type of elliptical motion where the eccentricity, or the degree of deviation from a perfect circle, is equal to 0. In elliptical motion, the eccentricity is greater than 0, causing the orbiting object to have a varying distance from the central body.

4. Do all planets and satellites follow circular orbits?

No, not all planets and satellites follow perfectly circular orbits. The eccentricity of their orbits varies, resulting in elliptical or even more complex paths. However, some objects, such as moons of Jupiter, have nearly circular orbits.

5. How does circular motion affect the speed of planets and satellites?

In circular motion, the speed of an orbiting object remains constant as it travels along its circular path. This speed is determined by the distance from the central body and the mass of the central body. Objects closer to the central body will have a higher speed compared to those farther away.

Suggested for: Circular motion in planets and satellites

Replies
19
Views
465
Replies
5
Views
662
Replies
8
Views
1K
Replies
8
Views
891
Replies
19
Views
1K
Replies
4
Views
952
Replies
7
Views
763
Replies
2
Views
754
Back
Top