Centripetal acceleration on Earth

In summary, 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. The centripetal acceleration is 3.37 x 10-2 m/s2.
  • #1
daysrunaway
19
0

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[tex]\pi[/tex]2v/T2
v = 2[tex]\pi[/tex]r/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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  • #2
Try using this formula- 2[tex]\pi[/tex]/[tex]\omega[/tex]

[tex]\omega[/tex]= 360/24*3600


Let me know if you got the answer.
 
  • #3
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.
 
  • #4
Is the mass of the satellite given?
 
  • #5
No, it isn't.
 
  • #6
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
 
  • #7
gravitational frce of Earth is utilised for centripetal force
 
  • #8
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?
 
  • #9
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= [tex]\omega[/tex]^2*r= (2[tex]\pi[/tex]/T)*r

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

Centripetal acceleration= [tex]\omega[/tex]^2*r= (2[tex]\pi[/tex]/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.
 

1. What is centripetal acceleration on Earth?

Centripetal acceleration on Earth is the acceleration an object experiences when it moves in a circular path around the Earth. It is always directed towards the center of the circular motion and is caused by the force of gravity pulling the object towards the Earth.

2. How is centripetal acceleration calculated on Earth?

The formula for calculating centripetal acceleration on Earth is a = v^2/r, where a is the acceleration, v is the velocity of the object, and r is the radius of the circular path. This formula can also be written as a = ω^2r, where ω is the angular velocity of the object.

3. What is the difference between centripetal acceleration and centrifugal force?

Centripetal acceleration is the acceleration an object experiences when moving in a circular path. It is always directed towards the center of the circular motion. Centrifugal force, on the other hand, is a fictitious force that appears to push an object away from the center of the circular path. It is a result of the object's inertia trying to keep it moving in a straight line.

4. Does centripetal acceleration affect all objects equally on Earth?

Yes, centripetal acceleration affects all objects equally on Earth regardless of their mass. This is because the acceleration is caused by the force of gravity, which is the same for all objects on Earth regardless of their mass.

5. What are some real-life examples of centripetal acceleration on Earth?

Some examples of centripetal acceleration on Earth include the motion of a satellite orbiting the Earth, the circular motion of a car around a racetrack, and the rotation of a Ferris wheel. Even the motion of the Earth itself around the Sun is an example of centripetal acceleration.

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