Centripetal acceleration on Earth

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Homework Help Overview

The problem involves calculating the centripetal acceleration of an object orbiting the Earth at a constant speed in a circular path very close to the Earth's surface, with a specified radius. The original poster expresses confusion regarding their calculated centripetal acceleration compared to a known value for the Earth's surface.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various equations related to centripetal acceleration and angular velocity. The original poster attempts to apply these equations but questions the validity of their results compared to known values. Some participants suggest alternative formulas and check the assumptions regarding the mass of the satellite and the period of orbit.

Discussion Status

The discussion is ongoing, with participants providing different equations and perspectives on the problem. There is a recognition of potential errors in the original poster's calculations, particularly regarding the velocity used in the centripetal acceleration formula. Multiple interpretations of the problem and its parameters are being explored.

Contextual Notes

There is a lack of clarity regarding the period of the orbit and the specific values for velocity, which are critical to the calculations. The original poster's confusion stems from discrepancies between their calculated acceleration and the expected value based on gravitational acceleration at the Earth's surface.

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Homework Statement


An object orbits the Earth at a constant speed in a circle of radius 6.38 x 106 m, very close to but not touching the Earth's surface. What is its centripetal acceleration?


Homework Equations


a = v2/r = 4\pi2v/T2
v = 2\pir/T

The Attempt at a Solution


I plugged in r = 6.38 x 106 m and T = 24.0 h = (24.0 h x 3600 s / 1 h) = 86,400 s into the equation above and found a = 3.37 x 10-2 m/s2. However, I looked up the centripetal acceleration on the Earth's surface and found out it is 0.006 m/s2. I can't understand why my answer is wrong. Can anyone point out the error in my logic?

Thanks!
 
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Try using this formula- 2\pi/\omega

\omega= 360/24*3600


Let me know if you got the answer.
 
Thanks for the suggestion RoughRoad but I don't really understand how I am supposed to use this equation.

I found that my calculation for the angular velocity (2piR/T) is approximately equal to the wikipedia value for angular velocity, so I really am confused now because I can't see what I'm doing wrong in just squaring that value and dividing by 6.38 x 106.
 
Is the mass of the satellite given?
 
No, it isn't.
 
try using the equation GMm/r^2=mv^2/r
m-mass of satellite
M-mass of earth
r-radius of orbit
v^2/r is the centripetal acceleration
 
gravitational frce of Earth is utilised for centripetal force
 
jyothsna pb said:
try using the equation GMm/r^2=mv^2/r
m-mass of satellite
M-mass of earth
r-radius of orbit
v^2/r is the centripetal acceleration

You are right. But what about the velocity?
 
you are asked to find the centripetal acceleration you just have to find the value of v^2/r which gives the centripetal acceleration
 
  • #10
jyothsna pb said:
you are asked to find the centripetal acceleration you just have to find the value of v^2/r which gives the centripetal acceleration


Oh yeah! How can I be so foolish. Thanks for helping!
 
  • #11
hope you got the answer
 
  • #12
u r welcome
 
  • #13
And reply to my visitor msg pls
 
  • #14
I didn't, though, and that's the source of my confusion. I know I have the right value for v but when I plug it into v^2/r, I get 3.37 x 10^-2 m/s^2. This answer is not equal to the answer I found for the actual acceleration, which is 0.006 m/s^2. My question is why is my answer different from the real value?
 
  • #15
what is the value of v?
 
  • #16
465.1 m/s
 
  • #17
there is some error in the velocity value
 
  • #18
we get acceleration value almost equal to g
 
  • #19
velocity in dis orbit must b approximately equal to 7.9*10^3m/s
 
  • #20
I don't understand why that is, especially since Wikipedia says it is 451 m/s
en.wikipedia.org/wiki/Earth
 
  • #21
it wud b easy if u jus show me dat particular part from d article
 
  • #22
It's on the sidebar, under "Physical Characteristics".
 
  • #23
it is equatorial rotation velocity of Earth while that mentioned here is the velocity of a satellite of Earth in an orbit very close to it both r different
 
  • #24
The centripetal acceleration must balance out the acceleration due to gravity (it is essentially at ground level, so 9.8 m/s^2).

*Force of gravity = mg
*Centripetal acceleration = a (not v^2/r, because you are solving for this total value, not any sub-values within a)

The centripetal acceleration when traveling in a constant circular orbit at ground level on Earth should be equal to gravity, or 9.8. You don't necessarily need velocity, radius, mass, time, etc.
 
  • #25
Try this-

Centripetal acceleration= \omega^2*r= (2\pi/T)*r

Let me know if you got the answer.
 
  • #26
RoughRoad said:
Try this-

Centripetal acceleration= \omega^2*r= (2\pi/T)*r

Let me know if you got the answer.
but the problem is that the period is unknown, i think
 
  • #27
Jokerhelper said:
but the problem is that the period is unknown, i think

The body moves with a constant speed, so it's T should be 24*3600. I may be wrong though.
 

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