What is the period of the second planet in this exercise?

Click For Summary
SUMMARY

The discussion centers on calculating the orbital period of a second planet in a distant galaxy using Kepler's Third Law, represented by the equation T² = constant * r³. The first planet has a period of 108 seconds and orbits at a distance of 1.8 x 10^12 m. The user calculated the constant as 2 x 10^-33 and found the period of the second planet to be approximately 38.18 seconds. However, it was concluded that the period of the first planet should be 10^8 seconds instead of 108 seconds, which aligns with the expected results.

PREREQUISITES
  • Understanding of Kepler's Third Law of planetary motion
  • Basic algebra for solving equations
  • Familiarity with scientific notation
  • Knowledge of orbital mechanics
NEXT STEPS
  • Study Kepler's Third Law in detail
  • Learn how to apply scientific notation in calculations
  • Explore the implications of orbital periods on planetary motion
  • Investigate the physical constraints of orbital mechanics in astrophysics
USEFUL FOR

Astronomy students, physics enthusiasts, and anyone interested in understanding planetary motion and orbital mechanics.

OierL
Messages
5
Reaction score
0

Homework Statement


In a distant galaxy, a planet orbits its sun at a distance of
c28fb096237b0a04a15c64e37a292086.gif
m with a period of 108 s. A second planet orbits the same sun at a distance of
6b8c23b8fdf999e3d16717c8daf58b5e.gif
m. What is the period of the second planet?

Select one:
a.
55c2bc2f030e754241487b390145364b.gif
s
b.
b1558216fac73da8cf7be349fc9c0524.gif
s
c.
aaaf30018536b0436cdd4a744f488597.gif
s
d.
605809b649a0376f8fa4b8c28d33f15f.gif
s
e.
6f428d14e6b8e2ae0744043ee5d58576.gif


Homework Equations


T^2=constant * r^3

The Attempt at a Solution


First of all I compute the constant whit the values of the first planet:
(108)^2 = const. * (1'8*10^12)^3 → const= 2*10^-33
Then, I compute the period of the second planet:
T^2 = 2*10^-33 * (9*10^11)^3 → T = √1458 = 38,18 s
This solution doesn't apear in the results I have to choose. What do I have wrong?

Thank you!
 

Attachments

  • c28fb096237b0a04a15c64e37a292086.gif
    c28fb096237b0a04a15c64e37a292086.gif
    561 bytes · Views: 611
  • 6b8c23b8fdf999e3d16717c8daf58b5e.gif
    6b8c23b8fdf999e3d16717c8daf58b5e.gif
    463 bytes · Views: 562
  • 55c2bc2f030e754241487b390145364b.gif
    55c2bc2f030e754241487b390145364b.gif
    489 bytes · Views: 358
  • b1558216fac73da8cf7be349fc9c0524.gif
    b1558216fac73da8cf7be349fc9c0524.gif
    460 bytes · Views: 351
  • aaaf30018536b0436cdd4a744f488597.gif
    aaaf30018536b0436cdd4a744f488597.gif
    584 bytes · Views: 348
  • 605809b649a0376f8fa4b8c28d33f15f.gif
    605809b649a0376f8fa4b8c28d33f15f.gif
    481 bytes · Views: 330
  • 6f428d14e6b8e2ae0744043ee5d58576.gif
    6f428d14e6b8e2ae0744043ee5d58576.gif
    659 bytes · Views: 343
Physics news on Phys.org
I suspect the period in the problem should read 10^8s, not 108 s. 108s is an unphysical orbital period for a planet at this distance.
 
  • Like
Likes   Reactions: Buzz Bloom and OierL
phyzguy said:
I suspect the period in the problem should read 10^8s, not 108 s. 108s is an unphysical orbital period for a planet at this distance.
You are right! If I do the exercise with 10^8s the result coincides with e.
Thank you very much!
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
1K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 13 ·
Replies
13
Views
8K