I want to know if I got this question right on my physics test

  • Thread starter Thread starter warfreak131
  • Start date Start date
  • Tags Tags
    Physics Test
Click For Summary
SUMMARY

The discussion centers on calculating the centripetal acceleration of a shuttle in orbit 400 km above Earth's surface, which completes an orbit every 90 minutes. The user correctly applies the formula for centripetal acceleration, a_{c}=\frac{v^2}{r}, and calculates the orbital velocity using v=\frac{2{\pi}r}{t}. The final result shows a centripetal acceleration of 9.2 m/s², equivalent to 0.92g, confirming the calculations align with the assumptions of gravity being 10 m/s².

PREREQUISITES
  • Understanding of centripetal acceleration and its formula
  • Knowledge of orbital mechanics and gravitational forces
  • Familiarity with unit conversions (km to m, minutes to seconds)
  • Basic proficiency in algebra and physics equations
NEXT STEPS
  • Study the derivation of centripetal acceleration in circular motion
  • Explore the effects of altitude on gravitational acceleration
  • Learn about orbital mechanics and satellite motion
  • Investigate the relationship between velocity, radius, and period in circular orbits
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and orbital dynamics, as well as educators looking for practical examples of centripetal acceleration calculations.

warfreak131
Messages
186
Reaction score
0

Homework Statement



A shuttle is in orbit 400km above the Earths surface, and circles it every 90 minutes. Find the centrip. acceleration and find it in terms of g. (He says for sake of ease, assume gravity = 10 m/s^2)

Earth radius = 6400 km

Homework Equations



[tex]a_{c}=\frac{v^2}{r}, v=\frac{2{\pi}r}{t}[/tex]

The Attempt at a Solution



First I converted everything to meters and seconds:

6400 km = 6400000m, 90 minutes = 5400 seconds

[tex]v=\frac{2{\pi}r}{t}[/tex]
[tex]v=\frac{2{\pi}(6400000+400000)}{5400}[/tex]
[tex]v=7900\frac{m}{s}[/tex]

[tex]a_{c}=\frac{v^2}{r}[/tex]
[tex]a_{c}=\frac{7900^2}{6800000}[/tex]
[tex]a_{c}=\frac{6.3{\cdot}10^7}{6800000}[/tex]
[tex]a_{c}=9.2\frac{m}{s^2}[/tex]

= .92g's
 
Physics news on Phys.org
Looks right, but I am not the one grading it.
 
damn, because i re did the question when posting it on this forum, but i remember that my answer on the test was nothign close to that

i hope i can get some extra credit.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
Replies
22
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
8K
Replies
8
Views
3K
Replies
43
Views
8K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 7 ·
Replies
7
Views
3K