What is the period of revolution for an artificial planet

Click For Summary
SUMMARY

The discussion focuses on calculating the period of revolution for an artificial planet designed as a band encircling a sun, similar to Earth's distance from the sun. The gravitational force required to simulate Earth's gravity (1 g or 9.8 m/s²) is derived using the equations of motion: g = GM/r² and a = v²/r. Given the mass of the sun (Msun = 1.98x10³⁰ kg) and its radius (Rsun = 6.96x10⁸ m), the participants aim to determine the necessary rotational speed and the corresponding period of revolution in Earth days.

PREREQUISITES
  • Understanding of gravitational force equations (g = GM/r²)
  • Basic knowledge of circular motion (a = v²/r)
  • Familiarity with astronomical units (Earth-Sun distance)
  • Concept of artificial gravity and its calculation
NEXT STEPS
  • Calculate the rotational speed required to achieve 1 g using the formula a = v²/r.
  • Determine the radius of the artificial planet's band based on the Earth-Sun distance.
  • Research the implications of centrifugal force in artificial gravity environments.
  • Explore the concept of orbital mechanics and its applications in science fiction scenarios.
USEFUL FOR

Students in physics, engineers designing space habitats, and science fiction writers exploring concepts of artificial gravity and planetary motion.

balletgirl
Messages
35
Reaction score
0

Homework Statement



A science fiction tale describes an artificail "planet" in the form of a band completely encircling a sun, the inhabitants living on the inside surface (where it is always noon). Imagine the sun is like our own, that the distance to the band is the same as the Earth-Sun distance (to make the climate temperature), and that the ring rotates quickly enough to produce an apparent gravity of one g as on Earth. What will be the period of revolution, this planet's year, in Earth days?

Msun = 1.98x10^30
Rsun= 6.96x10^8

Homework Equations


g= GM/r^2 a= v^2/r
g= GM/(r+h)^2


The Attempt at a Solution



I'm not sure how to start. Would the distance/height be 1 ly (9.5x10^15m), and doesn't one Earth g = 9.8m/s?
 
Physics news on Phys.org
If the ring spins fast enough to produce Earth's gravity (9.8 m/s^2) how fast must it be rotating?
 
balletgirl said:

Homework Statement



A science fiction tale describes an artificail "planet" in the form of a band completely encircling a sun, the inhabitants living on the inside surface (where it is always noon). Imagine the sun is like our own, that the distance to the band is the same as the Earth-Sun distance (to make the climate temperature), and that the ring rotates quickly enough to produce an apparent gravity of one g as on Earth. What will be the period of revolution, this planet's year, in Earth days?

Msun = 1.98x10^30
Rsun= 6.96x10^8

Homework Equations


g= GM/r^2 a= v^2/r
g= GM/(r+h)^2


The Attempt at a Solution



I'm not sure how to start. Would the distance/height be 1 ly (9.5x10^15m), and doesn't one Earth g = 9.8m/s?
To answer your first question:
"...the distance to the band is the same as the Earth-Sun distance..."

To answer your second:
One Earth g is 9.8m/s^2. How fast would it have to revolve to emulate that?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
6K
  • · Replies 6 ·
Replies
6
Views
8K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 13 ·
Replies
13
Views
8K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 11 ·
Replies
11
Views
3K