# Physics: dynamics in a circle

1. Feb 27, 2005

### oldunion

In the Bohr model of the hydrogen atom, an electron (mass=9.1x10^-31kg) orbits a proton at a distance of 5.3x10^-11m. The proton pulls on the electron with an electric force of 9.2x10^-8N.

I did F=mv^2/r to solve for velocity. units dont match but i got 2.315. Centripetal acceleration next? I know ive made a mistake thus far.

Just came across this question:
What is the acceleration due to gravity of the sun at the distance of the earth's orbit?

Thats all the info they give you, they want the answer in m/s^2 :uhh:

Last edited: Feb 27, 2005
2. Feb 28, 2005

### xelda

I think you calculated velocity correctly but you left off the scientific notation. Keep the velocity squared and use that to solve for the acceleration.

a = v^2 / r

This will give you the answer in m/s^2

3. Feb 28, 2005

### dextercioby

As for the second part,what formula would u have to use...?

Daniel.

4. Feb 28, 2005

### ramollari

Hey, by the way don't use again the formula $$g = \frac{v^2}{r_{sun}}$$ here, because you don't know the velocity! If $$R_{Earth}$$ is the radius of the Earth, and we know $$g$$, then express your answer in terms of $$g$$ and the ratio $$R_{Earth}/r_{sun}$$.

Last edited: Feb 28, 2005
5. Feb 28, 2005

### dextercioby

Yes,he does know the velocity.He can approximate the ellipse Earth is making around the sun and determine the velocity to approx 30Km/s...(Besides,even if you didn't know the physics to compute it,it's very well known,because it's tabulated )...

Daniel.

6. Feb 28, 2005

### ramollari

I don't think that is the purpose.
Wait a minute, why does the Earth velocity come up here? It won't do!

7. Feb 28, 2005

### dextercioby

You probably forgot the question.There it is:"What is the acceleration due to gravity of the sun at the distance of the earth's orbit?"

The acceleration is simply $\frac{v^{2}}{R}$,where v is the orbital velocity of the Earth (namely ~30Km/s) and R is the mean distance Earth-Sun (~150*10^{9}m}.

Daniel.

8. Feb 28, 2005

### ramollari

It is the earth moving, not the sun. So the formula you gave gives the centripetal acceleration of the Earth moving around the sun.
I understood the question like this: find the value of $$g$$ at the distance the sun is.

9. Feb 28, 2005

### dextercioby

Nope,it's the other way around.It should be the Earth graviting the sun."due to gravity of the sun" ring a bell...?"at the distance of the Earth's orbit"...Does this one ring a bell,then...?

Daniel.

10. Feb 28, 2005

### oldunion

i got the second part. but the first part is incorrect.

a got a v^2 of 5.358x10^12 then i divided this by the radius of 5.3x10^-11= 1.011x10^23 which is not the answer.

edit: nevermind im being retarded, they asked for it in revs/ second. is this a new equation or just dimensional analysis?

11. Feb 28, 2005

### dextercioby

Revolutions per second,well that's Hertz and normally,in SI revolutions doesn't have a unit...

Daniel.

12. Feb 28, 2005

### oldunion

so how would i go from meters/s^2 to revs/second

13. Feb 28, 2005

### dextercioby

Hold it,one unit id for angular velocity (Hz) and the other is for linear acceleration.There's no way to go from one to the other...There are different quantities...

Daniel.

14. Feb 28, 2005

### ramollari

$$a = \omega^2r$$

$$\omega = \sqrt{\frac{a}{r}}$$, where $$\omega$$ is in rad/s.

to go to rev/s, you are looking for the frequency:

$$f = \frac{\omega}{2\pi}$$.

15. Feb 28, 2005

### oldunion

no those units dont jive. you would end up with some fraction of #/T where t is the period. tried anyway, 5.99 or 9.42 incorrect

16. Feb 28, 2005

### dextercioby

Okay:What is velocity in m/s and what is the acceleration in m/s^{2}...?

Daniel.

17. Feb 28, 2005

### oldunion

v^2= 5.358x10^12
a=1.011x10^23

18. Feb 28, 2005

### dextercioby

The first # is correct.The second looks good as well...You're done...

Daniel.

19. Feb 28, 2005

### oldunion

yes but this doesnt solve my dilemma of how to find the rev/s

5.99x10^21 rad/s (assuming it is in rad/s for whatever reason) x 1 rev/2pirad=9.42x10^21

Last edited: Feb 28, 2005
20. Feb 28, 2005

### dextercioby

Yes,if that $5.99\times 10^{21} rad \ s^{-1}$ is the correct "omega",then te final answer is correct.

Daniel.

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