Radial acceleration around a planet

AI Thread Summary
The discussion focuses on calculating the radial acceleration of a point on a planet's surface due to its rotation. The planet has a radius of 7.31 x 10^6 m and a rotation period of 25.4 hours. The initial attempt yielded an incorrect acceleration value of 2.88 x 10^8 m/s², which was attributed to not converting hours to seconds. After correcting the time conversion and recalculating, the correct radial acceleration was found to be 0.0345 m/s², which was confirmed as accurate. The importance of unit conversion in physics calculations is emphasized throughout the discussion.
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top