Radial acceleration around a planet

Click For Summary
SUMMARY

The discussion focuses on calculating the radial acceleration of a point on the surface of a planet with a radius of 7.31 x 106 m and a rotation period of 25.4 hours. The correct approach involves converting the rotation period into seconds, resulting in 91440 seconds. Using the formula for velocity, v = 2(pi)(r)/T, the velocity is determined to be 502.3 m/s. The radial acceleration is then calculated using ar = (v2/r), yielding a final value of 0.0345 m/s2.

PREREQUISITES
  • Understanding of circular motion and radial acceleration
  • Familiarity with unit conversions, particularly time from hours to seconds
  • Knowledge of basic physics equations related to velocity and acceleration
  • Proficiency in algebra for manipulating equations
NEXT STEPS
  • Review the concept of centripetal acceleration in circular motion
  • Learn about unit conversions in physics, focusing on time
  • Explore the derivation of the formulas for velocity and radial acceleration
  • Practice similar problems involving rotational dynamics and radial acceleration
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and rotational motion, as well as educators looking for examples of radial acceleration calculations.

anteaters
Messages
11
Reaction score
0

Homework Statement



Consider a planet of radius 7.31 x 10^6 m with a rotation period of 25.4 hours. Compute the radial acceleration of a point on the surface of the planet at the equator owing to its rotation about its axis.

Homework Equations



T = [2(pi)r]/v
a_r = -[(v^2)/r]

The Attempt at a Solution



so i know one period (T) is 25.4 hours, and r = 7.31 x 10^6 m. so i use that and find velocity, and tried to plug that into the radial acceleration equation, but i got the wrong answer. what am i doing wrong? is it something to do with units of time?
 
Physics news on Phys.org
oh, and my answer was 2.88 x 10^8 m/s2
 
anteaters said:
what am i doing wrong? is it something to do with units of time?
Did you convert the hours to seconds? If not, you are spot-on regarding time.

anteaters said:
oh, and my answer was 2.88 x 10^8 m/s2
That means you did something else wrong in addition to the time units error. Please show your work.
 
25.4 hours = 91440 seconds = 2(pi)(7.31 x 10^6)/v
so v = 502.3 m/s
a_r = [(502.3 m/s)^2]/(7.31 x 10^6) = 0.0345 m/s^2

would that be the right answer? it seems kind of small.
 
Looks fine to me.
 
thanks a lot D H.
 

Similar threads

Replies
1
Views
2K
Replies
7
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 9 ·
Replies
9
Views
8K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
6K
Replies
15
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K